ltc1
groups_of_special_equivalent_strings
Problem Statement
Given a list of strings, group them into sets of equivalent strings. Two strings are equivalent if they are anagrams of each other.
Solution
The solution is to use a hashing function to convert each string into a unique identifier. The identifier is based on the frequency of each character in the string.
import collections
def group_strings(strs):
"""
Groups a list of strings into sets of equivalent strings.
Args:
strs (list): The list of strings to group.
Returns:
list: A list of lists of equivalent strings.
"""
# Create a dictionary to store the frequency of each character in each string.
char_counts = collections.defaultdict(list)
for s in strs:
char_counts[s] = collections.Counter(s)
# Create a dictionary to store the unique identifier for each string.
identifiers = {}
for s, char_count in char_counts.items():
identifiers[s] = tuple(char_count.values())
# Group the strings by their unique identifier.
groups = collections.defaultdict(list)
for s, identifier in identifiers.items():
groups[identifier].append(s)
# Return the list of groups.
return list(groups.values())Breakdown
The
group_stringsfunction takes a list of strings as input and returns a list of lists of equivalent strings.The function first creates a dictionary to store the frequency of each character in each string. The dictionary keys are the strings and the dictionary values are
Counterobjects that represent the frequency of each character in the string.The function then creates a dictionary to store the unique identifier for each string. The dictionary keys are the strings and the dictionary values are tuples that represent the frequency of each character in the string.
The function then groups the strings by their unique identifier. The dictionary keys are the unique identifiers and the dictionary values are lists of strings that have the same unique identifier.
The function finally returns the list of groups.
Real-World Applications
The group_strings function can be used in a variety of real-world applications. For example, the function can be used to:
Group email addresses by their domain name.
Group phone numbers by their area code.
Group text messages by their sender.
Group social media posts by their topic.
The function can also be used to improve the performance of search engines. By grouping similar strings together, search engines can more efficiently find the results that users are looking for.
k_th_smallest_prime_fraction
k_th_smallest_prime_fraction
Given an array of numbers, return the k-th smallest prime fraction.
Algorithm
Generate all prime fractions.
Sort the prime fractions in ascending order.
Return the k-th fraction in the sorted list.
Implementation
def k_th_smallest_prime_fraction(nums, k):
"""
Returns the k-th smallest prime fraction in the array nums.
Args:
nums: An array of numbers.
k: The index of the prime fraction to return.
Returns:
The k-th smallest prime fraction.
"""
# Generate all prime fractions.
prime_fractions = []
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
if i != 0 or j != len(nums) - 1:
prime_fractions.append((nums[i] / nums[j], (i, j)))
# Sort the prime fractions in ascending order.
prime_fractions.sort()
# Return the k-th fraction in the sorted list.
return prime_fractions[k - 1]Example
nums = [1, 2, 3, 5]
k = 3
result = k_th_smallest_prime_fraction(nums, k)
print(result) # (0.5, (1, 2))Applications
Prime fractions have applications in a variety of fields, including:
Number theory: Prime fractions are used to study the distribution of prime numbers.
Algebra: Prime fractions are used to solve Diophantine equations.
Computer science: Prime fractions are used in algorithms for finding the greatest common divisor and least common multiple of two numbers.
maximum_distance_in_arrays
Maximum Distance in Arrays
Problem Statement
Given m arrays, where each array is sorted in ascending order. Find the minimum distance between any two elements from different arrays.
Example
Input: arrays = [[1, 2, 3], [4, 5], [1, 2, 3]]
Output: 1
Explanation: The distance between 1 and 4 is 1.
Approach
The brute-force approach is to compute the distance between each pair of elements from different arrays and find the minimum distance. This approach has a time complexity of O(mn^2), where m is the number of arrays and n is the maximum length of an array.
A more efficient approach is to use a greedy algorithm. For each array, keep track of the index of the smallest element that has been visited so far. Then, find the minimum distance between the smallest element from each array. This approach has a time complexity of O(mn).
Implementation
def maximum_distance_in_arrays(arrays):
"""
Finds the minimum distance between any two elements from different arrays.
Parameters:
arrays: a list of sorted arrays.
Returns:
The minimum distance between any two elements from different arrays.
"""
# Initialize the minimum distance to infinity.
min_distance = float('inf')
# Keep track of the index of the smallest element that has been visited so far in each array.
min_idx = [0 for _ in range(len(arrays))]
# Iterate over the arrays.
for arr in arrays:
# Find the smallest element in the current array.
min_val = min(arr)
# Update the minimum distance if the current element is smaller.
min_distance = min(min_distance, min_val - arr[min_idx[arrays.index(arr)]])
# Increment the index of the smallest element in the current array.
min_idx[arrays.index(arr)] += 1
# Return the minimum distance.
return min_distanceReal World Applications
The maximum distance in arrays problem can be used in a variety of real-world applications, such as:
Scheduling: Find the minimum amount of time that it will take to complete a set of tasks, where each task must be completed by a different machine.
Warehousing: Find the minimum distance that a forklift must travel to retrieve a set of items from a warehouse.
Manufacturing: Find the minimum amount of time that it will take to assemble a set of components, where each component is produced by a different factory.
valid_tic_tac_toe_state
Problem Statement:
Given a 3x3 board filled with 'X', 'O', or ' ', determine if the current state of the board represents a valid Tic-Tac-Toe game.
Valid State Conditions:
There must be an equal number of 'X' and 'O' (or one more 'X' than 'O').
No more than one winner ('X' or 'O').
Winning moves must follow the game's rules (3 in a row, horizontally, vertically, or diagonally).
Implementation:
def valid_tic_tac_toe_state(board):
"""
Checks if a given Tic-Tac-Toe board represents a valid game state.
Parameters:
board (list[list[str]]): A 3x3 board filled with 'X', 'O', or ' '.
Returns:
bool: True if the board state is valid, False otherwise.
"""
# Check if board size is valid
if len(board) != 3 or any(len(row) != 3 for row in board):
return False
# Count 'X' and 'O' occurrences
x_count = 0
o_count = 0
for row in board:
for cell in row:
if cell == 'X':
x_count += 1
elif cell == 'O':
o_count += 1
# Check if 'X' and 'O' counts are valid
x_count_valid = (x_count == o_count) or (x_count == o_count + 1)
if not x_count_valid:
return False
# Check for horizontal wins
for row in board:
if all(cell == row[0] for cell in row):
if row[0] != ' ':
return True
# Check for vertical wins
for col in range(3):
if all(board[row][col] == board[0][col] for row in range(3)):
if board[0][col] != ' ':
return True
# Check for diagonal wins
if all(board[row][row] == board[0][0] for row in range(3)):
if board[0][0] != ' ':
return True
if all(board[row][2-row] == board[0][2] for row in range(3)):
if board[0][2] != ' ':
return True
# No winners found, check if the game is still in progress
for row in board:
for cell in row:
if cell == ' ':
return True
# No more empty cells, game is finished with no winners
return FalseExplanation:
Validate Board Size: Ensure the board has dimensions of 3x3.
Count 'X' and 'O': Count the occurrences of 'X' and 'O' to check if their numbers are valid.
Check Horizontal Wins: Iterate over each row and check if all cells have the same non-empty value (representing a win).
Check Vertical Wins: Similar to horizontal wins, but iterate over each column.
Check Diagonal Wins: Check both main diagonals to see if all cells have the same non-empty value.
Check for Game Continuation: If no wins are found, check if there are still empty cells, indicating the game is still in progress.
Return Result: Return True if the board state is valid or False otherwise.
Example:
board = [['X', 'O', 'X'],
['X', 'O', 'O'],
['O', 'X', 'X']]
result = valid_tic_tac_toe_state(board)
print(result) # Output: TruePotential Applications in Real World:
Tic-Tac-Toe game engine or referee
Game AI analysis and strategy optimization
Verifying game state integrity in online or offline games
design_circular_queue
Circular Queue Implementation & Explanation
Imagine a circle with a fixed number of slots. We want to store elements in these slots, but with a twist: we can only enter and exit the circle from a specific point. This is known as a circular queue.
Initialization:
class CircularQueue:
def __init__(self, k: int):
self.queue = [None] * k
self.head = 0
self.tail = 0
self.size = 0
self.max_size = kk: The initial size of the circular queue.queue: The underlying list that stores the elements of the queue.head: The index of the first element in the queue.tail: The index of the last element in the queue.size: The current number of elements in the queue.max_size: The maximum size of the circular queue.
Enqueue (Adding an Element):
def enqueue(self, value: int):
if self.is_full():
return
self.queue[self.tail] = value
self.tail = (self.tail + 1) % self.max_size # Wrap around if needed
self.size += 1Check if the queue is full by comparing
sizetomax_size.If not full, store the element at the
tailindex in thequeuelist.Advance the
tailindex by 1 and wrap around to the start if necessary, using the modulo operator.Increase the
sizeby 1.
Dequeue (Removing an Element):
def dequeue(self):
if self.is_empty():
return None
value = self.queue[self.head]
self.head = (self.head + 1) % self.max_size # Wrap around if needed
self.size -= 1
return valueCheck if the queue is empty by comparing
sizeto 0.If not empty, retrieve the element at the
headindex from thequeuelist.Advance the
headindex by 1 and wrap around to the start if necessary, using the modulo operator.Decrease the
sizeby 1.
Other Methods:
is_full(): Checks if the queue is at its capacity.is_empty(): Checks if the queue is empty.front(): Returns the value at the front of the queue without removing it.rear(): Returns the value at the rear of the queue without removing it.
Real-World Applications:
Buffer Management: Circular queues can be used to manage buffers in streaming applications, such as audio and video playback.
Resource Allocation: They can be used to allocate limited resources in a fair and efficient manner, ensuring that all users have access.
Scheduling: Circular queues can be used to schedule tasks in a round-robin fashion, giving each task an equal opportunity to execute.
non_decreasing_subsequences
Non-Decreasing Subsequences
Problem Statement: Given an array of integers, find the number of increasing subsequences that are present in the array.
Breakdown:
What is a subsequence?
A subsequence is a sequence that can be obtained by deleting elements from an original sequence while preserving the order of the remaining elements.
What is a non-decreasing subsequence?
A non-decreasing subsequence is a subsequence where the elements are in non-decreasing order.
Dynamic Programming Approach:
We can use dynamic programming to solve this problem. Let dp[i] be the number of non-decreasing subsequences that end at index i.
We can calculate dp[i] as follows:
If nums[i] >= nums[i-1], then dp[i] = dp[i-1] + 1 (we can extend the subsequence ending at index i-1)
Otherwise, dp[i] = 1 (we start a new subsequence ending at index i)
Code Implementation:
def count_non_decreasing_subsequences(nums):
"""
:param nums: List[int]
:return: int
"""
dp = [1] * len(nums)
for i in range(1, len(nums)):
for j in range(i):
if nums[i] >= nums[j]:
dp[i] = max(dp[i], dp[j] + 1)
return sum(dp)Example:
nums = [1, 2, 3, 4, 5]
result = count_non_decreasing_subsequences(nums)
print(result) # Output: 15Explanation:
There are 15 non-decreasing subsequences in the array, which are:
[1]
[1, 2]
[1, 2, 3]
[1, 2, 3, 4]
[1, 2, 3, 4, 5]
[1, 2, 4]
[1, 2, 4, 5]
[1, 3]
[1, 3, 4]
[1, 3, 4, 5]
[1, 4]
[1, 4, 5]
[2]
[2, 3]
[2, 3, 4]
[2, 3, 4, 5]
[2, 4]
[2, 4, 5]
[3]
[3, 4]
[3, 4, 5]
[4]
[4, 5]
[5]
Potential Applications in Real Life:
Natural language processing: Identifying patterns in sentences or documents.
Speech recognition: Identifying non-decreasing patterns in speech.
Time-series prediction: Predicting future values based on non-decreasing trends.
Financial analysis: Identifying patterns in stock prices or other financial data.
boundary_of_binary_tree
Problem Statement
Given a binary tree, return the boundary of the tree. The boundary of a binary tree is the concatenation of the left boundary, the leaves, and the right boundary.
The left boundary is the path from the root to the left-most node in the tree. The right boundary is the path from the root to the right-most node in the tree. The leaves are the nodes that have no children.
For example, the boundary of the following binary tree is:
1
/ \
2 3
/ \ \
4 5 6
[1, 2, 4, 5, 6, 3]Solution
We can solve this problem using a recursive approach. We start at the root of the tree and perform the following steps:
If the node is the root of the tree, add it to the boundary.
If the node is not the root of the tree, add it to the boundary if it is a left boundary or a right boundary.
Recursively call the function on the left and right children of the node.
The following Python code implements this algorithm:
def boundary_of_binary_tree(root):
boundary = []
# Add the root to the boundary.
boundary.append(root)
# Recursively call the function on the left and right children of the root.
left_boundary = boundary_of_binary_tree_left(root.left)
right_boundary = boundary_of_binary_tree_right(root.right)
# Add the left boundary to the boundary.
boundary += left_boundary
# Add the leaves to the boundary.
boundary += boundary_of_binary_tree_leaves(root)
# Add the right boundary to the boundary.
boundary += right_boundary
return boundary
def boundary_of_binary_tree_left(root):
boundary = []
# If the root is None, return an empty list.
if root is None:
return boundary
# If the root is a leaf, add it to the boundary.
if root.left is None and root.right is None:
boundary.append(root)
# Recursively call the function on the left child of the root.
left_boundary = boundary_of_binary_tree_left(root.left)
# If the left child is None, add the root to the boundary.
if root.left is None:
boundary.append(root)
# Add the left boundary to the boundary.
boundary += left_boundary
return boundary
def boundary_of_binary_tree_right(root):
boundary = []
# If the root is None, return an empty list.
if root is None:
return boundary
# If the root is a leaf, add it to the boundary.
if root.left is None and root.right is None:
boundary.append(root)
# Recursively call the function on the right child of the root.
right_boundary = boundary_of_binary_tree_right(root.right)
# If the right child is None, add the root to the boundary.
if root.right is None:
boundary.append(root)
# Add the right boundary to the boundary.
boundary += right_boundary
return boundary
def boundary_of_binary_tree_leaves(root):
boundary = []
# If the root is None, return an empty list.
if root is None:
return boundary
# If the root is a leaf, add it to the boundary.
if root.left is None and root.right is None:
boundary.append(root)
# Recursively call the function on the left and right children of the root.
left_leaves = boundary_of_binary_tree_leaves(root.left)
right_leaves = boundary_of_binary_tree_leaves(root.right)
# Add the left and right leaves to the boundary.
boundary += left_leaves + right_leaves
return boundaryComplexity Analysis
Time complexity: O(n), where n is the number of nodes in the binary tree.
Space complexity: O(n), where n is the number of nodes in the binary tree.
Applications
The boundary of a binary tree can be used to solve a variety of problems, such as:
Finding the perimeter of a binary tree.
Finding the distance between two nodes in a binary tree.
Checking if a binary tree is a complete binary tree.
next_greater_element_iii
Problem Statement:
Given an integer n, find the next larger integer that is permutations of its digits.
Example:
Input: 12
Output: 21Explanation:
The next larger permutation of 12 is 21.
Solution:
Step 1: Find the Pivot
Start from the rightmost digit and move left until you find a digit that is smaller than the digit to its right.
This digit is called the pivot.
Step 2: Find the Swap
Continue moving left until you find the smallest digit that is greater than the pivot.
This digit is called the swap.
Step 3: Swap the Pivot and the Swap
Swap the pivot and the swap.
Step 4: Reverse the Digits to the Right of the Pivot
Reverse the order of the digits to the right of the pivot.
Python Implementation:
def next_greater_element(n):
# Convert n to a string
n = str(n)
# Find the pivot
i = len(n) - 2
while i >= 0 and n[i] >= n[i + 1]:
i -= 1
# No pivot found, return -1
if i < 0:
return -1
# Find the swap
j = len(n) - 1
while j >= 0 and n[j] <= n[i]:
j -= 1
# Swap the pivot and the swap
n[i], n[j] = n[j], n[i]
# Reverse the digits to the right of the pivot
n = n[:i + 1] + n[i + 1:][::-1]
# Convert the string back to an integer
return int(n)Explanation:
The next_greater_element function takes an integer n as input and returns the next larger permutation of n if it exists, or -1 otherwise.
The function first converts n to a string. It then iterates over the digits of n from right to left until it finds a digit that is smaller than the digit to its right. This digit is the pivot.
The function then continues iterating over the digits of n from right to left until it finds the smallest digit that is greater than the pivot. This digit is the swap.
The function then swaps the pivot and the swap and reverses the order of the digits to the right of the pivot. Finally, the function converts the modified string back to an integer and returns it.
Real-World Applications:
The next greater element problem can be used in various real-world applications, such as:
Data encryption: Next greater element can be used as a building block in cryptographic algorithms to generate pseudorandom numbers.
Scheduling algorithms: Next greater element can be used to optimize the scheduling of tasks in a system.
Resource allocation: Next greater element can be used to allocate resources efficiently in a distributed system.
koko_eating_bananas
Problem Statement:
Koko loves to eat bananas. There are piles of bananas, where each pile has a specific number of bananas. Koko can eat only one banana at a time, and she can choose any pile of bananas to eat from.
We want to know how many bananas Koko will eat before she gets tired. Koko gets tired after eating a specific number of bananas, k.
Input:
piles: An array of integers representing the number of bananas in each pile.
h: The number of hours Koko has to eat bananas.
k: The number of bananas Koko will eat before she gets tired.
Output:
The maximum number of bananas that Koko can eat before getting tired.
Example 1:
Input: piles = [3, 6, 7, 11], h = 8, k = 4
Output: 6Explanation: Koko can eat a maximum of 6 bananas before getting tired. She can eat 2 bananas from the first pile, 2 bananas from the second pile, and 2 bananas from the third pile.
Example 2:
Input: piles = [30, 11, 23, 4, 20], h = 5, k = 30
Output: 23Explanation: Koko can eat a maximum of 23 bananas before getting tired. She can eat all 11 bananas from the second pile and 12 bananas from the third pile.
Time Complexity: O(N * log(max(piles))), where N is the number of piles and max(piles) is the maximum number of bananas in a pile.
Space Complexity: O(1), as no extra space is allocated.
Implementation in Python:
def minEatingSpeed(piles, h, k):
# Binary search for the minimum eating speed
left, right = 1, max(piles)
while left < right:
mid = (left + right) // 2
# Calculate the number of hours Koko needs to eat all the bananas at this speed
hours = 0
for pile in piles:
hours += math.ceil(pile / mid)
# If Koko can eat all the bananas within h hours, reduce the search space
if hours <= h:
right = mid
# Otherwise, increase the search space
else:
left = mid + 1
# Return the minimum eating speed
return leftExplanation of the Solution:
We start by initializing the binary search range to [1, max(piles)]. This means that Koko's eating speed can be anywhere between 1 banana per hour and the maximum number of bananas in any pile.
We then perform a binary search until the range is reduced to a single value.
For each mid value in the search range, we calculate the number of hours it would take Koko to eat all the bananas at that speed.
If Koko can eat all the bananas within h hours, we reduce the search space to [left, mid].
Otherwise, we increase the search space to [mid + 1, right].
The binary search continues until the left and right pointers meet, and we return the left pointer as the minimum eating speed.
Real-World Applications:
This problem can be applied to a wide range of real-world scenarios, such as:
Optimizing production lines: By determining the optimal speed at which a production line should operate, manufacturers can maximize efficiency and minimize production costs.
Scheduling appointments: Doctors or dentists can use a similar approach to determine the optimal number of patients to schedule in a day to avoid delays and ensure patient satisfaction.
Resource allocation: Project managers can use this algorithm to determine the optimal way to allocate resources to a team to complete a project within a specified time frame.
maximum_width_of_binary_tree
Problem Statement
Given a binary tree, write a function to find the maximum width of the tree. The maximum width of a tree is the maximum number of nodes on any level.
Solution
We can use a breadth-first search (BFS) to find the maximum width of a binary tree. The BFS algorithm involves traversing the tree level by level, starting from the root node. At each level, we store the number of nodes in an array. The maximum width of the tree is the maximum number of nodes in any of the arrays.
Here's a step-by-step breakdown of the BFS algorithm:
Initialize a queue with the root node.
While the queue is not empty:
Remove the first node from the queue and store it in a variable called
current_node.If the current node has a left child, add it to the queue.
If the current node has a right child, add it to the queue.
Increment the counter for the current level.
After processing all the nodes at the current level, store the counter in an array.
Repeat steps 2 and 3 until the queue is empty.
The maximum width of the tree is the maximum number of nodes in any of the arrays.
Here's a simplified example of how the BFS algorithm works:
1
/ \
2 3
/ \ \
4 5 6The BFS algorithm would traverse the tree as follows:
Start at the root node (1).
Add the left child (2) and the right child (3) to the queue.
The counter for the current level is 2.
Remove the first node from the queue (1).
Add the left child (4) and the right child (5) of node 2 to the queue.
The counter for the current level is now 4.
Remove the first node from the queue (2).
Add the right child (6) of node 3 to the queue.
The counter for the current level is now 2.
Remove the first node from the queue (3).
The queue is empty, so the BFS algorithm is complete.
The maximum width of the tree is the maximum number of nodes in any of the arrays, which is 4.
Code Implementation
Here's a Python implementation of the BFS algorithm to find the maximum width of a binary tree:
def maximum_width_of_binary_tree(root):
"""
Finds the maximum width of a binary tree.
Args:
root: The root node of the binary tree.
Returns:
The maximum width of the binary tree.
"""
# Initialize a queue with the root node.
queue = [root]
# Initialize the maximum width to 0.
max_width = 0
# While the queue is not empty.
while queue:
# Initialize the counter for the current level to 0.
level_width = 0
# For each node in the current level.
for node in queue:
# Increment the counter for the current level.
level_width += 1
# If the node has a left child, add it to the queue.
if node.left:
queue.append(node.left)
# If the node has a right child, add it to the queue.
if node.right:
queue.append(node.right)
# Update the maximum width if the current level is wider.
max_width = max(max_width, level_width)
# Remove all the nodes from the current level from the queue.
while queue:
queue.pop(0)
# Return the maximum width of the binary tree.
return max_widthReal-World Applications
The maximum width of a binary tree can be used in a variety of real-world applications, including:
Layouting out binary trees on a computer screen. The maximum width of a binary tree can be used to determine how wide the tree should be displayed on a computer screen.
Determining the complexity of a binary tree. A binary tree with a large maximum width is likely to be more complex than a binary tree with a small maximum width.
Balancing binary trees. The maximum width of a binary tree can be used to balance the tree, which can improve the performance of operations such as searching and insertion.
longest_palindromic_subsequence
Longest Palindromic Subsequence
Given a string s, the longest palindromic subsequence is the longest substring that reads the same forward and backward.
DP Solution
We can solve this problem using dynamic programming (DP). Let dp[i][j] be the length of the longest palindromic subsequence of the substring s[i:j+1]. We can fill in the DP table as follows:
If
i == j, thendp[i][j] = 1.If
i + 1 == j, thendp[i][j] = 2ifs[i] == s[j], and1otherwise.Otherwise,
dp[i][j] = max(dp[i+1][j], dp[i][j-1]) + (s[i] == s[j]).
Python Implementation
def longest_palindromic_subsequence(s):
"""
Returns the length of the longest palindromic subsequence of the string s.
"""
n = len(s)
dp = [[0 for _ in range(n)] for _ in range(n)]
# Fill in the DP table.
for i in range(n):
dp[i][i] = 1
for i in range(n-1):
if s[i] == s[i+1]:
dp[i][i+1] = 2
else:
dp[i][i+1] = 1
for length in range(3, n+1):
for i in range(n-length+1):
j = i + length - 1
if s[i] == s[j]:
dp[i][j] = dp[i+1][j-1] + 2
else:
dp[i][j] = max(dp[i+1][j], dp[i][j-1])
return dp[0][n-1]Time Complexity
The time complexity of the DP solution is O(n^2), where n is the length of the string.
Space Complexity
The space complexity of the DP solution is O(n^2), where n is the length of the string.
Real-World Applications
The longest palindromic subsequence problem has applications in various areas, such as:
Computational biology: Finding palindromic sequences in DNA and RNA is important for gene identification and sequencing.
Information retrieval: Palindromic sequences can be used to improve search algorithms by identifying common substrings between documents.
Natural language processing: Palindromic sequences can be used to identify anagrams and spelling errors.
push_dominoes
Problem statement:
There are N dominoes in a line, and we place each domino vertically upright.
In the beginning, we simultaneously push some of the dominoes either to the left or to the right.
After each second, each domino that is falling to the left or right will push the adjacent domino on the same side.
Determine the final state of the dominoes. If there are multiple solutions, print any of them.
Solution:
Initialization:
Create an array
dominoesto store the initial state of the dominoes. Each element in the array can be 'L' (left), 'R' (right), or '.' (upright).Initialize a variable
push_lefttoFalseto indicate that no domino has been pushed to the left yet.
Iterate over the dominoes:
For each domino at index
i:If the domino is upright ('.'):
Check if the domino to the left has been pushed to the left.
If so, push this domino to the left as well.
Otherwise, check if the domino to the right has been pushed to the right.
If so, push this domino to the right as well.
If the domino is already pushed to the left or right, continue.
Return the final state:
Return the
dominoesarray as the final state.
Simplified explanation:
Think of the dominoes as a line of vertical sticks.
You can push the sticks to the left or right.
If you push a stick to the left, it will fall and push the stick next to it to the left (if it's upright).
Similarly, if you push a stick to the right, it will fall and push the stick next to it to the right (if it's upright).
The goal is to find the final state of the dominoes after all the dominoes have fallen.
Code implementation:
def push_dominoes(dominoes):
# Initialize variables
push_left = False
dominoes = list(dominoes)
# Iterate over the dominoes
for i in range(len(dominoes)):
# If the domino is upright ('.'):
if dominoes[i] == '.':
# Check if the domino to the left has been pushed to the left
if i > 0 and dominoes[i - 1] == 'L':
dominoes[i] = 'L'
push_left = True
# Otherwise, check if the domino to the right has been pushed to the right
elif i < len(dominoes) - 1 and dominoes[i + 1] == 'R':
dominoes[i] = 'R'
# If the domino is already pushed to the left or right, continue
elif dominoes[i] == 'L' or dominoes[i] == 'R':
continue
# If any domino was pushed to the left, iterate again to handle cascading effects
if push_left:
for i in range(1, len(dominoes)):
if dominoes[i] == '.' and dominoes[i - 1] == 'L':
dominoes[i] = 'L'
# Return the final state
return ''.join(dominoes)Potential applications in real world:
Modeling the spread of viruses or diseases.
Simulating the flow of traffic.
Analyzing the dynamics of social networks.
widest_pair_of_indices_with_equal_range_sum
Problem Statement: Given an array of integers, find the widest pair of indices (i, j) with the same range sum. The range sum is defined as the sum of all integers in the array from index i to index j inclusive.
Solution: The problem can be optimally solved using a prefix sum array. The prefix sum array for an array nums is an array pre where pre[i] contains the sum of the first i elements in nums.
To find the widest pair of indices with equal range sum, we can iterate through all pairs of indices and compute the range sum for each pair. However, this approach would have a time complexity of O(N^2), which is not optimal.
A more efficient approach is to use a hash map to store the prefix sums and their corresponding indices. For each prefix sum, we store the smallest index it appears in the hash map.
When we compute the range sum for a pair of indices (i, j), we can check if the prefix sum for index i-1 is already stored in the hash map. If it is, then the range sum for (i, j) is the difference between pre[j] and pre[i-1]. We also update the hash map to store the prefix sum for index j with the smallest index it appears.
The following Python code implements this algorithm:
def widest_pair_of_indices_with_equal_range_sum(nums):
# Create a prefix sum array
pre = [0] * len(nums)
pre[0] = nums[0]
for i in range(1, len(nums)):
pre[i] = pre[i-1] + nums[i]
# Create a hash map to store prefix sums and their corresponding indices
sums = {}
sums[0] = 0
# Find the widest pair of indices with equal range sum
max_len = 0
max_i = 0
max_j = 0
for i in range(len(nums)):
if pre[i] in sums:
j = sums[pre[i]]
if i - j > max_len:
max_len = i - j
max_i = i
max_j = j
sums[pre[i]] = i
return max_i, max_jTime Complexity: The time complexity of this algorithm is O(N), where N is the length of the input array.
Applications in the Real World: This algorithm can be used in a variety of real-world applications, such as:
Finding the longest subarray with a given sum
Finding the number of subarrays with a given sum
Compressing data by removing duplicate subarrays
all_possible_full_binary_trees
Problem:
Given an integer n, construct all possible full binary trees with n nodes.
Solution:
Concept:
A full binary tree is a tree where every node has either zero or two children.
To construct all possible full binary trees with n nodes, we can use recursion.
Recursive Approach:
Base case: If
nis 0, return an empty list (no trees).Recursion:
For each possible left subtree with
inodes (0 <= i < n):Recursively construct all possible full binary trees with
inodes.For each possible right subtree with
n - i - 1nodes:Recursively construct all possible full binary trees with
n - i - 1nodes.Combine the left and right subtrees to form a full binary tree with
nnodes.
Return the list of all possible full binary trees with
nnodes.
Python Implementation:
def all_possible_full_binary_trees(n):
if n == 0:
return []
result = []
for i in range(n):
left_subtrees = all_possible_full_binary_trees(i)
right_subtrees = all_possible_full_binary_trees(n - i - 1)
for left in left_subtrees:
for right in right_subtrees:
root = TreeNode(None)
root.left = left
root.right = right
result.append(root)
return resultTime Complexity:
The time complexity of this solution is O(2^n * n^2).
The number of recursive calls is bounded by 2^n (for the number of possible combinations of left and right subtrees).
The total number of operations within each recursive call is bounded by n^2 (for combining the subtrees).
Real-World Applications:
Data structures: Full binary trees are used as a way to represent binary heaps, which are efficient data structures for storing and retrieving data.
Search algorithms: Full binary trees can be used to implement binary search trees, which are efficient algorithms for searching for data in a sorted collection.
Trie: Full binary trees are used as a data structure for implementing tries, which are efficient for storing and retrieving strings.
all_paths_from_source_to_target
Problem Statement:
Given a directed graph where each edge has a positive weight, find all the paths from a given source node to a given target node.
Intuition:
We can use a Depth-First Search (DFS) to explore all the possible paths from the source to the target. Whenever we reach the target node, we have found a path and we add it to our result list.
Implementation:
def all_paths_from_source_to_target(graph, source, target):
"""
Finds all the paths from a given source node to a given target node in a directed graph.
Args:
graph: The directed graph represented as a dictionary of nodes to their adjacent nodes.
source: The source node.
target: The target node.
Returns:
A list of all the paths from the source to the target.
"""
# Initialize the result list.
result = []
# Create a stack to store the current path.
stack = [source]
# While the stack is not empty, continue searching for paths.
while stack:
# Pop the current node from the stack.
current_node = stack.pop()
# Add the current node to the path.
path.append(current_node)
# If the current node is the target node, then we have found a path.
if current_node == target:
# Add the path to the result list.
result.append(path)
else:
# For each adjacent node, add it to the stack.
for adjacent_node in graph[current_node]:
stack.append(adjacent_node)
# Return the result list.
return resultExample:
Consider the following directed graph:
graph = {
'A': ['B', 'C'],
'B': ['D'],
'C': ['D'],
'D': []
}If we want to find all the paths from node 'A' to node 'D', we can call the following function:
paths = all_paths_from_source_to_target(graph, 'A', 'D')The function will return the following list of paths:
[['A', 'B', 'D'], ['A', 'C', 'D']]Real-World Applications:
This algorithm can be used in a variety of real-world applications, such as:
Finding the shortest path between two points on a map.
Finding all the possible routes between two cities.
Finding all the possible ways to complete a task.
partition_array_into_disjoint_intervals
Leetcode Problem:
Given an array of intervals, partition them into a minimum number of disjoint intervals.
Example:
Input: [[1,2],[2,3],[3,4],[1,3]] Output: [[1,4]]
Solution:
1. Sort the intervals by their starting points:
This helps group the intervals with overlapping start points together.
intervals.sort(key=lambda x: x[0])2. Initialize a stack with the first interval:
The stack will store the disjoint intervals.
stack = [intervals[0]]3. Iterate through the remaining intervals:
For each interval, check if it overlaps with the interval on top of the stack.
for interval in intervals[1:]:4. If it overlaps with the top interval:
Update the top interval to include the current interval.
stack[-1][1] = max(stack[-1][1], interval[1])5. If it does not overlap with the top interval:
Push the current interval onto the stack.
stack.append(interval)Simplified Explanation:
We start by sorting the intervals. This helps us group overlapping intervals together. We then use a stack to store the disjoint intervals.
We iterate through the intervals, and for each interval, we check if it overlaps with the top interval in the stack. If it overlaps, we update the top interval to include the current interval. If it does not overlap, we push the current interval onto the stack.
Real-World Application:
This algorithm can be used in a variety of real-world applications, such as:
Scheduling tasks to avoid conflicts
Allocating resources to multiple users without overlaps
Managing inventory to avoid overstocking
Example Implementation:
def partition_intervals(intervals):
"""
Partitions an array of intervals into a minimum number of disjoint intervals.
Args:
intervals: A list of intervals represented as tuples of (start, end).
Returns:
A list of disjoint intervals.
"""
intervals.sort(key=lambda x: x[0])
stack = [intervals[0]]
for interval in intervals[1:]:
if interval[0] <= stack[-1][1]:
stack[-1][1] = max(stack[-1][1], interval[1])
else:
stack.append(interval)
return stackExample Usage:
intervals = [[1,2],[2,3],[3,4],[1,3]]
disjoint_intervals = partition_intervals(intervals)
print(disjoint_intervals) # [[1,4]]keys_and_rooms
Problem:
There are N rooms and you start in room 0. Each room has a key that opens a different room or a lock that prevents you from leaving that room. Given an array rooms where rooms[i] is the key that opens the ith room or a lock that prevents you from leaving that room, determine whether you can visit all the rooms.
Solution:
1. Understand the Problem
You have a set of rooms and each room has a key to open a different room or a lock preventing you from leaving. You need to find out if you can visit all the rooms starting from room 0.
2. Plan the Solution
We can use a Queue to keep track of the rooms we need to visit. We start by adding room 0 to the Queue. Then we iterate through the Queue and for each room, we check if it has a key or a lock. If it has a key, we add the corresponding room to the Queue. If it has a lock, we skip it. We continue this process until we have visited all the rooms or the Queue is empty.
3. Implement the Solution
def can_visit_all_rooms(rooms):
"""
Determine whether you can visit all the rooms starting from room 0.
Parameters:
rooms: list of lists, where rooms[i] is the key that opens the ith room or a lock that prevents you from leaving that room
Returns:
bool: True if you can visit all the rooms, False otherwise
"""
# Create a Queue to keep track of the rooms we need to visit
queue = [0]
# Keep track of the visited rooms
visited = set()
# While there are still rooms to visit
while queue:
# Get the first room from the Queue
room = queue.pop(0)
# Add the room to the visited set
visited.add(room)
# Check if the room has a key or a lock
if rooms[room]:
# Add the corresponding room to the Queue
queue.extend(rooms[room])
# Return True if we have visited all the rooms, False otherwise
return len(visited) == len(rooms)4. Analyze the Complexity
The time complexity of the solution is O(N), where N is the number of rooms. We visit each room at most once, and we add each key to the Queue at most once. Therefore, the overall complexity is O(N).
Applications:
This problem can be applied to real-world scenarios such as:
Escape rooms: Determining if you can escape a room given the keys and locks available.
Maze traversal: Finding a path through a maze given the keys and locks encountered along the way.
Network connectivity: Verifying if all nodes in a network can communicate with each other.
implement_rand10_using_rand7
Implement rand10() using rand7()
Problem:
You are given a method called rand7() that generates a random integer between 1 and 7 (inclusive). Implement a method rand10() that generates numbers between 1 and 10 using only rand7().
Solution:
Explanation:
To generate a random number between 1 and 10 using rand7(), we can create a mapping from possible outcomes of rand7() to the range 1-10.
Divide the possible outcomes of
rand7()(1-7) into two sets: {1, 2, 3, 4} and {5, 6, 7}.When
rand7()generates 1-4, use it as the random number for 1-4.When
rand7()generates 5-7, generate another random number usingrand7(). If this new number is 1-4, add 5 to the original number to get a random number in the range 6-9. Otherwise, discard the original number and repeat the process.
Algorithm:
Call
rand7().If the number is 1-4, return it as the random number.
If the number is 5-7, call
rand7()again.If the new number is 1-4, add 5 to the original number to get a random number in the range 6-9.
If the new number is 5-7, discard the original number and go back to step 2.
Python Code:
import random
def rand10():
while True:
num1 = rand7()
num2 = rand7()
if num1 <= 4:
return num1
elif num2 <= 4:
return num1 + 5Real-World Applications:
Rolling dice, simulating probabilities, generating random samples in scientific experiments.
coin_change_ii
Problem Statement:
Given a target amount of money and a set of coin denominations, count the total number of possible ways to make up that amount using the given coins.
Simplified Explanation:
Imagine you have a piggy bank filled with coins of different denominations, like pennies, nickels, dimes, and quarters. You want to know how many different ways you can make up a certain amount of money using only these coins.
Optimal Solution:
Dynamic Programming is the best approach for this problem. Here's how it works:
Create a DP Array: Start with an array of length equal to the target amount plus one. Each element represents the number of ways to make up that amount.
Initialize the Array: Set the element at index 0 to 1 (one way to make up 0 amount is not using any coins). All other elements are initialized to 0.
Iterate Over Denominations: For each coin denomination, iterate over all the elements in the DP array. If the current amount minus the denomination is greater than or equal to 0, then add the number of ways to make up the current amount minus the denomination to the current amount.
Return the Last Element: The last element in the DP array represents the number of ways to make up the target amount.
Real-World Application:
This problem has various applications in real-world scenarios, such as:
Currency Exchange: Determine the number of ways to exchange a certain amount of money into different currencies with given exchange rates.
Inventory Management: Calculate the number of ways to package items into boxes of different sizes to minimize waste.
Scheduling: Determine the number of ways to schedule tasks within a given time frame with constraints.
Example Implementation in Python:
def coin_change_ii(amount, coins):
"""Counts the number of ways to make up the amount using the given coins.
Args:
amount: The target amount of money.
coins: The list of coin denominations.
Returns:
The number of ways to make up the amount using the given coins.
"""
# Create the DP array.
dp = [0] * (amount + 1)
dp[0] = 1 # Initialize the element at index 0 to 1.
# Iterate over the coin denominations.
for coin in coins:
# Iterate over the DP array.
for i in range(amount + 1):
# If the current amount minus the denomination is greater than or equal to 0,
# then add the number of ways to make up the current amount minus the
# denomination to the current amount.
if i - coin >= 0:
dp[i] += dp[i - coin]
# Return the last element in the DP array.
return dp[amount]Example Usage:
# Example amount and coin denominations.
amount = 10
coins = [1, 5, 10]
# Compute the number of ways to make up the amount using the given coins.
num_ways = coin_change_ii(amount, coins)
# Print the result.
print("The number of ways to make up", amount, "using the given coins is:", num_ways)Output:
The number of ways to make up 10 using the given coins is: 4sum_of_subarray_minimums
Problem Statement:
Given an array of integers, return the sum of the minimum value in every contiguous subarray.
Example:
For the array [3, 1, 2, 4], the contiguous subarrays are:
[3] with minimum value 3
[1, 2] with minimum value 1
[2, 4] with minimum value 2
[4] with minimum value 4
Therefore, the sum of the minimum value in every contiguous subarray is 3 + 1 + 2 + 4 = 10.
Optimal Solution:
Sliding Window with Monotonic Stack:
Initialize a stack and a variable
sumto store the sum of the minimum values.Iterate through the array:
For each element, pop elements from the stack while they are greater than the current element. This ensures that the stack always contains a decreasing sequence of elements.
Push the current element onto the stack.
Add the minimum value on the stack (which is the element at the top of the stack) to
sum.
Return
sum.
Python Implementation:
def sum_of_subarray_minimums(arr):
stack = []
sum = 0
for num in arr:
while stack and stack[-1] > num:
stack.pop()
stack.append(num)
sum += stack[-1]
return sumExplanation:
The
stackmaintains a decreasing sequence of elements.When a smaller element is encountered, it is pushed onto the stack, ensuring that the minimum value is always at the top.
The
sumvariable is updated with the minimum value on the stack for each element.The final
sumrepresents the sum of the minimum values in all contiguous subarrays.
Real-World Application:
This algorithm can be useful in situations where we need to find the minimum value in a series of consecutive measurements or observations. For example, in financial data analysis, it can be used to determine the minimum value of a stock price over a given period.
binary_subarrays_with_sum
Problem Statement:
Given a binary array, find the number of non-empty subarrays that have a sum equals to k.
Example:
Input: nums = [1,1,1], k = 2
Output: 2Explanation:
The subarray [1,1] has a sum of 2.
The subarray [1,1] has a sum of 2.
Solution:
We can use a prefix sum array to solve this problem. The prefix sum array stores the cumulative sum of the elements in the binary array.
Create a Prefix Sum Array:
Start with a prefix sum array of size n, where n is the length of the binary array.
For each element
nums[i]in the binary array:If
nums[i]is 1, add 1 to the corresponding element in the prefix sum array.Otherwise, add 0.
Count Subarrays with Sum
k:Iterate through the prefix sum array:
For each element
prefix[i], check ifprefix[i] - prefix[j] == kfor anyj < i.If so, increment the count of non-empty subarrays with sum
k.
Python Implementation:
def binary_subarrays_with_sum(nums, k):
count = 0
prefix = [0] * len(nums)
prefix[0] = 1 if nums[0] == 1 else 0
for i in range(1, len(nums)):
prefix[i] = prefix[i - 1] + (1 if nums[i] == 1 else 0)
for i in range(len(nums)):
for j in range(i):
if prefix[i] - prefix[j] == k:
count += 1
return countTime Complexity: O(n^2), where n is the length of the binary array.
Space Complexity: O(n), for the prefix sum array.
Applications:
This problem has applications in signal processing, data compression, and other areas where counting subarrays with specific properties is important.
flip_string_to_monotone_increasing
Introduction
The flip_string_to_monotone_increasing problem is a coding interview question that asks you to find the minimum number of flips needed to make a string of 0s and 1s monotone increasing. A string is monotone increasing if all the 0s come before all the 1s.
Problem Statement
Given a string of 0s and 1s, find the minimum number of flips needed to make the string monotone increasing.
Example
Input: "001110"
Output: 1
Input: "010110"
Output: 2Solution
The solution to this problem is to find the longest prefix of 0s and the longest suffix of 1s. The minimum number of flips needed is equal to the length of the string minus the length of the prefix and the length of the suffix.
Python Implementation
def flip_string_to_monotone_increasing(string):
"""
Flips a string of 0s and 1s to make it monotone increasing.
Args:
string: The string to flip.
Returns:
The minimum number of flips needed.
"""
# Find the longest prefix of 0s.
prefix_length = 0
for i in range(len(string)):
if string[i] == '0':
prefix_length += 1
else:
break
# Find the longest suffix of 1s.
suffix_length = 0
for i in range(len(string) - 1, -1, -1):
if string[i] == '1':
suffix_length += 1
else:
break
# Return the minimum number of flips needed.
return len(string) - prefix_length - suffix_length
# Example
string = "001110"
result = flip_string_to_monotone_increasing(string)
print(result) # Output: 1Applications
This problem has applications in data cleaning and data analysis. For example, you could use this algorithm to clean up a dataset that contains inconsistent data.
network_delay_time
Problem Statement
Given an array of network tower heights, determine the minimum duration it will take for a signal to reach the last tower.
Example 1:
Input: [1, 2, 3, 4, 5]
Output: 4Explanation:
The signal can directly reach tower 2 from tower 1, tower 3 from tower 2, and so on.
Example 2:
Input: [1, 2, 4, 7, 3, 5]
Output: 6Explanation:
The signal can directly reach tower 2 from tower 1, but it takes 2 units of time to reach tower 3 from tower 2.
Solution
The key observation is that the signal can only travel from a higher tower to a lower tower. Therefore, we can use a stack to keep track of the towers that have been visited.
Here's the step-by-step algorithm:
Push the first tower into the stack.
While the stack is not empty, pop the top tower and check the height of the next tower.
If the next tower is lower than the popped tower, increment the duration by 1 and push the next tower into the stack.
Otherwise, continue to the next tower.
Repeat steps 2-4 until the signal reaches the last tower.
Python Implementation
def network_delay_time(towers):
stack = [towers[0]]
duration = 0
while stack:
curr = stack.pop()
next_tower = curr + 1
if next_tower <= len(towers) and towers[next_tower - 1] < towers[curr - 1]:
duration += 1
stack.append(next_tower)
return durationComplexity Analysis
The time complexity of this solution is O(N), where N is the number of towers. The algorithm iterates over the towers once, and each iteration takes constant time.
The space complexity is also O(N), as the stack can store up to N towers at any given time.
Real-World Applications
This algorithm can be applied to any situation where a signal needs to travel from one point to another, such as:
Communication networks: Determining the minimum delay for a signal to travel from a source to a destination.
Traffic management: Calculating the minimum time it takes for a vehicle to travel from one road segment to another.
Supply chain management: Determining the minimum time it takes for a product to be delivered from a warehouse to a store.
number_of_provinces
Problem: Given a 2D grid representing a map of a country, where each cell represents a province, determine the number of unique provinces in the map. Two provinces are considered distinct if they share no common border, even diagonally.
Solution: Union Find (Disjoint Set Union)
This problem can be solved efficiently using the Union Find data structure. Union Find maintains a collection of disjoint sets, and we can perform two main operations:
Union(set1, set2): Merge two sets into one.
Find(element): Find the representative element of the set that contains the given element.
Algorithm:
Initialize a Union Find data structure with each cell in the grid as a separate set.
Iterate over the grid and for each cell:
Check if its neighbors (up, down, left, right, excluding diagonals) belong to the same set.
If they do, continue (no new province found).
If not, perform Union(cell_set, neighbor_set) to merge the sets.
Return the number of sets in the Union Find data structure.
Implementation:
class UnionFind:
def __init__(self):
self.parents = {} # Maps each element to its parent
self.sizes = {} # Maps each element to its set size
def find(self, element):
if element not in self.parents:
self.parents[element] = element
self.sizes[element] = 1
parent = self.parents[element]
while parent != element:
parent = self.parents[parent]
self.parents[element] = parent
return parent
def union(self, set1, set2):
parent1, parent2 = self.find(set1), self.find(set2)
if parent1 != parent2:
# Merge the smaller set into the larger one
if self.sizes[parent1] > self.sizes[parent2]:
self.parents[parent2] = parent1
self.sizes[parent1] += self.sizes[parent2]
else:
self.parents[parent1] = parent2
self.sizes[parent2] += self.sizes[parent1]
def find_provinces(grid):
if not grid:
return 0
uf = UnionFind()
for i, row in enumerate(grid):
for j, cell in enumerate(row):
if cell == 1:
uf.union((i, j), (i - 1, j)) # Check top neighbor
uf.union((i, j), (i + 1, j)) # Check bottom neighbor
uf.union((i, j), (i, j - 1)) # Check left neighbor
uf.union((i, j), (i, j + 1)) # Check right neighbor
return len(set(uf.find(cell) for row in grid for cell in row if cell == 1))Complexity:
Time complexity: O(N), where N is the number of cells in the grid.
Space complexity: O(N), for the Union Find data structure.
Real-World Applications:
Union Find can be used in a variety of real-world applications, such as:
Identifying connected components in a graph
Detecting cycles in a graph
Solving problems involving disjoint sets
construct_binary_tree_from_preorder_and_postorder_traversal
Problem Statement:
Given the preorder and postorder traversals of a binary tree, reconstruct the original binary tree.
Input:
Preorder traversal: A list of nodes visited in the order of root, left subtree, right subtree.
Postorder traversal: A list of nodes visited in the order of left subtree, right subtree, root.
Output:
The original binary tree.
Best & Performant Solution:
Iterative Approach:
Create a stack: Initialize an empty stack.
Push the root note on the stack: Since we know the root node is always the first element in the preorder traversal, push it onto the stack.
Initialize two pointers: prev, which will point to the previously visited node, and i to traverse the postorder traversal.
While the stack is not empty:
Push the next node from the postorder traversal onto the stack.
If the top of the stack is the same as prev, it means we have finished processing the left subtree of the last visited node.
Pop the top of the stack and assign it to prev.
Make the top of the stack (i.e., the last element pushed) as the right child of the prev node.
Update prev to point to the right child.
Push the left child of the prev node onto the stack.
Time Complexity: O(n) Space Complexity: O(n)
Python Implementation:
def construct_binary_tree(preorder, postorder):
stack = []
root = TreeNode(preorder[0])
stack.append(root)
prev = None
i = len(postorder) - 1
while stack and i >= 0:
curr = TreeNode(postorder[i])
if not stack[-1].left or stack[-1].left == prev:
stack[-1].left = curr
stack.append(curr)
prev = curr
elif stack[-1].right == prev:
stack[-1].right = curr
prev = curr
stack.pop()
i -= 1
return rootExample:
Input:
preorder = [3, 9, 20, 15, 7]
postorder = [9, 15, 7, 20, 3]Output:
3
/ \
9 20
/ \ / \
2 7 15 3Applications in Real World:
Data Compression: Preorder and postorder traversals can be used to represent binary trees in a compressed form, which can be useful for storing and transmitting large trees efficiently.
Tree Serialization: Binary trees can be serialized (converted into a sequence of data) and deserialized (reconstructed from the sequence) using preorder and postorder traversals.
Tree Analysis: Preorder and postorder traversals provide information about the structure and content of a binary tree, which can be useful for analysis and debugging.
search_in_a_sorted_array_of_unknown_size
Problem Statement:
Given a sorted array of unknown size, find the target element in the array.
Approach:
The key idea is to use binary search. However, since the array is of unknown size, we need to first find the length of the array.
Step 1: Find the Length of the Array
Start by setting the start and end indices to 0 and 1, respectively.
While the array element at the end index is not empty:
Double the end index.
This process quickly finds the length of the array (let's call it n).
Step 2: Perform Binary Search
Now that we know the length of the array, we can perform binary search:
Start with start and end indices of 0 and
n-1, respectively.While start <= end:
Calculate the mid index.
Compare the target element with the array element at the mid index.
If equal, return the mid index.
If the target is less than the mid element, set end to mid-1.
Otherwise, set start to mid+1.
Code Implementation:
def search_in_a_sorted_array_of_unknown_size(nums, target):
# Find the length of the array
start, end = 0, 1
while nums[end] is not None:
end *= 2
# Perform binary search
while start <= end:
mid = (start + end) // 2
if nums[mid] == target:
return mid
elif nums[mid] < target:
start = mid + 1
else:
end = mid - 1
return -1 # Target not foundExample:
# nums is a sorted array of unknown size
nums = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
target = 5
result = search_in_a_sorted_array_of_unknown_size(nums, target)
print(result) # Output: 4Applications:
This algorithm can be used in any scenario where you have a sorted dataset of unknown size and need to quickly find a specific element. For example:
Searching for a specific record in a database
Finding the index of a specific keyword in a large text document
Locating a file in a file system with a very large number of files
out_of_boundary_paths
Problem Statement:
Given a m x n grid with obstacles, you are starting at the top left corner.
Return the number of different paths that you can take to reach the bottom right corner without hitting any obstacles.
Constraints:
1 <= m, n <= 100obstacles.length <= 50obstacles[i]is in the range[0, m - 1] x [0, n -1].The top left and bottom right corners are not obstacles.
Example 1:
Input: grid = [[0,0,0],[0,1,0],[0,0,0]]
Output: 2
Explanation: There are two different paths:
1. (0,0) -> (0,1) -> (0,2)
2. (0,0) -> (1,0) -> (2,0)Example 2:
Input: grid = [[0,1],[0,0]]
Output: 1
Explanation: There is only one path:
(0,0) -> (0,1) -> (1,1)Intuition:
The number of paths from the top left corner to any other cell in the grid is equal to the sum of the number of paths from the cells above and to the left of that cell.
If a cell is an obstacle, then there are no paths to that cell, so the number of paths to that cell is 0.
If a cell is the top left corner, then there is only one path to that cell, so the number of paths to that cell is 1.
Steps (in Python):
Create a 2D array
dpof sizem x n, wheredp[i][j]represents the number of paths from the top left corner to cell(i, j).Initialize
dp[0][0]to 1, since there is only one path to the top left corner.Iterate over the grid from left to right and top to bottom, and for each cell
(i, j), calculatedp[i][j]as follows:if grid[i][j] == 1: # If the cell is an obstacle, set dp[i][j] to 0. dp[i][j] = 0 else: # Otherwise, set dp[i][j] to the sum of dp[i-1][j] and dp[i][j-1]. dp[i][j] = dp[i-1][j] + dp[i][j-1]Return
dp[m-1][n-1], which is the number of paths from the top left corner to the bottom right corner.
Simplified Explanation:
Imagine you are standing at the top left corner of a grid. You want to find out how many different paths you can take to reach the bottom right corner.
You can either move right or down. If you move right, you will end up in the cell to the right of your current cell. If you move down, you will end up in the cell below your current cell.
You can keep track of the number of paths to each cell in a grid. For example, the number of paths to the top left corner is 1 (because there is only one way to get there).
You can then use this information to calculate the number of paths to any other cell in the grid. For example, the number of paths to the cell to the right of the top left corner is 2 (because you can either move right from the top left corner or down from the top left corner).
You can continue this process until you reach the bottom right corner of the grid. The number of paths to the bottom right corner is the answer to the problem.
Potential Applications:
This problem can be used to solve a variety of problems in the real world, such as:
Finding the shortest path through a maze
Finding the number of ways to get from one place to another on a map
Finding the number of ways to solve a puzzle
Complete Code Implementation:
def unique_paths_with_obstacles(grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
m, n = len(grid), len(grid[0])
dp = [[0] * n for _ in range(m)]
for i in range(m):
for j in range(n):
if grid[i][j] == 1: # If the cell is an obstacle, set dp[i][j] to 0.
dp[i][j] = 0
elif i == 0 or j == 0: # If the cell is the top left corner or on the first row/column, set dp[i][j] to 1.
dp[i][j] = 1
else: # Otherwise, set dp[i][j] to the sum of dp[i-1][j] and dp[i][j-1].
dp[i][j] = dp[i-1][j] + dp[i][j-1]
return dp[m-1][n-1]prime_palindrome
Problem Statement
A prime palindrome is a prime number that is also a palindrome. For example, 11 is a prime palindrome because it is a prime number and it reads the same backward as it does forward.
Given an integer n, return the number of prime palindromes that are less than or equal to n.
Solution
We can use the following steps to solve this problem:
Generate a list of all prime numbers less than or equal to n.
For each prime number, check if it is a palindrome.
Count the number of prime palindromes.
Here is a Python implementation of this solution:
def count_prime_palindromes(n):
"""Counts the number of prime palindromes less than or equal to n."""
# Generate a list of all prime numbers less than or equal to n.
primes = [2]
for i in range(3, n + 1):
is_prime = True
for prime in primes:
if i % prime == 0:
is_prime = False
break
if is_prime:
primes.append(i)
# Count the number of prime palindromes.
count = 0
for prime in primes:
if prime == reversed(prime):
count += 1
return countExample
The following example shows how to use the count_prime_palindromes function:
n = 100
count = count_prime_palindromes(n)
print(count) # Output: 25In this example, the count_prime_palindromes function returns 25, which is the number of prime palindromes less than or equal to 100.
Applications
This problem has applications in cryptography and data security. For example, prime palindromes can be used to generate secure keys for encryption and decryption.
image_overlap
Image Overlap
Given two images, represented as binary matrices, calculate the area of their overlap.
Problem Breakdown
We have two binary matrices, representing two images.
Each matrix contains 1s and 0s, where 1 represents a pixel in the image and 0 represents a background pixel.
We need to find the area where both images overlap, which means both matrices have 1s in the same positions.
The area of overlap is the number of 1s in the overlapping region.
Solution
To find the area of overlap, we can iterate through both matrices simultaneously and count the number of positions where both matrices have 1s:
def image_overlap(img1, img2):
# Get the dimensions of the images
rows1, cols1 = len(img1), len(img1[0])
rows2, cols2 = len(img2), len(img2[0])
# Ensure both images have the same dimensions
if rows1 != rows2 or cols1 != cols2:
raise ValueError("Images must have the same dimensions")
# Initialize the overlap area counter
overlap = 0
# Iterate through the images and count overlapping pixels
for i in range(rows1):
for j in range(cols1):
if img1[i][j] == 1 and img2[i][j] == 1:
overlap += 1
return overlapApplications in Real World
Image processing
Computer vision
Medical imaging
Object detection
Visual tracking
candy_crush
Candy Crush
Problem Statement:
You are given an m x n board, where each cell contains a candy of a certain type. The candies are arranged in a 2D grid.
A move consists of choosing a group of candies that are connected vertically or horizontally, and then removing them from the board. If you remove at least 3 candies of the same type in one move, they will explode and disappear.
After the candies explode, the candies above them will fall down and fill the empty spaces. If there are no candies above them, new candies will appear from the top.
Your task is to determine the maximum number of candies that can be removed in a single move.
Example:
Input:
board = [[1,2,3],[4,5,6],[7,8,9]]Output:
6Explanation: The following candies can be removed in one move:
[1,2,3]
[4,5,6]This will result in the following board:
[7,8,9]Simplified Explanation:
Imagine a game of Candy Crush, where you have a grid of candies. Each move, you can choose a group of candies that are touching each other horizontally or vertically, and then click on them to make them explode. If you click on at least 3 candies of the same type, they will all disappear.
After you click on the candies, the candies above them will fall down and fill the empty spaces. If there are no candies above them, new candies will appear from the top.
Your task is to figure out the maximum number of candies that you can click on and make disappear in a single move.
Real-World Applications:
Candy Crush is a popular mobile game that has been downloaded over 2 billion times. The game is based on the concept of matching candies and making them disappear.
The algorithms used to solve the Candy Crush problem can be applied to other real-world problems, such as:
Image processing: Identifying and removing objects from images.
Data mining: Clustering data points and identifying patterns.
Logistics: Optimizing the distribution of goods.
Code Implementation:
Here is a Python implementation of the Candy Crush algorithm:
def candy_crush(board):
"""
Finds the maximum number of candies that can be removed in a single move.
Args:
board: A 2D list of candies.
Returns:
The maximum number of candies that can be removed in a single move.
"""
# Check if the board is empty.
if not board:
return 0
# Get the dimensions of the board.
m, n = len(board), len(board[0])
# Initialize the maximum number of candies to 0.
max_candies = 0
# Iterate over the rows and columns of the board.
for i in range(m):
for j in range(n):
# Check if the current candy can be removed.
candies = set()
candies.add(board[i][j])
if i+1 < m and board[i+1][j] == board[i][j]:
candies.add(board[i+1][j])
if i-1 >= 0 and board[i-1][j] == board[i][j]:
candies.add(board[i-1][j])
if j+1 < n and board[i][j+1] == board[i][j]:
candies.add(board[i][j+1])
if j-1 >= 0 and board[i][j-1] == board[i][j]:
candies.add(board[i][j-1])
# If the current candy can be removed, update the maximum number of candies.
if len(candies) >= 3:
max_candies = max(max_candies, len(candies))
# Return the maximum number of candies.
return max_candiesExample Usage:
# Create a Candy Crush board.
board = [[1,2,3],[4,5,6],[7,8,9]]
# Find the maximum number of candies that can be removed in a single move.
max_candies = candy_crush(board)
# Print the maximum number of candies.
print(max_candies)Output:
6encode_and_decode_tinyurl
Problem Statement:
Given a long URL, design a URL shortening service that encodes it into a shorter string and decodes it back to the original URL, like tinyurl.
Solution:
The solution involves two main steps:
Encoding: Convert the long URL into a shorter string using a unique identifier.
Decoding: Convert the shorter string back to the original long URL using the unique identifier.
Implementation:
Encoding:
Create a hash table (e.g., a dictionary in Python) to store the mapping between long URLs and short URLs.
For a new long URL, generate a unique identifier (e.g., a random string or a hash of the URL).
Store the mapping of the long URL to the unique identifier in the hash table.
def encode(long_url):
if long_url in url_map:
return url_map[long_url]
# Generate a unique identifier
unique_id = generate_unique_id()
# Store the mapping in the hash table
url_map[long_url] = unique_id
return unique_idDecoding:
Lookup the long URL corresponding to the short URL (unique identifier) in the hash table.
If found, return the long URL.
If not found, return an error.
def decode(short_url):
if short_url not in url_map:
return "Invalid short URL"
return url_map[short_url]Real-World Applications:
URL shortening services are widely used in:
Social media: Shortening long URLs in tweets or posts.
Email marketing: Shortening URLs in email campaigns to track clicks.
Analytics: Tracking website traffic by shortening URLs and monitoring clicks.
E-commerce: Shortening long product URLs to make them easier to share.
Security: Hiding sensitive information within long URLs by shortening them.
lonely_pixel_i
Problem: Find the maximum product of two numbers in a list.
Efficient Solution:
Python Implementation:
def max_product(arr):
"""Returns the maximum product of two numbers in a list."""
# Initialize maximum and minimum products.
max_prod = -float('inf')
min_prod = float('inf')
# Initialize previous maximum and minimum products.
prev_max = 1
prev_min = 1
for num in arr:
# Update maximum and minimum products.
curr_max = max(num, num * prev_max, num * prev_min)
curr_min = min(num, num * prev_max, num * prev_min)
# Update previous maximum and minimum products.
prev_max = curr_max
prev_min = curr_min
# Update maximum product.
max_prod = max(max_prod, curr_max)
return max_prodExplanation:
Initialize maximum and minimum products: Since the product can be positive or negative, we initialize the maximum and minimum products to the smallest and largest possible values, respectively.
Initialize previous maximum and minimum products: We initialize the previous maximum and minimum products to 1. This is because multiplying any number by 1 doesn't change its value.
Iterate over the list: For each number in the list, we calculate the current maximum and minimum products.
Calculate current maximum and minimum products: We calculate the current maximum product by taking the maximum of the following three values:
The number itself.
The product of the number and the previous maximum product.
The product of the number and the previous minimum product.
We calculate the current minimum product by taking the minimum of the following three values:
The number itself.
The product of the number and the previous maximum product.
The product of the number and the previous minimum product.
Update previous maximum and minimum products: We update the previous maximum and minimum products with the current maximum and minimum products, respectively.
Update maximum product: We update the maximum product with the current maximum product.
Return maximum product: After iterating over the list, we return the maximum product.
Applications:
This algorithm can be used in real-world applications such as:
Finance: Calculating the maximum product of stock prices.
Machine learning: Finding the maximum product of features for classification or regression models.
Computer graphics: Calculating the maximum product of colors for image blending.
soup_servings
Problem Statement:
There is a restaurant with a list of soups to be served. Customers come in and order a certain soup, and the restaurant wants to know how many servings of a particular soup they have left.
Given a list of soup types and a list of ordered soups, write a function that takes these two lists as input and calculates the remaining servings of each soup type.
Python Implementation:
def soup_servings(soup_types, ordered_soups):
"""
Calculates the remaining servings of each soup type.
Args:
soup_types (list): List of soup types.
ordered_soups (list): List of ordered soups.
Returns:
list: List of remaining servings of each soup type.
"""
# Initialize a dictionary to store the remaining servings of each soup type.
soup_servings = {}
for soup_type in soup_types:
soup_servings[soup_type] = 0
# Iterate over the ordered soups and decrement the remaining servings of each type.
for soup_type in ordered_soups:
soup_servings[soup_type] -= 1
# Return the list of remaining servings.
return soup_servingsExample:
soup_types = ["Soup A", "Soup B", "Soup C"]
ordered_soups = ["Soup A", "Soup B", "Soup A"]
remaining_servings = soup_servings(soup_types, ordered_soups)
print(remaining_servings) # Output: {'Soup A': -1, 'Soup B': -1, 'Soup C': 0}In this example, the restaurant has three types of soup: "Soup A", "Soup B", and "Soup C". Three customers come in and order a soup each: "Soup A", "Soup B", and "Soup A". The function calculates the remaining servings of each soup type, which are -1 for "Soup A" and "Soup B", and 0 for "Soup C".
Real-World Applications:
This function can be used in a restaurant to keep track of the remaining servings of each soup type. This information can be used to plan the menu and ensure that there is enough soup to meet the demand.
Performance Optimization:
The function can be optimized by storing the remaining servings of each soup type in a dictionary. This eliminates the need to iterate over the list of soup types to find the remaining servings of a particular soup type.
split_concatenated_strings
Problem Statement
Given a list of strings, each containing two concatenated words, split each string into its two component words.
Example:
Input: ["ab", "cd", "ef"]
Output: [["a", "b"], ["c", "d"], ["e", "f"]]Solution
One approach to solving this problem is to use the split() method. The split() method takes a delimiter as an argument and splits the string into a list of substrings at the delimiter. In this case, we can use the empty string as the delimiter, which will split the string at every character.
def split_concatenated_strings(strings):
result = []
for string in strings:
result.append(list(string))
return resultExample:
strings = ["ab", "cd", "ef"]
result = split_concatenated_strings(strings)
print(result)Output:
[['a', 'b'], ['c', 'd'], ['e', 'f']]Complexity Analysis
Time complexity: O(n), where n is the total length of all strings in the input list.
Space complexity: O(n), since we create a new list for each string in the input list.
Real-World Applications
This problem has several real-world applications, including:
Text processing: Splitting text into individual words or phrases is a common task in natural language processing.
Data cleaning: Removing unnecessary whitespace or other characters from data can be done by splitting the data into individual components.
Security: Splitting user input into individual characters can help prevent malicious strings from being executed.
binary_tree_pruning
LeetCode Problem: Binary Tree Pruning
Problem Statement:
Given a binary tree, prune it such that all subtrees with a sum of values less than a specified threshold are removed.
Example:
Input:
1
/ \
2 3
/ \
4 5
Threshold = 3
Output:
1
/ \
2 None
/ \
4 NoneSolution:
1. Recursive DFS:
Traverse the tree using Depth-First Search (DFS).
For each node:
Calculate the sum of values in its left and right subtrees.
If the sum is less than the threshold, prune both subtrees (set them to
None).If the sum is greater than or equal to the threshold, recursively apply the DFS to both subtrees.
Simplified Python Code:
def prune_tree(root, threshold):
# Check if the root is None
if not root:
return None
# Calculate the sum of values in left and right subtrees
left_sum = prune_tree(root.left, threshold)
right_sum = prune_tree(root.right, threshold)
# Check if the current node's sum is less than the threshold
if root.val + left_sum + right_sum < threshold:
return None
# Recursively prune the left and right subtrees
root.left = prune_tree(root.left, threshold)
root.right = prune_tree(root.right, threshold)
# Return the root node
return root2. Iterative BFS:
Traverse the tree using Breadth-First Search (BFS).
Maintain a queue of nodes to visit.
For each node in the queue:
Calculate the sum of values in its left and right subtrees.
If the sum is less than the threshold, prune both subtrees and remove the node from the queue.
If the sum is greater than or equal to the threshold, enqueue its left and right subtrees.
Simplified Python Code:
def prune_tree_bfs(root, threshold):
# Create a queue and enqueue the root
queue = [root]
# While the queue is not empty
while queue:
# Dequeue the first node
node = queue.pop(0)
# Calculate the sum of values in left and right subtrees
left_sum = node.left.val if node.left else 0
right_sum = node.right.val if node.right else 0
# Check if the current node's sum is less than the threshold
if node.val + left_sum + right_sum < threshold:
# Prune both subtrees
node.left = node.right = None
else:
# Enqueue the left and right subtrees
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
# Return the root node
return rootReal-World Applications:
Binary tree pruning can be used in various real-world applications, such as:
Data compression: By removing unnecessary or redundant subtrees, binary tree pruning can reduce the size of a tree-structured data without compromising its functionality.
Image processing: Pruning can be applied to binary trees representing images to simplify them and remove noise or unwanted details.
Decision Trees: Pruning can be applied to decision trees to prevent overfitting and improve their generalization performance.
short_encoding_of_words
Problem: Given a list of words, encode them in a way that they can be decoded back to the original list of words. The encoding should be as short as possible.
Solution:
Step 1: Sort the words alphabetically.
This step helps reduce the size of the encoded string because words that are close together in alphabetical order will have shorter encodings.
Step 2: Assign each word a unique integer code.
The integer codes should be assigned in ascending order, starting from 1.
Step 3: Encode the list of words.
The encoded string is a concatenation of the integer codes of the words, separated by a delimiter. The delimiter can be any character that is not used in any of the words.
Step 4: Decode the encoded string.
The encoded string can be decoded back into the original list of words by splitting the string into individual integer codes, and then looking up the corresponding words using the integer codes.
Example:
Input: ['apple', 'banana', 'cake', 'dog']
Output: '1 2 3 4'
Explanation:
Sort the words: ['apple', 'banana', 'cake', 'dog']
Assign integer codes: 1: apple, 2: banana, 3: cake, 4: dog
Encode the list: '1 2 3 4'
Decode the encoded string: ['apple', 'banana', 'cake', 'dog']
Code Implementation:
def short_encode_words(words):
# Sort the words alphabetically
words.sort()
# Assign each word a unique integer code
word_to_code = {}
for i, word in enumerate(words, 1):
word_to_code[word] = i
# Encode the list of words
encoded_string = ' '.join(str(word_to_code[word]) for word in words)
return encoded_string
def decode_short_encoded_words(encoded_string):
# Split the encoded string into individual integer codes
codes = [int(code) for code in encoded_string.split()]
# Decode the integer codes into words
words = [code_to_word[code] for code in codes]
return words
# Example
words = ['apple', 'banana', 'cake', 'dog']
encoded_string = short_encode_words(words)
decoded_words = decode_short_encoded_words(encoded_string)
print(decoded_words) # ['apple', 'banana', 'cake', 'dog']Applications in Real World:
Short encoding of words has applications in data compression, such as in text files, databases, and XML documents. By encoding words with shorter codes, the overall size of the data can be reduced.
lonely_pixel_ii
Problem Statement:
Given an array of integers, find the maximum sum of non-adjacent elements in the array.
Example:
Input: [1, 2, 3, 1]
Output: 4 (maximum sum is 2 + 1)Solution:
There are two subproblems to consider:
Take the first element: If you take the first element, you can't take the second element. So, the maximum sum excluding the second element is the maximum sum of the subarray starting from the third element.
Skip the first element: If you skip the first element, you can take the second element. So, the maximum sum including the second element is the maximum sum of the subarray starting from the third element, plus the second element.
To solve the problem, we can recursively compute the maximum sum for each subarray, starting from the third element. We can store the maximum sums in an array to avoid recomputation.
Simplified Explanation:
Imagine you have a row of coins, and you want to collect the maximum amount of money without taking any adjacent coins. You can either start with the first coin, or you can skip it and start with the second coin.
If you start with the first coin, you can't take the second coin. So, you need to find the maximum amount of money you can collect from the rest of the coins, starting with the third coin.
If you skip the first coin, you can take the second coin. So, you need to find the maximum amount of money you can collect from the rest of the coins, starting with the third coin, and add the second coin to that amount.
You can keep doing this recursively until you reach the end of the row of coins. The maximum amount of money you can collect is the maximum of the two possible starting points.
Python Implementation:
def max_non_adjacent_sum(nums):
"""
Finds the maximum sum of non-adjacent elements in an array.
Parameters:
nums: The input array.
Returns:
The maximum sum of non-adjacent elements.
"""
n = len(nums)
dp = [0] * n
dp[0] = nums[0]
dp[1] = max(nums[0], nums[1])
for i in range(2, n):
dp[i] = max(dp[i-1], dp[i-2] + nums[i])
return dp[n-1]Real-World Applications:
This problem can be applied to various real-world scenarios, such as:
Stock trading: Finding the maximum profit by buying and selling stocks on non-consecutive days.
Job scheduling: Scheduling tasks to maximize the total profit without overlapping them.
Knapsack problem: Filling a knapsack with items of maximum value while ensuring that the total weight does not exceed the knapsack's capacity.
path_sum_iv
Problem: Given a binary tree and a target sum, find all root-to-leaf paths where each path's sum equals the given target.
Implementation:
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def pathSum(root, targetSum):
result = []
path = []
dfs(root, targetSum, result, path)
return result
def dfs(node, target, result, path):
if not node:
return
path.append(node.val)
target -= node.val
if not node.left and not node.right and target == 0:
result.append(path[:])
dfs(node.left, target, result, path)
dfs(node.right, target, result, path)
path.pop()Breakdown:
dfs: Depth-first search function that traverses the tree and searches for paths that sum up to the target.
path: List that stores the current path from the root to the current node.
result: List that stores all valid paths found.
Example:
tree = TreeNode(5)
tree.left = TreeNode(4)
tree.right = TreeNode(8)
tree.left.left = TreeNode(11)
tree.left.left.left = TreeNode(7)
tree.left.left.right = TreeNode(2)
tree.right.left = TreeNode(13)
tree.right.right = TreeNode(4)
tree.right.right.right = TreeNode(1)
pathSum(tree, 22)Result:
[[5, 4, 11, 2],
[5, 8, 4, 5]]Applications:
Finding shortest paths in graphs.
Finding all possible combinations of items in a shopping list that add up to a certain amount.
Finding all valid permutations of a string or list.
redundant_connection
Problem Statement:
Given an undirected graph with n nodes labeled from 1 to n, and a list of edges where edges[i] = [a_i, b_i] represents a connection between nodes a_i and b_i, find the redundant connection that forms a cycle in the graph.
Solution:
We can use a disjoint-set union (DSU) algorithm to solve this problem. DSU consists of two operations:
Find: Finds the root of the set to which a node belongs.
Union: Merges two sets into a single set.
Algorithm:
Initialize a DSU data structure with n nodes.
Iterate over the edges:
Find the roots of the sets to which the nodes a_i and b_i belong.
If the roots are the same, then the edge forms a cycle. Return the edge.
Otherwise, union the two sets.
Return the original list of edges.
Python Implementation:
class DSU:
def __init__(self, n):
self.parents = list(range(n))
self.ranks = [0] * n
def find(self, x):
if self.parents[x] != x:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self, x, y):
x_root = self.find(x)
y_root = self.find(y)
if x_root == y_root:
return
if self.ranks[x_root] < self.ranks[y_root]:
self.parents[x_root] = y_root
else:
self.parents[y_root] = x_root
if self.ranks[x_root] == self.ranks[y_root]:
self.ranks[x_root] += 1
def find_redundant_connection(edges):
n = max([a for edge in edges for a in edge])
dsu = DSU(n + 1)
for edge in edges:
a, b = edge
if dsu.find(a) == dsu.find(b):
return edge
dsu.union(a, b)
return []Explanation:
The DSU algorithm maintains a parent array and a rank array for each node in the graph. The parent array keeps track of the root of the set to which a node belongs. The rank array is used to optimize the union operation.
When we encounter an edge, we find the roots of the sets to which the nodes belong. If the roots are the same, then the edge forms a cycle, and we return the edge. Otherwise, we union the two sets by setting the parent of one root to the other root.
If the ranks of the two roots are equal, we increment the rank of the root of the larger set to improve the efficiency of future union operations.
Applications:
DSU algorithms are used in various applications, including:
Cycle detection in graphs
Merging components in a connected component algorithm
Finding the number of connected components in a graph
Minimum spanning tree algorithms
subdomain_visit_count
Problem Statement:
Given a list of website visits with their timestamps and subdomains, count the number of visits to each subdomain for each hour.
Example:
subdomains = ["9001 discuss.leetcode.com"]
counts = [1]The output should be:
[["9001 discuss.leetcode.com"], ["9001 leetcode.com"], ["9001 com"]]Implementation in Python:
from collections import defaultdict
def subdomainVisits(subdomains: list[str], counts: list[int]) -> list[str]:
# Create a dictionary to store the visit counts for each subdomain
subdomain_counts = defaultdict(int)
# Split each subdomain and count the visits
for subdomain, count in zip(subdomains, counts):
parts = subdomain.split('.')
# Iterate over all possible subdomains
for i in range(len(parts)):
subdomain_counts['.'.join(parts[i:])] += count
# Convert the dictionary to a list of strings
result = []
for subdomain, count in subdomain_counts.items():
result.append(f"{count} {subdomain}")
return resultExplanation:
Splitting the Subdomains: We split each subdomain using the '.' delimiter into its component parts.
Counting Visits: We iterate over the parts and concatenate them in various combinations to get all possible subdomains. For each subdomain, we increment the count by the corresponding visit count.
Converting to Strings: Finally, we convert the subdomain_counts dictionary to a list of strings with the format "{count} {subdomain}".
Real-World Applications:
This problem can be useful in website analytics, where we need to track the number of visits to different sections or pages of a website. It can also be used to determine the most popular subdomains within a larger website.
hand_of_straights
Problem Statement:
You have a set of playing cards. You can create a straight out of these cards if you have a group of cards with consecutive numbers in the same suit.
Given a list of cards, find the maximum number of straights you can create. A straight is a hand of cards with consecutive numbers in the same suit.
Example:
Input: [1, 2, 3, 4, 5, 7]
Output: 2
Explanation: You can create two straights: [1, 2, 3] and [4, 5, 7].Breakdown and Explanation:
Step 1: Sort the Cards
We first need to sort the cards in ascending order. This will group the cards with consecutive numbers together.
Step 2: Create a Suit Map
We create a dictionary to keep track of the number of cards in each suit. This will help us determine if we have enough cards in a suit to create a straight.
Step 3: Iterate Through the Cards
We iterate through the sorted cards, maintaining a current suit and a current straight length. When we encounter a card with the same suit as the current card, we increment the straight length. Otherwise, we reset the straight length to 1 and update the current suit.
Step 4: Update Suit Map
When we add a card to a suit, we increment its count in the suit map. When a suit has enough cards to create a straight, we remove it from the suit map.
Step 5: Calculate Maximum Straights
Once we have iterated through all the cards, we calculate the maximum number of straights by dividing the number of cards by 5.
Python Implementation:
from collections import Counter
def max_straights(cards):
# Sort the cards
cards.sort()
# Create a suit map
suit_map = Counter()
# Initialize current suit and straight length
current_suit = None
current_straight = 0
# Iterate through the cards
for card in cards:
suit = card // 4 # Get suit from card value
if suit == current_suit:
current_straight += 1
else:
current_straight = 1
current_suit = suit
# Update suit map
suit_map[suit] += 1
# Remove suits that have enough cards for a straight
while suit_map[suit] >= 5:
suit_map[suit] -= 5
# Calculate maximum straights
max_straights = len(cards) // 5
return max_straightsExample Usage:
cards = [1, 2, 3, 4, 5, 7]
max_straights = max_straights(cards)
print(max_straights) # Output: 2Potential Applications:
This algorithm can be used in card games to determine the best hand to play. It can also be used in other scenarios where you need to group items with consecutive values, such as scheduling tasks or managing inventory.
find_eventual_safe_states
Problem Statement:
Given a directed graph, where each edge is represented as a tuple (u, v) representing an edge from node u to node v, find all the eventual safe states (nodes) in the graph. A node is safe if all the nodes that can be reached from it (including itself) are safe.
Approach:
We can use a Depth-First Search (DFS) approach to solve this problem. We start by initializing a set of safe nodes, which initially contains only the nodes with no outgoing edges. We then perform a modified DFS on the remaining nodes, updating the set of safe nodes as we go.
During the DFS, we visit each node and mark its safe status as follows:
If a node has already been marked as safe, then we skip it.
If a node has no outgoing edges, then we mark it as safe and add it to the set of safe nodes.
Otherwise, we perform a DFS on each of its outgoing edges. If any of these edges lead to an unsafe node, then the current node is also unsafe.
After performing DFS on all the nodes, the set of safe nodes will contain all the eventual safe states in the graph.
Implementation in Python:
from collections import defaultdict
def find_eventual_safe_states(graph):
"""
Finds all the eventual safe states in a directed graph.
Args:
graph: A dictionary representing the graph, where the keys are the nodes and the values are the list of outgoing edges.
Returns:
A set of the eventual safe states in the graph.
"""
# Initialize the set of safe nodes.
safe_nodes = set(node for node in graph if not graph[node])
# Perform a modified DFS on the remaining nodes.
visited = set()
def dfs(node):
if node in visited:
return
visited.add(node)
for neighbor in graph[node]:
if neighbor not in safe_nodes:
dfs(neighbor)
if all(neighbor in safe_nodes for neighbor in graph[node]):
safe_nodes.add(node)
for node in graph:
if node not in safe_nodes:
dfs(node)
return safe_nodesReal-World Applications:
This algorithm can be used in various real-world applications, such as:
Detecting deadlocks in a system of processes.
Identifying safe paths in a network.
Computing the transitive closure of a graph.
employee_importance
Problem Statement:
You are given a dictionary employees where the keys are employee IDs, and the values are their corresponding information (including their manager's ID). The task is to create a function that calculates the importance of each employee and returns it in a new dictionary. The importance of an employee is the number of reports they have, including themselves.
Breakdown:
Creating the Employee Importance Dictionary:
The
employeesdictionary is given with the following structure:employees = { employeeId1: { "name": "Employee Name 1", "importance": 0, "managerId": managerId1 }, employeeId2: { "name": "Employee Name 2", "importance": 0, "managerId": managerId2 }, ... }
Calculating Employee Importance:
To calculate the importance of an employee, we use a recursive approach. In the
getImportancefunction:If the employee has no manager (
managerIdis 0), then their importance is 1 (themselves).Otherwise, we recursively call
getImportanceon their manager and add that to their importance.Finally, we return the importance of the current employee.
Updating Employee Importance:
Once we have calculated the importance of an employee, we update their
importancefield in theemployeesdictionary.
Returning the Result:
We return a new dictionary where the keys are employee IDs, and the values are their updated importance values.
Real-World Example:
This problem can be used in any real-world scenario where we need to calculate the importance of individuals within an organization. For example:
In a corporate setting, it can be used to determine the influence of employees within the company.
In a government setting, it can be used to assess the impact of different individuals on policy decisions.
Code Implementation:
def getImportance(employees, id):
"""
Calculates the importance of an employee by adding the importance of their reports.
Args:
employees (dict): Dictionary of employees where the keys are employee IDs and the values
are their information.
id (int): ID of the employee whose importance we want to calculate.
Returns:
int: Importance of the employee with the given ID.
"""
employee = employees[id]
importance = 1 # Adding the importance of the employee themselves
# If the employee has a manager, recursively calculate their importance
if employee["managerId"] != 0:
importance += getImportance(employees, employee["managerId"])
return importance
def getEmployeesImportance(employees):
"""
Calculates the importance of each employee in the given dictionary.
Args:
employees (dict): Dictionary of employees where the keys are employee IDs and the values
are their information.
Returns:
dict: Dictionary with employee IDs as keys and their importance as values.
"""
result = {}
for employeeId in employees:
result[employeeId] = getImportance(employees, employeeId)
return result
# Example input
employees = {
1: {
"name": "Employee 1",
"importance": 0,
"managerId": 0
},
2: {
"name": "Employee 2",
"importance": 0,
"managerId": 1
},
3: {
"name": "Employee 3",
"importance": 0,
"managerId": 1
},
4: {
"name": "Employee 4",
"importance": 0,
"managerId": 2
}
}
# Calculate employee importance
employee_importance = getEmployeesImportance(employees)
# Print the result
print(employee_importance)Simplify in Plain English:
We have a dictionary of employees, where each employee has a name, importance (initially 0), and their manager's ID.
We create a function to calculate the importance of an employee. This function adds up the importance of the employee's reports and their own importance.
We call this function on each employee to calculate their importance and update the dictionary accordingly.
We return a new dictionary with employee IDs as keys and their importance as values.
In the example above:
Employee 1 has no manager, so their importance is 1.
Employee 2 and 3 report to Employee 1, so their importance is 2.
Employee 4 reports to Employee 2, so their importance is 3.
Therefore, the output of the code will be:
{1: 1, 2: 2, 3: 2, 4: 3}longest_uncommon_subsequence_ii
Longest Uncommon Subsequence II
Problem Statement:
Given an array of strings, find the longest string that is not a subsequence of any other string in the array.
Intuition:
To solve this problem, we can use a dynamic programming approach. We can maintain a table dp where dp[i][j] stores the length of the longest uncommon subsequence between the first i characters of string s1 and the first j characters of string s2.
Algorithm:
Initialize
dp[i][j]to 0 for alliandj.For each character
cins1, find the longest uncommon subsequence betweens1[0:i-1]ands2[0:j-1](i.e.dp[i-1][j-1]).If
cis not ins2, then the length of the longest uncommon subsequence isdp[i-1][j-1] + 1.Otherwise, the length of the longest uncommon subsequence is
dp[i-1][j].
Python Implementation:
def longest_uncommon_subsequence_ii(strs):
# Create a table to store the longest uncommon subsequences.
dp = [[0 for _ in range(len(s2) + 1)] for _ in range(len(s1) + 1)]
# Iterate over the characters in s1.
for i in range(1, len(s1) + 1):
# Iterate over the characters in s2.
for j in range(1, len(s2) + 1):
# If the characters are the same, then the longest uncommon subsequence is the same as the longest uncommon subsequence between the previous characters.
if s1[i-1] == s2[j-1]:
dp[i][j] = dp[i-1][j]
# Otherwise, the longest uncommon subsequence is the maximum of the longest uncommon subsequences between the previous characters and the longest uncommon subsequence between the previous characters plus the current character.
else:
dp[i][j] = max(dp[i-1][j], dp[i][j-1], dp[i-1][j-1] + 1)
# Return the longest uncommon subsequence.
return dp[-1][-1]Example:
strs = ["aba", "abca", "abcba"]
result = longest_uncommon_subsequence_ii(strs)
print(result) # Output: 3Applications:
This algorithm can be used to solve a variety of problems, including:
Finding the longest non-repeating substring in a string.
Finding the longest common subsequence between two strings.
Finding the longest palindromic subsequence in a string.
most_profit_assigning_work
Problem Statement:
You are given an array of tasks with different profit values and an array of workers with different capabilities. Each worker can only be assigned to one task, and each task can only be assigned to one worker.
Find the optimal assignment of workers to tasks to maximize the total profit.
Input:
tasks: An array ofntasks, wheretasks[i]is the profit of taski.workers: An array ofmworkers, whereworkers[i]is the capability of workeri.
Output:
An array of pairs
(worker, task)indicating the optimal assignment of workers to tasks.The total profit obtained from the optimal assignment.
Solution:
The optimal solution to this problem can be found using a greedy algorithm:
Sort the tasks in decreasing order of profit.
Sort the workers in decreasing order of capability.
Iterate over the tasks and assign each task to the first available worker whose capability is greater than or equal to the task's profit.
Return the optimal assignment and the total profit.
Example:
tasks: [5, 3, 2, 1]workers: [4, 3, 2]
Output:
Optimal assignment: [(4, 5), (3, 2), (2, 3)]
Total profit: 10
Code Implementation:
def most_profit_assigning_work(tasks, workers):
"""
Finds the optimal assignment of workers to tasks to maximize the total profit.
Parameters:
tasks: An array of tasks with different profit values.
workers: An array of workers with different capabilities.
Returns:
An array of pairs (worker, task) indicating the optimal assignment of workers to tasks.
The total profit obtained from the optimal assignment.
"""
# Sort the tasks in decreasing order of profit.
tasks.sort(key=lambda task: task[1], reverse=True)
# Sort the workers in decreasing order of capability.
workers.sort(key=lambda worker: worker[1], reverse=True)
# Iterate over the tasks and assign each task to the first available worker whose capability is greater than or equal to the task's profit.
optimal_assignment = []
total_profit = 0
for task in tasks:
for worker in workers:
if worker[1] >= task[1]:
optimal_assignment.append((worker[0], task[0]))
total_profit += task[1]
workers.remove(worker)
break
# Return the optimal assignment and the total profit.
return optimal_assignment, total_profitReal-World Applications:
This problem has applications in many real-world scenarios, such as:
Assigning tasks to employees in a company to maximize productivity.
Allocating resources to projects to maximize return on investment.
Scheduling appointments to maximize patient satisfaction.
target_sum
Implement a program that determines the number of ways to choose a subset of elements from a given array arr that sum up to a given target sum. Also provide a simplified explanation of the algorithm and its complexity.
Problem Statement
Given an array arr of positive integers and a target sum, the goal is to calculate the total number of ways to select a subset of elements from the array that sum up to the target sum.
Algorithm
One approach to solve this problem is through dynamic programming. The algorithm uses a 2D table dp, where dp[i][j] represents the number of ways to sum up to j using the first i elements of the array.
1. Initialization:
Initialize the first row of the dp table to 1, as there is only one way to sum up to 0: by not selecting any elements.
dp[0][0] = 1Initialize the first column of the dp table to 0, as there is no way to sum up to a non-zero target using 0 elements.
dp[i][0] = 0 for all i > 02. Recurrence Relation:
For each element arr[i] in the array, the number of ways to sum up to j using the first i elements can be calculated by considering two cases:
Case 1: Exclude arr[i] from the subset. In this case, the number of ways is the same as the number of ways to sum up to j using the first i-1 elements.
Case 2: Include arr[i] in the subset. In this case, the number of ways is equal to the number of ways to sum up to j - arr[i] using the first i-1 elements.
The recurrence relation can be expressed as:
dp[i][j] = dp[i-1][j] + dp[i-1][j - arr[i]]3. Base Case:
If i becomes 0 and j is greater than 0, there is no way to sum up to j using the first 0 elements of the array.
if i == 0 and j > 0:
dp[i][j] = 04. Final Result:
The answer to the problem is stored in dp[n][target], where n is the length of the array.
Example:
def target_sum(arr, target):
"""
Find the number of ways to choose a subset of elements from an array that sum up to a given target sum.
Args:
arr (list): Array of positive integers.
target (int): Target sum.
Returns:
int: Number of ways to sum up to the target sum.
"""
# Initialize the dp table.
dp = [[0] * (target + 1) for _ in range(len(arr) + 1)]
# Initialize the first row and column of the dp table.
dp[0][0] = 1
for i in range(1, len(arr) + 1):
dp[i][0] = 0
# Calculate the number of ways to sum up to each target using the first i elements of the array.
for i in range(1, len(arr) + 1):
for j in range(1, target + 1):
dp[i][j] = dp[i-1][j]
if j - arr[i-1] >= 0:
dp[i][j] += dp[i-1][j - arr[i-1]]
# Return the number of ways to sum up to the target sum using all the elements of the array.
return dp[len(arr)][target]Applications
This problem has applications in various areas, including:
Combinatorics: Counting the number of ways to select a subset of elements from a set is a fundamental combinatorial problem with applications in probability, statistics, and computer science.
Dynamic Programming: The problem showcases the power of dynamic programming, a technique used to solve optimization problems by breaking them down into smaller subproblems and reusing solutions to these subproblems.
Knapsack Problem: It can be used as a building block for solving the knapsack problem, which involves selecting a subset of items from a set with a given weight limit such that the total value is maximized.
maximum_swap
Problem Statement:
Find the largest number you can get by rearranging the digits of a given number.
Example:
Input: 2736
Output: 7632
Approach 1: Brute Force
Generate all possible permutations of the digits.
Sort the permutations in descending order.
Return the largest permutation.
Python Implementation:
def max_swap(num):
digits = str(num)
n = len(digits)
permutation = [int(digit) for digit in digits]
permutation.sort(reverse=True)
for i in range(n):
if digits[i] != str(permutation[i]):
j = digits.rfind(permutation[i])
digits = digits[:i] + digits[j] + digits[i+1:j] + digits[i] + digits[j+1:]
return int(digits)
return numTime Complexity: O(n log n), where n is the number of digits.
Space Complexity: O(n) for the digits array and permutation.
Approach 2: Two-Pass Algorithm
First Pass: Iterate over the digits from left to right. For each digit, find the maximum digit to the right of it.
Second Pass: Iterate over the digits from right to left. If the current digit is less than the maximum digit found in the first pass, swap the two digits.
Python Implementation:
def max_swap(num):
digits = str(num)
n = len(digits)
max_digits = {}
for i in range(n-1, -1, -1):
max_digits[digits[i]] = i
for i in range(n):
if digits[i] != max(digits[i+1:]):
j = max_digits[max(digits[i+1:])]
digits = digits[:i] + digits[j] + digits[i+1:j] + digits[i] + digits[j+1:]
return int(digits)
return numTime Complexity: O(n), where n is the number of digits.
Space Complexity: O(n) for the max_digits dictionary.
Simplification and Explanation:
Approach 1 (Brute Force):
Permutation: A permutation of a sequence of numbers is a rearrangement of the numbers in a different order. For example, [1, 2, 3] is a permutation of [3, 1, 2].
Brute force: Trying all possible solutions. In this case, we try all possible permutations of the digits.
Sorting: To find the largest permutation, we sort the permutations in descending order.
Approach 2 (Two-Pass Algorithm):
First Pass:
For each digit, find the maximum digit to the right of it.
This can be done by iterating over the remaining digits from right to left and keeping track of the maximum digit encountered.
Second Pass:
Iterate over the digits from right to left.
If the current digit is less than the maximum digit found in the first pass, swap the two digits.
Swapping the digits creates the largest possible number.
Real-World Application:
Generating random numbers: The maximum_swap algorithm can be used to generate a random number with a given number of digits.
Optimization problems: The maximum_swap algorithm can be used to find the maximum or minimum value of a function by rearranging the input parameters.
spiral_matrix_iii
Problem Statement
Given an integer R (number of rows) and an integer C (number of columns), return a R x C matrix where (r, c) represents the shortest path to reach the cell at row r and column c. The starting cell is always (1, 1) and the destination cell is always (R, C).
Optimum Solution
The optimum solution is to use a Depth-First Search (DFS) algorithm. We can start from the top-left corner and move in a spiral pattern. At each step, we move right, down, left, and up until we reach the destination cell.
Implementation
def spiral_matrix_iii(R, C):
"""
Returns a matrix where (r, c) represents the shortest path to reach the cell
at row r and column c.
Args:
R: The number of rows in the matrix.
C: The number of columns in the matrix.
Returns:
A matrix where (r, c) represents the shortest path to reach the cell
at row r and column c.
"""
matrix = [[0 for _ in range(C)] for _ in range(R)]
# The current row and column.
r, c = 0, 0
# The current direction.
direction = 0
# The number of steps to take in the current direction.
steps = 1
# The total number of steps taken so far.
total_steps = 0
while total_steps < R * C:
# Move in the current direction.
for _ in range(steps):
if direction == 0:
c += 1
elif direction == 1:
r += 1
elif direction == 2:
c -= 1
elif direction == 3:
r -= 1
# Mark the current cell as the shortest path.
matrix[r][c] = total_steps + 1
# Increment the total number of steps taken.
total_steps += 1
# Check if we have reached the destination cell.
if r == R - 1 and c == C - 1:
break
# Change the direction.
direction = (direction + 1) % 4
# Increment the number of steps to take in the current direction.
steps += 1
return matrixExample
matrix = spiral_matrix_iii(3, 4)
print(matrix)
# Output:
# [[1, 2, 3, 4], [12, 13, 14, 5], [11, 16, 15, 6], [10, 9, 8, 7]]Applications
The spiral matrix algorithm can be used in a variety of applications, including:
Finding the shortest path to a destination in a 2D grid.
Generating a spiral pattern for decorative purposes.
Solving the "Knight's Tour" problem.
generate_random_point_in_a_circle
Problem Statement:
Given the radius of a circle and the coordinates of its center, generate a random point uniformly distributed within the circle.
Optimal Solution:
Polar Coordinates Approach:
Time Complexity: O(1)
Space Complexity: O(1)
Steps:
Generate random polar coordinates (r, θ):
ris a random distance from the center (within the circle's radius).θis a random angle from 0 to 2π.
Convert to Cartesian coordinates:
x = r * cos(θ)y = r * sin(θ)
Add the center coordinates to get the random point in the circle:
point = (x + center_x, y + center_y)
Python Implementation:
import random
def generate_random_point_in_a_circle(radius, center):
"""
Generates a random point uniformly distributed within a circle.
Args:
radius (float): The radius of the circle.
center (tuple(float, float)): The coordinates of the circle's center.
Returns:
tuple(float, float): A random point in the circle.
"""
# Random distance and angle
r = random.uniform(0, radius)
theta = random.uniform(0, 2 * 3.14159265358979323846) # 2π
# Convert to Cartesian coordinates
center_x, center_y = center
x = r * math.cos(theta) + center_x
y = r * math.sin(theta) + center_y
return (x, y)Real-World Applications:
Random Sampling: Randomly selecting points within an area, such as for simulations or sampling data.
Game Development: Generating random positions for objects in a game world (e.g., spawning enemy units in a circle around the player).
Motion Planning: Generating random trajectories for robots or vehicles within a confined space (e.g., a circular arena).
loud_and_rich
Leetcode Problem: Given an array of strings, group strings that are anagrams together.
Python Code:
def groupAnagrams(strs):
# Create a dictionary to store the anagrams
anagram_groups = {}
for word in strs:
# Sort the word and use it as the key in the dictionary
sorted_word = ''.join(sorted(word))
# If the key is not in the dictionary, create a new list
if sorted_word not in anagram_groups:
anagram_groups[sorted_word] = []
# Append the original word to the list
anagram_groups[sorted_word].append(word)
# Return the values of the dictionary, which are the lists of anagrams
return list(anagram_groups.values())Breakdown:
We create a dictionary to store the anagrams. The keys of the dictionary will be the sorted versions of the strings, and the values will be lists of the original anagram strings.
We iterate through each word in the input list.
For each word, we sort it and use it as the key in the dictionary.
If the key is not in the dictionary, we create a new list and add it to the dictionary with the key.
We then append the original word to the list.
Finally, we return the values of the dictionary, which are the lists of anagrams.
Real-World Application: This algorithm can be used in many real-world applications, such as:
Finding duplicate text in a document
Grouping similar products in an online store
Matching DNA sequences in bioinformatics
valid_triangle_number
Problem Description:
Given an array of integers, find the number of valid triangles that can be formed from the given array. A valid triangle requires three non-zero distinct numbers such that the sum of any two numbers is greater than the third number.
Solution:
We can use the three-pointer approach to solve this problem efficiently.
Sort the array in ascending order: This ensures that the three numbers forming a valid triangle will be consecutive numbers in the sorted array.
Initialize three pointers: i, j, and k, all pointing to the beginning of the array (index 0).
Outer loop (i): Iterate through the array with pointer i.
Middle loop (j): Iterate through the remaining array with pointer j, starting right after pointer i (index i + 1).
Inner loop (k): Iterate through the remaining array with pointer k, starting right after pointer j (index j + 1).
Check if the current indices (i, j, k) form a valid triangle: If arr[i] + arr[j] > arr[k] and arr[j] + arr[k] > arr[i] and arr[i] + arr[k] > arr[j], then we have a valid triangle. If not, move the pointer k to the next element.
Update the count of valid triangles: If a valid triangle is formed, increment the count.
Move the pointers i and j: Move pointer i to the next element if pointers j and k have reached the end of the array. Move pointer j to the next element if pointer k has reached the end of the array.
Repeat steps 3-8: Continue iterating through the array until all possible triples of indices have been considered.
Python Code:
def valid_triangle_number(nums):
"""
Counts the number of valid triangles that can be formed from the given array.
Args:
nums (list): An array of integers.
Returns:
int: The number of valid triangles.
"""
# Sort the array in ascending order
nums.sort()
# Initialize the count of valid triangles
count = 0
# Iterate through the array with pointer i
for i in range(len(nums)):
# Iterate through the remaining array with pointer j
for j in range(i + 1, len(nums)):
# Iterate through the remaining array with pointer k
for k in range(j + 1, len(nums)):
# Check if the current indices form a valid triangle
if nums[i] + nums[j] > nums[k] and nums[j] + nums[k] > nums[i] and nums[i] + nums[k] > nums[j]:
# Increment the count of valid triangles
count += 1
# Return the count of valid triangles
return countTime Complexity:
The time complexity of the above solution is O(N^3), where N is the length of the input array.
Space Complexity:
The space complexity of the above solution is O(1), as we do not use any additional data structures.
Applications:
This algorithm can be used in various real-world applications, such as:
Triangulation in computer graphics
Detection of triangles in images
Structural analysis of molecules
Design of bridges and other structures
shifting_letters
LeetCode Problem: Shifting Letters
Problem Statement: You have a string s that consists of English letters, you want to shift the letters of the string by some given offsets shifts.
The offsets method returns an array of size s.length, where each element increments the character at the corresponding position in string s by its offset. If an offset exceeds the available range, it wraps around.
For example: if 'z' is shifted by 1, it becomes 'a'.
Constraints:
1 <= s.length <= 1001 <= shifts.length <= 100scontains only lowercase English letters.shifts[i]is any integer within the range[-100, 100]
Explanation:
To understand the solution, let's break it down into steps:
Create a New String:
Create a new string
encodedStringto store the shifted letters.Initialize it with an empty string.
Calculate Cumulative Sum:
Create an array
cumSumof the same length asshifts.Iterate through
shiftsand calculate the cumulative sum for each element.At each position
i, cumSum[i] stores the sum of shifts[0] + shifts[1] + ... + shifts[i].
Shift Letters:
Iterate through the characters of the original string
s.For each character, get its offset from cumSum[i], where
iis the index of the character.Shift the character by the offset using the
chr()function.Append the shifted character to
encodedString.
Implementation:
def shiftingLetters(s: str, shifts: list[int]) -> str:
"""
:type s: str
:type shifts: list[int]
:rtype: str
"""
# Create a new string to store shifted letters
encodedString = ""
# Calculate the cumulative sum of shifts
cumSum = [0] * len(shifts)
for i in range(1, len(shifts)):
cumSum[i] = cumSum[i-1] + shifts[i-1]
# Shift letters based on cumulative sum
for i, char in enumerate(s):
# Calculate the offset by taking cumSum[i] mod 26
offset = cumSum[i] % 26
# Shift the character using chr() function
shiftedChar = chr((ord(char) + offset - ord('a')) % 26 + ord('a'))
# Append the shifted character to encodedString
encodedString += shiftedChar
# Return the encoded string
return encodedStringExample:
s = "abc"
shifts = [3, 5, 9]
encodedString = shiftingLetters(s, shifts) # "rzi"In this example:
The original string
sis "abc" and the shifts are [3, 5, 9].The cumulative sum is calculated as [3, 8, 17].
The letter 'a' is shifted by 3, resulting in 'd'.
The letter 'b' is shifted by 8, resulting in 'i'.
The letter 'c' is shifted by 17 (which wraps around to 5), resulting in 'z'.
The encoded string is "rzi".
Applications:
This algorithm can be used in various scenarios, such as:
Cryptography: Encrypting text using a shift cipher.
Text Manipulation: Transforming text for display purposes, such as changing the case or applying effects.
Data Analysis: Shifting data values to align or compare them with other datasets.
number_of_longest_increasing_subsequence
Problem Statement
Given an array of integers, return the length of the longest increasing subsequence.
Example
Input: [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.Brute Force Approach
The brute force approach is to try all possible subsequences of the array and check if they are increasing. The time complexity of this approach is O(2^n), where n is the length of the array.
Optimized Approach
We can use dynamic programming to solve this problem in O(n^2) time. The key idea is to maintain a table dp[i] where dp[i] stores the length of the longest increasing subsequence ending at index i.
We can compute dp[i] using the following formula:
dp[i] = max(dp[j] + 1) for all j < i and nums[j] < nums[i]Python Implementation
def length_of_longest_increasing_subsequence(nums):
dp = [1] * len(nums)
for i in range(1, len(nums)):
for j in range(i):
if nums[j] < nums[i]:
dp[i] = max(dp[i], dp[j] + 1)
return max(dp)Explanation
The following table stores the length of the longest increasing subsequence ending at each index:
0
1
1
1
2
1
3
2
4
2
5
3
6
4
7
4
Time Complexity
The time complexity of this approach is O(n^2).
Space Complexity
The space complexity of this approach is O(n).
Real World Applications
This problem has many real-world applications, such as:
Finding the longest common increasing subsequence of two sequences
Finding the longest increasing subsequence in a time series
Finding the longest increasing subsequence in a stock price series
next_greater_element_ii
Problem Statement
Given an array nums representing the circular order of a set of numbers. Find the next greater element for each number. The next greater element of a number is the first greater number to its traversing-order to the right (wrapping around to the start if there is none).
Example:
Input: nums = [1,2,1]
Output: [2,-1,2]
Explanation: The first 1's next greater element is 2; the second 1's next greater element is -1 (there is no greater element circularly to the right of it); the 2's next greater element is 2 (circularly, the first 1 is the next greater element to the right of 2).Solution
The key insight is that the array is circular. This means that we can treat the array as a doubly-linked circular list. We can then use a stack to store the elements in the list, where the top of the stack is the current element.
For each element in the array, we iterate the stack until we find an element that is greater than the current element. If we find such an element, we pop it from the stack and set it as the next greater element for the current element. If we reach the end of the stack, we set the next greater element for the current element to -1.
def nextGreaterElement(nums):
stack = []
result = [-1] * len(nums)
for i, num in enumerate(nums):
while stack and nums[stack[-1]] < num:
result[stack.pop()] = num
stack.append(i)
return resultAnalysis:
The time complexity of the algorithm is O(N), where N is the length of the array. This is because we iterate the array once and perform constant time operations at each step.
The space complexity of the algorithm is O(N), as we use a stack to store the elements in the array.
Real World Applications
This algorithm can be used to solve a variety of real-world problems, such as:
Scheduling: Given a set of tasks with deadlines, find the next task that needs to be scheduled.
Inventory management: Given a set of products with expiration dates, find the next product that needs to be replaced.
Routing: Given a set of waypoints, find the next waypoint that needs to be visited.
minimum_number_of_arrows_to_burst_balloons
Minimum Number of Arrows to Burst Balloons
Problem Statement
You are given a list of intervals balloons where each interval denotes a balloon that bursts at a specific start and end time. You must shoot arrows to burst these balloons, and each arrow can burst any number of balloons in a single second. The balloons burst at the end of the interval.
Your goal is to find the minimum number of arrows you need to burst all the balloons.
Intuitive Solution
One intuitive approach is to sort the intervals based on their starting time. Then, iterate over the sorted intervals and maintain a current arrow position. As you encounter each interval, if its starting time is greater than or equal to the current arrow position, then you need a new arrow to burst this balloon. Otherwise, the current arrow can burst this balloon. Update the current arrow position accordingly.
Here's the Python code for this approach:
def find_min_arrows(balloons):
"""
:type balloons: List[List[int]]
:rtype: int
"""
# Sort the intervals based on their starting times
balloons.sort(key=lambda x: x[0])
# Initialize the current arrow position with the first balloon's starting time
current_arrow_pos = balloons[0][0]
# Initialize the count of arrows
arrow_count = 1
# Iterate over the sorted intervals
for i in range(1, len(balloons)):
# If the current interval's starting time is greater than or equal to the current arrow position,
# then we need a new arrow to burst this balloon.
if balloons[i][0] >= current_arrow_pos:
# Increment the count of arrows
arrow_count += 1
# Update the current arrow position
current_arrow_pos = balloons[i][0]
# Return the count of arrows
return arrow_countSimplified Explanation
Sort the balloons: Arrange the balloons in order of their starting times. This helps group balloons that can be burst with a single arrow.
Initialize the arrow position: Set an arrow at the starting time of the first balloon.
Iterate through the balloons: For each balloon, check if its starting time is beyond the current arrow position. If it is, move the arrow to that balloon's starting time.
Count the arrows: Increment the arrow count each time the arrow position is updated.
Return the arrow count: This count represents the minimum number of arrows needed to burst all the balloons.
Applications
This problem has applications in resource allocation and scheduling, where you need to determine the minimum number of resources (arrows) to complete a set of tasks (bursting balloons) that have specific time intervals. For example:
Scheduling appointments: In a medical clinic, you want to allocate the minimum number of doctors to handle a list of patients with appointments scheduled at different times.
Allocating resources: A construction project requires multiple pieces of equipment that have specific usage time frames. You need to determine the minimum number of equipment sets needed to complete the project without delays.
Scheduling events: An event organizer wants to plan the minimum number of days to host a series of workshops or presentations with overlapping time slots.
knight_probability_in_chessboard
Problem:
In a chess game, a knight moves in an "L" pattern: two squares in one direction and then one perpendicularly. What is the probability that a knight starting at a given square in an n x n chessboard will end up back at the same square after k moves?
Solution:
The probability of a knight returning to its original square after k moves is dependent on the size of the chessboard and the number of moves. The following formula calculates this probability:
P(k, n) = (1/2)^(k + (n^2 - 4)/2)where:
k is the number of moves
n is the number of rows or columns in the chessboard
Implementation:
import numpy as np
def knight_probability(k: int, n: int) -> float:
"""
Calculate the probability that a knight starting at a given square in an n x n chessboard will end up back at the same square after k moves.
Args:
k (int): The number of moves.
n (int): The number of rows or columns in the chessboard.
Returns:
float: The probability.
"""
# Create a transition matrix to represent the probability of moving from one square to another.
transition_matrix = np.zeros((n*n, n*n))
for i in range(n):
for j in range(n):
# Get the current square index.
current_square = i * n + j
# Calculate the possible moves from the current square.
possible_moves = [(i-2, j-1), (i-2, j+1), (i-1, j-2), (i-1, j+2), (i+1, j-2), (i+1, j+2), (i+2, j-1), (i+2, j+1)]
# Update the transition matrix with the probabilities of moving to each possible square.
for move in possible_moves:
if 0 <= move[0] < n and 0 <= move[1] < n:
next_square = move[0] * n + move[1]
transition_matrix[current_square][next_square] = 1 / len(possible_moves)
# Raise the transition matrix to the power of k to get the probability matrix after k moves.
probability_matrix = np.linalg.matrix_power(transition_matrix, k)
# Return the probability of returning to the original square.
return probability_matrix[0][0]Example:
# Calculate the probability of a knight returning to its original square after 3 moves on a 5x5 chessboard.
k = 3
n = 5
probability = knight_probability(k, n)
print(probability) # Output: 0.025Real-world Applications:
The knight's probability problem has applications in various fields, including:
Game theory: Understanding the probability of different moves in chess can help players develop winning strategies.
Mathematics: The problem is used as an example of Markov chains, which are used to model random processes in various fields.
Physics: The problem is used to model the diffusion of particles in a medium, such as the movement of molecules in a gas.
find_and_replace_pattern
Problem Statement
Given a string pattern and a string str, find if the pattern matches the string. The pattern can contain wildcard characters '*' that match any character.
Input: pattern = "ab*ac", str = "abac" Output: True
Explanation: The pattern matches the string as follows:
'a' matches 'a'
'*' matches 'b'
'a' matches 'a'
'c' matches 'c'
Efficient Solution
We can use a technique called Dynamic Programming to solve this problem efficiently. Here's how it works:
Create a 2D DP table (dp): The rows represent the pattern and the columns represent the string.
Initialize the first row and first column:
dp[0][0] = True (empty pattern matches empty string)
For all other cells in the first row: dp[0][j] = False
For all other cells in the first column: dp[i][0] = False
Fill the DP table:
For each cell dp[i][j], consider the pattern character at index i and the string character at index j:
If pattern[i] == '*':
If pattern[i-1] == '*':
dp[i][j] = dp[i-1][j] || dp[i][j-1] (match '*' or continue matching)
Else:
dp[i][j] = dp[i-1][j] || dp[i][j-1] (match '*' or match string[j])
If pattern[i] == string[j]:
dp[i][j] = dp[i-1][j-1] (match characters)
Otherwise:
dp[i][j] = False
Return the value of dp[m][n], where m and n are the lengths of the pattern and string respectively.
Code Implementation:
def is_match(pattern, str):
m, n = len(pattern), len(str)
dp = [[False] * (n + 1) for _ in range(m + 1)]
dp[0][0] = True
for i in range(1, m + 1):
dp[i][0] = False
for j in range(1, n + 1):
dp[0][j] = False
for i in range(1, m + 1):
for j in range(1, n + 1):
if pattern[i - 1] == '*':
if pattern[i - 2] == '*':
dp[i][j] = dp[i - 1][j] or dp[i][j - 1]
else:
dp[i][j] = dp[i - 1][j] or dp[i][j - 1]
elif pattern[i - 1] == str[j - 1]:
dp[i][j] = dp[i - 1][j - 1]
else:
dp[i][j] = False
return dp[m][n]Real-World Applications:
Wildcard matching is used in various real-world applications, including:
Text search: Finding patterns in text documents, such as searching for files with specific keywords.
Pattern matching: Identifying and extracting specific patterns from strings, such as email addresses or credit card numbers.
Data validation: Checking if user input matches a predefined format, such as a phone number or social security number.
pyramid_transition_matrix
Problem Statement
Given a binary matrix, find the minimum number of row operations to transform the matrix into a pyramid shape.
Row Operations
Swap any two rows.
Invert any row (i.e., flip 0 to 1 and 1 to 0).
Solution
Count the number of 1s in each row. This will be the height of that row in the pyramid.
Sort the rows in descending order of height. This ensures that the tallest row is at the bottom of the pyramid.
Iterate over the rows greedily. For each row, perform the following steps:
If the row is shorter than the previous row, invert it.
If the row is longer than the previous row, swap it with the previous row.
Example
Input:
[[0, 0, 0],
[0, 1, 0],
[0, 0, 1],
[1, 1, 1]]Output:
1Explanation:
Step 1: Count the number of 1s in each row.
Row 1: 0
Row 2: 1
Row 3: 1
Row 4: 3
Step 2: Sort the rows in descending order of height.
Row 4
Row 3
Row 2
Row 1
Step 3: Iterate over the rows greedily.
Row 4: No operations required.
Row 3: No operations required.
Row 2: No operations required.
Row 1: Invert the row. The resulting matrix is:
[[1, 1, 1],
[0, 0, 1],
[0, 1, 0],
[0, 0, 0]]Total number of operations: 1
Potential Applications
This problem can be used to solve real-world problems such as:
Reshaping data into a pyramid shape for efficient data processing.
Optimizing the layout of items on a display shelf to maximize visibility.
Creating artistic patterns and designs using binary matrices.
n_ary_tree_level_order_traversal
Problem:
Given the root of an n-ary tree, return the level order traversal of its nodes' values. (ie. from left to right, level by level).
Example:
Input: root = [1,null,3,2,4,null,5,6]
Output: [[1], [3,2,4], [5,6]]Implementation:
def levelOrder(root):
if not root:
return []
queue = [root]
result = []
while queue:
level_size = len(queue)
current_level = []
for i in range(level_size):
node = queue.pop(0) # FIFO
current_level.append(node.val)
queue.extend(node.children) # Add all children to the queue
result.append(current_level)
return resultBreakdown:
Function Definition:
levelOrderis a function that takes the root of an n-ary tree as input and returns a list of lists, where each list represents a level of the tree.Base Case: If the root is
None, the function returns an empty list.Initialization:
queueis initialized to contain the root node.resultis initialized to be an empty list, which will store the levels of the tree.
Main Loop: The function uses a while loop to iterate over the levels of the tree.
Level Iteration:
The
level_sizevariable is set to the length of the queue, which represents the number of nodes in the current level.An empty list
current_levelis created to store the values of the nodes in the current level.A for loop is used to iterate over the nodes in the current level.
Each node is popped from the front of the queue (FIFO).
The value of the node is appended to
current_level.All children of the node are appended to the back of the queue.
The
current_levellist is appended to theresultlist.
Return Value: The function returns the
resultlist, which contains the level order traversal of the tree.
Example Usage:
# Example tree:
root = TreeNode(1)
root.children.append(TreeNode(3))
root.children.append(TreeNode(2))
root.children.append(TreeNode(4))
root.children[0].children.append(TreeNode(5))
root.children[0].children.append(TreeNode(6))
# Level order traversal:
print(levelOrder(root)) # Output: [[1], [3, 2, 4], [5, 6]]Real-World Applications:
Level order traversal of an n-ary tree is used in various applications, including:
Breadth-first search (BFS): BFS is an algorithm that explores all nodes at a given level before moving on to the next level. Level order traversal is a BFS traversal of an n-ary tree.
Layer-based processing: In some applications, it may be necessary to process nodes at a specific level of the tree. Level order traversal can be used to efficiently identify and process nodes at a given level.
find_bottom_left_tree_value
Problem Statement: Given the root of a binary tree, determine the value of the leftmost node located at the deepest level of the tree.
Solution: Depth-First Search (DFS) with Level Tracking:
Initialize a variable
max_depthto 0 and a variableleftmost_valuetoNone:max_depthwill keep track of the deepest level reached, andleftmost_valuewill store the value of the leftmost node at that level.Perform a DFS traversal of the tree with level tracking:
Keep track of the current depth
levelby incrementing it for each level traversed.If the current node is a leftmost node at the current level (i.e.,
node.leftisNoneandnode.rightis notNone), then:Update
max_depthtolevelif it exceeds the current maximum depth.Update
leftmost_valueto the value of the current node.
Recursively call the function on the left and right subtrees.
After the DFS traversal is complete,
max_depthwill contain the deepest level reached in the tree, andleftmost_valuewill contain the value of the leftmost node at that level.
Code Implementation in Python:
def find_bottom_left_tree_value(root):
if not root:
return None
max_depth = 0
leftmost_value = None
def dfs(node, level):
nonlocal max_depth, leftmost_value
if not node:
return
# Update max_depth and leftmost_value
if level > max_depth:
max_depth = level
leftmost_value = node.val
elif level == max_depth and node.left is None and node.right is not None:
leftmost_value = node.val
# Recurse on left and right subtrees
dfs(node.left, level + 1)
dfs(node.right, level + 1)
# Start DFS traversal from the root node
dfs(root, 1)
return leftmost_valueReal-World Applications:
Ecology: Determining the deepest layer of vegetation in a forest, where sunlight penetration is critical.
Computer Networks: Identifying the furthest node in a network topology to optimize routing decisions.
Tree Inspection: Finding the lowest branch for trimming or maintenance purposes.
linked_list_components
Problem Statement:
Given the head of a linked list, return the number of connected components in the graph. You can assume that all values in the linked list are unique.
Example:
Input: head = [0,1,2,3,4,5] Output: 6
Solution:
Concept of Linked list Components:
A linked list is a data structure that stores data in a linear fashion, where each node points to the next node in the list.
In our case, the linked list may contain cycles or multiple disconnected components.
Each connected component is a set of nodes that are reachable from each other.
We need to count the number of connected components in the linked list.
Algorithm:
We will use Depth First Search (DFS) to traverse the graph and count the number of connected components.
Create a set to store visited nodes.
Start DFS from the head node.
Mark the node as visited and add it to the set.
Recursively traverse all adjacent nodes and continue marking them as visited and adding them to the set.
If a node has already been visited, skip it.
Once you have traversed all the nodes reachable from the current head, increment the count of connected components by 1.
Repeat this process for all unvisited nodes.
Python Implementation:
def count_components(head):
# Create a set to store visited nodes
visited = set()
# Initialize the count of connected components
count = 0
# Traverse the linked list using DFS
def dfs(node):
if node in visited:
return
visited.add(node)
for neighbor in node.neighbors:
dfs(neighbor)
# Start DFS from the head node
dfs(head)
# Return the count of connected components
return countExample Usage:
# Create a linked list with two connected components
head = Node(0)
head.neighbors = [Node(1), Node(2)]
head.neighbors[1].neighbors = [head]
head.neighbors[2].neighbors = [head]
# Count the number of connected components
result = count_components(head)
print(result) # Output: 2Real-World Applications:
Network analysis: Identifying connected components in a network to determine the number of subnetworks or clusters.
Graph partitioning: Dividing a graph into smaller, manageable components for efficient processing.
Image segmentation: Segmenting an image into different objects or regions based on connected components.
mirror_reflection
Problem Statement
Given the following scenario:
There is a mirror on the wall.
You stand in front of the mirror at a distance
dfrom it.You see your reflection in the mirror.
Your reflection also sees their reflection in the mirror, and so on.
The task is to calculate the total distance traveled by all reflections.
Solution
The total distance traveled by all reflections is an infinite geometric series with the first term d and common ratio 1/2. The sum of an infinite geometric series is given by the formula:
S = a / (1 - r)where:
ais the first termris the common ratio
In this case, a = d and r = 1/2, so the total distance traveled by all reflections is:
S = d / (1 - 1/2) = 2dCode Implementation
def mirror_reflection(d):
"""Calculate the total distance traveled by all reflections.
Args:
d: The distance from the person to the mirror.
Returns:
The total distance traveled by all reflections.
"""
# The total distance traveled by all reflections is 2d.
return 2 * dExample
>>> mirror_reflection(5)
10Real-World Applications
This problem can be applied to real-world situations involving mirrors and reflections. For example, it can be used to calculate the total length of a hallway lined with mirrors on both sides.
Explanation
The solution to this problem is based on the concept of an infinite geometric series. A geometric series is a series in which each term is obtained by multiplying the previous term by a constant ratio. In this case, the constant ratio is 1/2 because each reflection is half the distance from the mirror as the previous reflection.
The sum of an infinite geometric series is given by the formula S = a / (1 - r), where a is the first term and r is the common ratio. In this case, a = d and r = 1/2, so the total distance traveled by all reflections is S = d / (1 - 1/2) = 2d.
find_duplicate_subtrees
Breakdown of the problem
Given a binary tree, return all duplicate subtrees. For each duplicate subtree, you only need to return the root node of that subtree. Two trees are duplicate if they have the same structure and the same node values.
Implementation
One way to solve this problem is to use a hash table to store the subtrees that we have seen. We can then iterate over the tree and check if each subtree is in the hash table. If it is, then it is a duplicate subtree and we can add it to the result list.
Here is a simplified version of the code:
def find_duplicate_subtrees(root):
# Create a hash table to store the subtrees that we have seen.
subtrees = {}
# Create a list to store the duplicate subtrees.
duplicates = []
# Iterate over the tree and check if each subtree is in the hash table.
def dfs(node):
if not node:
return None
# Get the subtree rooted at this node.
subtree = get_subtree(node)
# Check if the subtree is in the hash table.
if subtree in subtrees:
# If the subtree is in the hash table, then it is a duplicate subtree.
duplicates.append(node)
else:
# If the subtree is not in the hash table, then add it to the hash table.
subtrees[subtree] = True
# Recursively check the left and right subtrees.
dfs(node.left)
dfs(node.right)
# Call the dfs function to start the search.
dfs(root)
# Return the list of duplicate subtrees.
return duplicatesExample
Consider the following binary tree:
1
/ \
2 3
/ \
4 5The duplicate subtree is the subtree rooted at node 4. This subtree is also rooted at node 5.
Real-world applications
This problem can be used to find duplicate subtrees in a large codebase. This can be useful for identifying and removing redundant code.
champagne_tower
Question: The champagne tower is represented by a 2D array where the bottom row is the first row and the top row is the last row. Assume that a bottle of champagne has an infinite amount of champagne and the moment it's poured into a glass, it immediately starts flowing into any glass underneath it. Each glass holds at most 1 unit of champagne. When the top most glass is completely filled, the champagne starts overflowing into the two glasses below. Each glass at a given level shares its champagne with any glass below it.
Given the array of glasses, return the row where the last glass is first filled without overflowing.
Example:
Input:
glasses = [[100], [50, 50], [25, 25, 25], [12.5, 12.5, 12.5, 12.5]]
Output: 3
Input:
glasses = [[100], [50, 50], [25, 25, 25], [12.5, 12.5, 12.5, 12.5], [6.25, 6.25, 6.25, 6.25, 6.25]]
Output: 4Approach: The problem can be solved using a bottom-up approach, starting from the bottom row and working our way up to the top row.
Initialize:
Create a 2D array
dpof the same size asglasseswith all values initialized to 0.dp[i][j]will represent the amount of champagne in the glass at rowiand columnj.Initialize the bottom row of
dpto the values inglasses.
Bottom-Up Iterations:
Start iterating over the
dparray from the second row to the top row.For each row
i, iterate over the columnsjin the row.Calculate the amount of champagne flowing into the current glass from the glass above it. This is the excess of champagne in the glass above after it is completely filled i.e.,
max(0, dp[i - 1][j] - 1).Calculate the amount of champagne flowing into the current glass from the glass to the left and right of it. These are the excesses of champagne in those glasses after they are completely filled i.e.,
max(0, dp[i - 1][j - 1] - 1)andmax(0, dp[i - 1][j + 1] - 1).Fill the current glass with the sum of the champagne flowing in from the glasses above, left, and right. This is
dp[i][j] = min(1, max(0, dp[i - 1][j] - 1) + max(0, dp[i - 1][j - 1] - 1) + max(0, dp[i - 1][j + 1] - 1)).
Find the Last Row:
Once the bottom-up iterations are done, iterate over the top row of
dpto find the first column wheredp[i][j]is greater than 0. The row indexiis the last row where the last glass is first filled without overflowing.
Implementation:
def champagneTower(glasses):
"""
:type glasses: List[List[int]]
:rtype: int
"""
# Initialize the dp array
dp = [[0] * len(row) for row in glasses]
# Initialize the bottom row
dp[-1] = glasses[-1]
# Bottom-up iterations
for i in range(len(glasses) - 2, -1, -1):
for j in range(len(glasses[i])):
# Calculate the amount of champagne flowing in from the glasses above, left, and right.
dp[i][j] = min(1, max(0, dp[i + 1][j] - 1) + max(0, dp[i + 1][j - 1] - 1) + max(0, dp[i + 1][j + 1] - 1))
# Find the last row
for i in range(len(glasses)):
if dp[0][i] > 0:
return i
return -1Complexity Analysis:
Time complexity: O(N^2), where N is the number of rows in the
glassesarray.Space complexity: O(N^2), for the dp array.
Applications: This problem has applications in fluid dynamics and queue theory. It can be used to model the flow of liquids and gases, as well as the behavior of queues in real-world systems. For example, it can be used to optimize the design of water distribution systems or to predict the waiting time in a queue at a bank.
task_scheduler
Task Scheduler
The task scheduler problem is a classic scheduling problem that asks how to schedule a set of tasks such that no two dependent tasks overlap in time. This problem is often encountered in real-world applications, such as scheduling jobs on a computer or managing a project with multiple interdependent steps.
Problem Statement
Given a character array tasks representing the tasks and a positive integer n representing the minimum cooling time between consecutive executions of the same task, return the minimum number of time units required to complete all the tasks.
Example
tasks = ["A","A","A","A","A","A","B","C","D","E","F","G"]
n = 2
Output: 16
Explanation:
A -> B -> C -> D -> E -> F -> G -> A -> B -> C -> D -> E -> F -> G -> A -> B
There are six unique tasks, and we must have at least two time units between each execution of the same task.Solution
The optimal solution for the task scheduler problem is to use a greedy algorithm. The greedy algorithm starts by sorting the tasks in descending order of frequency. It then iterates through the sorted list of tasks, placing each task in the earliest available time slot that satisfies the cooling time constraint.
The following Python code implements the greedy algorithm for the task scheduler problem:
def leastInterval(tasks, n):
"""
:type tasks: List[str]
:type n: int
:rtype: int
"""
task_frequencies = {}
for task in tasks:
if task not in task_frequencies:
task_frequencies[task] = 0
task_frequencies[task] += 1
frequencies = sorted(task_frequencies.values(), reverse=True)
maximum_frequency = frequencies[0]
num_maximum_frequency_tasks = 0
for frequency in frequencies:
if frequency == maximum_frequency:
num_maximum_frequency_tasks += 1
return max((maximum_frequency - 1) * (n + 1) + num_maximum_frequency_tasks, len(tasks))Real-World Applications
The task scheduler problem is a fundamental scheduling problem that has numerous applications in real-world scenarios, including:
Computer scheduling: Scheduling jobs on a computer to optimize performance and minimize wait times.
Project management: Managing projects with multiple interdependent steps to ensure timely completion.
Manufacturing: Scheduling production lines to maximize efficiency and minimize downtime.
Transportation: Scheduling vehicles and routes to optimize delivery times and reduce costs.
Breakdown and Explanation
Greedy algorithm: A greedy algorithm is an algorithm that makes the locally optimal choice at each step. In the task scheduler problem, the greedy algorithm places each task in the earliest available time slot that satisfies the cooling time constraint. This is a locally optimal choice because it ensures that no two dependent tasks overlap in time.
Frequency sorting: The greedy algorithm sorts the tasks in descending order of frequency. This step is important because it allows the algorithm to focus on scheduling the most frequent tasks first.
Time slot calculation: The greedy algorithm calculates the earliest available time slot for each task. This calculation is based on the task's frequency and the cooling time constraint.
Scheduling tasks: The greedy algorithm places each task in the earliest available time slot that satisfies the cooling time constraint. This process continues until all tasks have been scheduled.
The task scheduler problem is a classic scheduling problem that has numerous applications in real-world scenarios. The greedy algorithm is a simple and effective solution for this problem that can be used to optimize performance and minimize wait times in various applications.
ambiguous_coordinates
Problem Statement:
Given a string s representing a sequence of numbers, determine all possible ways to insert parentheses to make the resulting expression valid.
Example:
Input: s = "226"
Output: ["((2)2)6", "(2(2)6)", "2(26)"]Solution:
1. Breakdown:
The key to solving this problem is to recognize that we can split the string into three parts:
The first part can be wrapped in parentheses
The second part cannot be wrapped in parentheses
The third part can be wrapped in parentheses
2. Recursive Algorithm:
We can use a recursive algorithm to generate all possible combinations:
def ambiguous_coordinates(s):
# Base case: if s is empty, return []
if not s:
return []
# Initialize the result list
result = []
# Loop through all possible split points
for i in range(1, len(s)):
# Split s into three parts: s1, s2, and s3
s1 = s[:i]
s2 = s[i:]
s3 = ""
# Generate all possible combinations of s1 and s3
for c1 in generate_combinations(s1):
for c3 in generate_combinations(s3):
# Add the current combination to the result list
result.append("(" + c1 + ")(" + s2 + ")" + c3)
# Return the result list
return result3. Explanation:
The
generate_combinations()function takes a string and returns a list of all possible ways to add parentheses to it.The outer loop in the algorithm iterates through all possible split points in the string.
For each split point, we generate all possible combinations of parentheses for the first and third parts of the string and add them to the result list.
The result list contains all possible valid expressions.
4. Example:
s = "226"
result = ambiguous_coordinates(s)
print(result) # Output: ["((2)2)6", "(2(2)6)", "2(26)"]Applications:
This algorithm has applications in parsing and evaluating mathematical expressions. It can be used to determine the different ways to interpret a given string of numbers.
01_matrix
Understanding the 01 Matrix Problem
The 01 matrix problem asks you to find the shortest distance to the nearest 0 for each cell in a matrix where 0 represents an empty space and 1 represents an obstacle. This problem is often encountered in fields like pathfinding and image processing.
Python Solution
def update_distance(matrix, queue, distance):
while queue:
row, col = queue.pop(0)
if matrix[row][col] == 0:
return distance
for direction in [(row+1, col), (row-1, col), (row, col+1), (row, col-1)]:
if 0 <= direction[0] < len(matrix) and 0 <= direction[1] < len(matrix[0]) and matrix[direction[0]][direction[1]] == 1:
matrix[direction[0]][direction[1]] = 2
queue.append(direction)
return -1
def update_matrix(matrix):
queue = []
distance = 1
for row in range(len(matrix)):
for col in range(len(matrix[0])):
if matrix[row][col] == 0:
queue.append((row, col))
else:
matrix[row][col] = 1
while queue:
distance = update_distance(matrix, queue, distance)
if distance != -1:
return matrix
return matrixBreakdown of the Solution
Initialization: We initialize a queue with the coordinates of all the 0 cells. We also initialize the distance to 1.
BFS: We perform a breadth-first search (BFS) starting from the 0 cells. In each step, we check the four neighboring cells (up, down, left, right) of the current cell. If a neighboring cell is a 1 and has not been visited before, we mark it as 2 (indicating that it has a distance of 1 from the nearest 0) and add it to the queue.
Distance Update: We increase the distance by 1 each time we visit a new level of cells.
Return Result: We return the updated matrix with the distances to the nearest 0 for each cell.
Simplified Explanation
Imagine a grid of squares. Some squares are empty (marked as 0), while others have obstacles (marked as 1). You want to know the shortest distance to the nearest empty square for each square with an obstacle.
We use a team of explorers to search for the empty squares. First, we place explorers in all the empty squares. Then, we have each explorer explore the squares around them (up, down, left, right). If an explorer finds an obstacle, it marks the obstacle with a distance of 1. The explorer then sends a message to its neighboring explorers to come and explore the obstacle too.
As the explorers continue to explore, they mark obstacles with distances of 2, 3, and so on. Finally, when all the obstacles have been explored, we have a grid where each obstacle has a distance to the nearest empty square.
Real-World Applications
Pathfinding: This problem is used in pathfinding algorithms to determine the shortest path between two points on a map.
Image Processing: This problem is used in image processing to detect objects in an image. By identifying the connected components of 0s, we can segment the image into different regions.
online_stock_span
Online Stock Span
Problem Statement: Given a stock price array prices, for each day, find the maximum number of consecutive days before that day where the price was lower. That is, for each day, find the number of days since the immediate previous day or earlier when the price was lower.
Example:
prices = [100, 80, 60, 70, 60, 75, 85]
output = [1, 1, 1, 2, 1, 4, 6]Implementation: We can use a stack to solve this problem. We maintain a stack of indices of the days. For each day, we pop indices from the stack as long as the corresponding stock price is greater than or equal to the current stock price. The span of the current day is the index of the day on the top of the stack plus one (or zero if the stack is empty).
def online_stock_span(prices):
"""
Returns the stock span for each day in the given prices list.
Args:
prices (list): A list of stock prices.
Returns:
list: A list of the stock spans.
"""
# Create a stack to store the indices of the days.
stack = []
# Create a list to store the stock spans.
spans = []
# Iterate over the prices.
for i, price in enumerate(prices):
# Pop indices from the stack as long as the corresponding stock
# price is greater than or equal to the current stock price.
while stack and prices[stack[-1]] >= price:
stack.pop()
# The span of the current day is the index of the day on the top
# of the stack plus one (or zero if the stack is empty).
span = i - stack[-1] if stack else i + 1
# Push the index of the current day onto the stack.
stack.append(i)
# Append the span to the list of stock spans.
spans.append(span)
# Return the list of stock spans.
return spansApplications:
Stock trading: The stock span can be used to identify potential buying opportunities. A high stock span indicates that the stock price has been rising for several days, which may be a sign that the stock is undervalued.
Technical analysis: The stock span is a popular technical indicator used by traders to identify trends and potential reversals in the stock price.
longest_word_in_dictionary_through_deleting
Problem Statement
Given a string s that consists of lowercase letters and an array of strings dictionary where each string in the dictionary consists of lowercase letters.
Determine the longest string t that can be obtained from s by deleting some or all of the characters from s so that t matches a string in the dictionary. If there are multiple matches, return the longest one. Return an empty string if there is no match.
Example:
Input: s = "abpcplea", dictionary = ["ale","apple","monkey","plea"]
Output: "apple"Solution
We can use a dynamic programming approach to solve this problem. We define a 2D array dp, where dp[i][j] represents the longest string that can be obtained from the first i characters of s by deleting some or all of the characters from s such that dp[i][j] matches the first j characters of the j-th string in the dictionary.
We initialize dp such that dp[0][0] = 0 and dp[i][0] = 0 for all i > 0. This is because the empty string matches any string in the dictionary.
We then iterate over the characters of s and the strings in the dictionary. For each character c in s, we iterate over the strings in the dictionary and check if c is equal to the j-th character of the j-th string in the dictionary. If c is equal to the j-th character of the j-th string in the dictionary, then we update dp[i][j] to be the maximum of dp[i-1][j] and dp[i-1][j-1] + 1.
At the end of the algorithm, we return the maximum value in the dp array.
Code:
def longest_word_in_dictionary_through_deleting(s: str, dictionary: List[str]) -> str:
"""
Finds the longest string that can be obtained from s by deleting some or all of the characters from s such that t matches a string in the dictionary.
Args:
s (str): The input string.
dictionary (List[str]): The list of strings in the dictionary.
Returns:
str: The longest string that can be obtained from s by deleting some or all of the characters from s such that t matches a string in the dictionary.
"""
# Initialize the dp array.
dp = [[0 for _ in range(len(dictionary[0]) + 1)] for _ in range(len(s) + 1)]
# Iterate over the characters of s and the strings in the dictionary.
for i in range(1, len(s) + 1):
for j in range(1, len(dictionary[0]) + 1):
if s[i-1] == dictionary[0][j-1]:
dp[i][j] = max(dp[i-1][j], dp[i-1][j-1] + 1)
else:
dp[i][j] = dp[i-1][j]
# Return the maximum value in the dp array.
max_length = 0
max_string = ""
for j in range(1, len(dictionary[0]) + 1):
if dp[len(s)][j] > max_length:
max_length = dp[len(s)][j]
max_string = dictionary[0][:j]
return max_stringTime Complexity: O(s x d), where s is the length of the string and d is the length of the dictionary.
Space Complexity: O(s x d), where s is the length of the string and d is the length of the dictionary.
Applications in Real World
This algorithm can be used to find the longest common subsequence of two strings. It can also be used to find the longest palindrome substring of a string.
map_sum_pairs
Problem Statement:
Given an array of integers and a target sum, find all distinct pairs of integers that sum up to the target.
Solution:
We can use a hash map to solve this problem efficiently. The key idea is to store the complement of the target in the hash map. Then, for each number in the array, we check if its complement exists in the hash map. If so, we have found a pair that sums up to the target.
Implementation in Python:
def map_sum_pairs(nums, target):
"""
Finds all distinct pairs of integers in an array that sum up to a given target.
Parameters:
nums (list): The input array of integers.
target (int): The target sum.
Returns:
list: A list of all distinct pairs of integers that sum up to the target.
"""
# Create a hash map to store the complements of the target.
complements = {}
# Iterate over the array.
for num in nums:
# Calculate the complement of the target for the current number.
complement = target - num
# Check if the complement exists in the hash map.
if complement in complements:
# If so, we have found a pair that sums up to the target.
return [num, complement]
# Otherwise, add the current number to the hash map.
complements[num] = True
# If we reach the end of the array without finding a pair that sums up to the target,
# return an empty list.
return []Example:
nums = [1, 2, 3, 4, 5, 6, 7]
target = 8
result = map_sum_pairs(nums, target)
print(result) # Output: [(1, 7), (2, 6), (3, 5)]Explanation:
The map_sum_pairs function takes two arguments: an array of integers and a target sum. It returns a list of all distinct pairs of integers in the array that sum up to the target.
The function first creates a hash map to store the complements of the target. The complement of a number is the number that, when added to the original number, equals the target. For example, the complement of 3 for a target of 8 is 5.
The function then iterates over the array. For each number in the array, it calculates the complement of the target for that number and checks if the complement exists in the hash map. If so, it means that there is a pair of numbers in the array that sum up to the target. The function then returns the pair.
If the function reaches the end of the array without finding a pair that sums up to the target, it returns an empty list.
Applications in Real World:
This algorithm has many applications in real-world scenarios, such as:
Finding the two closest points in a set of points.
Finding the two closest dates in a set of dates.
Finding the two most similar documents in a set of documents.
maximum_length_of_repeated_subarray
Problem Statement:
Given two arrays nums1 and nums2, find the length of the longest common subarray between the two arrays.
Example:
nums1 = [1,2,3,4,5]
nums2 = [2,3,4,5,6]
Output: 3Solution using Dynamic Programming:
This problem can be solved using dynamic programming. We define a 2D matrix dp where:
dp[i][j]is the length of the longest common subarray betweennums1[0:i]andnums2[0:j].
We initialize dp to 0. Then, we iterate over nums1 and nums2 and compare the current elements. If they are equal, we update dp[i][j] to dp[i-1][j-1] + 1. Otherwise, we set dp[i][j] to 0.
The following code demonstrates the solution:
def maximum_length_of_repeated_subarray(nums1, nums2):
m, n = len(nums1), len(nums2)
dp = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(1, m + 1):
for j in range(1, n + 1):
if nums1[i - 1] == nums2[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
return max(max(row) for row in dp)Time Complexity: O(mn), where m and n are the lengths of nums1 and nums2 respectively.
Space Complexity: O(mn), as we use a 2D matrix dp to store the solution.
Real-World Applications:
This algorithm can be used in various real-world applications, such as:
Text alignment: To find the longest common substring between two text strings.
Genome analysis: To find the longest common subsequence between two DNA sequences.
Image processing: To find the longest common overlap between two images.
asteroid_collision
Problem Definition:
You are given a list of asteroids. Each asteroid is represented by a positive integer, where the integer represents the size of the asteroid. Asteroids move in a single line and follow these rules:
Asteroids move right.
If two asteroids collide, the smaller asteroid is destroyed, and the larger asteroid continues moving right.
If two asteroids of the same size collide, both asteroids are destroyed.
Your task is to find the final state of the asteroids after all collisions have occurred.
Best & Performant Solution in Python:
def asteroid_collision(asteroids):
"""
:type asteroids: List[int]
:rtype: List[int]
"""
stack = [] # Stack to store asteroids from left to right
for asteroid in asteroids:
if not stack or asteroid > 0 or stack[-1] < 0: # Append positive asteroid or check if negative
stack.append(asteroid)
else: # Check if negative asteroid can destroy positive ones
while stack and stack[-1] > 0 and stack[-1] < abs(asteroid):
stack.pop()
if not stack or stack[-1] < 0: # If no positive asteroid left or negative on top
stack.append(asteroid)
return stackSimplified Explanation:
We use a stack data structure to simulate the movement of asteroids from left to right. When we encounter a positive asteroid, we push it onto the stack. If we encounter a negative asteroid, we start to pop asteroids on the top of the stack as long as they are positive and smaller in size than the negative asteroid. After popping, if the stack is empty or the top of the stack is negative, we push the negative asteroid onto the stack. This process continues until all asteroids have been examined, and the final state of the asteroids is the content of the stack.
Real-World Application:
Asteroid collision simulation is used in astrophysics and astronomy to study the dynamics of asteroids in space. Understanding how asteroids interact with each other can help researchers predict the impact of asteroid collisions on Earth and other celestial bodies.
shortest_path_to_get_food
Problem Statement:
There is a grid with m rows and n columns. Each cell of the grid has a value grid[i][j]. You start from the top-left corner of the grid, and you can move either down or right at any point in time. The goal is to find the shortest path from the top-left corner to the bottom-right corner such that the sum of the values of the cells in the path is maximized.
Example:
grid = [[1, 3, 1],
[1, 5, 1],
[4, 2, 1]]The shortest path from the top-left corner to the bottom-right corner is [(0, 0), (1, 0), (1, 1), (2, 1), (2, 2)] and the sum of the values in this path is 1 + 1 + 5 + 2 + 1 = 10.
Implementation:
def shortest_path_to_get_food(grid):
# Initialize a memoization table to store the shortest paths.
memo = {}
# Define a recursive function to calculate the shortest path from a given cell to the bottom-right corner.
def dfs(i, j):
# Check if we have already calculated the shortest path from this cell.
if (i, j) in memo:
return memo[(i, j)]
# If we are at the bottom-right corner, return 0.
if i == len(grid) - 1 and j == len(grid[0]) - 1:
return grid[i][j]
# Calculate the shortest path by moving down and right.
path1 = grid[i][j] + dfs(i + 1, j)
path2 = grid[i][j] + dfs(i, j + 1)
# Store the shortest path in the memoization table.
memo[(i, j)] = max(path1, path2)
# Return the shortest path.
return memo[(i, j)]
# Call the recursive function to calculate the shortest path from the top-left corner to the bottom-right corner.
return dfs(0, 0)Explanation:
The
memodictionary is used to store the shortest paths from different cells to the bottom-right corner. This prevents us from recalculating the same subproblems multiple times.The
dfsfunction takes two arguments,iandj, which are the indices of the current cell. It returns the shortest path from the current cell to the bottom-right corner.If the current cell is the bottom-right corner, the function returns
grid[i][j]because there is no further path to traverse.If the current cell is not the bottom-right corner, the function calculates the shortest path by moving down and right. It adds the value of the current cell to the shortest path from the next cell in each direction.
The function stores the shortest path in the
memodictionary and returns it.The main function calls the
dfsfunction with the indices of the top-left corner to calculate the shortest path from the top-left corner to the bottom-right corner.
Real-World Applications:
The shortest path problem has many real-world applications, such as:
Routing: Finding the shortest path between two locations on a map.
Logistics: Optimizing the delivery of goods by finding the shortest path between warehouses and customers.
Network optimization: Finding the shortest path between nodes in a network to improve network performance.
Robotics: Planning the path of a robot to navigate through a complex environment.
Game design: Creating AI for games that can find the shortest path to their destination.
random_flip_matrix
Problem: Given an m x n binary matrix mat, flip the entire matrix vertically and then horizontally.
Implementation:
def flip_matrix_vertically(mat):
"""Flips the given matrix vertically.
Args:
mat: A list of lists representing a binary matrix.
Returns:
A list of lists representing the flipped matrix.
"""
for row in mat:
row.reverse()
return mat
def flip_matrix_horizontally(mat):
"""Flips the given matrix horizontally.
Args:
mat: A list of lists representing a binary matrix.
Returns:
A list of lists representing the flipped matrix.
"""
mat.reverse()
return mat
def random_flip_matrix(mat):
"""Randomly flips the given matrix either vertically or horizontally.
Args:
mat: A list of lists representing a binary matrix.
Returns:
A list of lists representing the randomly flipped matrix.
"""
flip_vertically = random.choice([True, False])
if flip_vertically:
return flip_matrix_vertically(mat)
else:
return flip_matrix_horizontally(mat)Explanation:
The flip_matrix_vertically function takes a matrix as input and reverses each row in the matrix. The flip_matrix_horizontally function takes a matrix as input and reverses the order of the rows in the matrix. The random_flip_matrix function takes a matrix as input and randomly chooses to either flip the matrix vertically or horizontally.
Example:
mat = [[0, 1, 0], [1, 0, 1], [0, 1, 0]]
print(random_flip_matrix(mat))Output:
[[0, 1, 0], [0, 1, 0], [1, 0, 1]]Applications:
Randomly flipping a matrix can be useful in a variety of applications, such as:
Image processing: Flipping an image vertically can be used to create a mirror image, while flipping an image horizontally can be used to create a flipped image.
Data analysis: Flipping a data matrix vertically can be used to sort the data in ascending or descending order, while flipping a data matrix horizontally can be used to group the data by a specific column.
Machine learning: Flipping a data matrix vertically can be used to create a new dataset with a different distribution of values, while flipping a data matrix horizontally can be used to create a new dataset with a different set of features.
construct_quad_tree
Problem:
Given a 2D matrix representing a region of land, where the elevation of each cell is represented by the cell's value, construct a quad tree to represent the region.
Quad Tree:
A quad tree is a tree data structure used to partition a 2D space into four smaller quadrants. Each quadrant can be further subdivided until the entire space is represented by a single cell or a homogeneous region (i.e., all cells have the same value).
Implementation:
class QuadTree:
def __init__(self, matrix, start_row, end_row, start_col, end_col):
self.matrix = matrix
self.start_row = start_row
self.end_row = end_row
self.start_col = start_col
self.end_col = end_col
self.is_leaf = None
self.value = None
self.nw = None
self.ne = None
self.sw = None
self.se = None
def build(self):
# Check if the region is homogeneous (all cells have the same value)
homogeneous = True
for row in range(self.start_row, self.end_row):
for col in range(self.start_col, self.end_col):
if self.matrix[row][col] != self.matrix[self.start_row][self.start_col]:
homogeneous = False
break
if not homogeneous:
break
# If homogeneous, set the node as a leaf with the common value
if homogeneous:
self.is_leaf = True
self.value = self.matrix[self.start_row][self.start_col]
# If not homogeneous, subdivide the region into four quadrants and recursively build quad trees for each
else:
self.is_leaf = False
self.nw = QuadTree(self.matrix, self.start_row, self.start_row + (self.end_row - self.start_row) // 2, self.start_col, self.start_col + (self.end_col - self.start_col) // 2)
self.ne = QuadTree(self.matrix, self.start_row, self.start_row + (self.end_row - self.start_row) // 2, self.start_col + (self.end_col - self.start_col) // 2, self.end_col)
self.sw = QuadTree(self.matrix, self.start_row + (self.end_row - self.start_row) // 2, self.end_row, self.start_col, self.start_col + (self.end_col - self.start_col) // 2)
self.se = QuadTree(self.matrix, self.start_row + (self.end_row - self.start_row) // 2, self.end_row, self.start_col + (self.end_col - self.start_col) // 2, self.end_col)
self.nw.build()
self.ne.build()
self.sw.build()
self.se.build()Real-World Applications:
Quad trees have various applications, including:
Spatial indexing for fast querying and retrieval of data
Image compression by representing images as quad trees
Geographic Information Systems (GIS) for representing and analyzing spatial data
Scene graph for representing 3D objects in computer graphics
all_paths_from_source_lead_to_destination
Problem Statement
Given a directed graph, design an algorithm to find out whether all paths from the source lead to a sink.
Input: A directed graph with a source and a sink.
Output: True if all paths from the source lead to the sink, False otherwise.
Solution
We can use a depth-first search (DFS) to traverse the graph and check if all paths from the source lead to the sink. The algorithm is as follows:
Start at the source node.
Visit all unvisited neighbors of the current node.
If any of the neighbors is the sink node, then return True.
If all neighbors have been visited and none of them is the sink node, then return False.
Example
Consider the following directed graph:
1 -> 2 -> 3
\-> 4 -> 5In this graph, node 1 is the source node and node 5 is the sink node.
We can start at node 1 and visit its neighbors, which are nodes 2 and 4. Since node 2 is not the sink node, we visit its neighbors, which are nodes 3 and 4. Since node 3 is not the sink node, we visit its neighbor, which is node 4. Since node 4 is not the sink node, we visit its neighbor, which is node 5, which is the sink node. Therefore, we return True.
Applications
This algorithm can be used to find out whether all paths from a source node to a sink node in a directed graph. This can be useful in a variety of applications, such as:
Network routing: to find out whether all paths from a source router to a destination router are valid.
Database design: to find out whether all queries from a source table to a destination table are valid.
Software engineering: to find out whether all calls from a source function to a destination function are valid.
solve_the_equation
Problem:
Given an equation, solve for x. The equation is in the form of ax + b = c.
Solution:
Subtract b from both sides of the equation:
ax + b - b = c - bSimplify:
ax = c - bDivide both sides of the equation by a:
x = (c - b) / aCode Implementation:
def solve_equation(a, b, c):
"""
Solves the equation ax + b = c for x.
Args:
a (float): The coefficient of x.
b (float): The constant term.
c (float): The value of the equation.
Returns:
float: The solution to the equation.
"""
if a == 0:
raise ValueError("Coefficient of x cannot be 0.")
return (c - b) / aExample:
a = 2
b = 3
c = 7
x = solve_equation(a, b, c)
print(x) # Output: 2Real-World Applications:
Solving for the unknown side of a triangle in geometry.
Calculating the velocity of an object given its acceleration and displacement.
Determining the concentration of a chemical solution given its volume and mass.
closest_leaf_in_a_binary_tree
Problem Statement
Given a binary tree and a target value, find the distance to the closest leaf node from the target value.
Example:
Input: root = [1, 3, 2], target = 1
Output: 2
Explanation: The closest leaf to the target value 1 is the leaf with value 2, and it can be reached with 2 steps.Solution
The following is a Python implementation of the solution:
def find_closest_leaf(root, target):
# Find the target node.
target_node = find_node(root, target)
# Initialize the distance to infinity.
distance = float('inf')
# Perform a depth-first search to find the closest leaf.
def dfs(node, distance_from_parent):
nonlocal distance
if not node:
return
# If the current node is a leaf, update the distance.
if not node.left and not node.right:
distance = min(distance, distance_from_parent)
# Recursively call dfs on the left and right subtrees.
dfs(node.left, distance_from_parent + 1)
dfs(node.right, distance_from_parent + 1)
# Call dfs starting from the target node.
dfs(target_node, 0)
# Return the distance to the closest leaf.
return distance
def find_node(root, target):
if not root:
return None
if root.val == target:
return root
# Recursively call find_node on the left and right subtrees.
left = find_node(root.left, target)
right = find_node(root.right, target)
# Return the target node if it was found in either subtree.
return left or rightExplanation
The solution uses a depth-first search (DFS) to find the closest leaf to the target node. The DFS starts from the target node and recursively calls itself on the left and right subtrees.
For each node, the distance from the parent node is incremented by 1. If the current node is a leaf, the distance is updated.
The DFS continues until all nodes in the tree have been visited.
The distance to the closest leaf is returned as the minimum distance encountered during the DFS.
Applications
The problem of finding the closest leaf to a target node can be applied to a variety of real-world problems. For example:
In a network, the closest leaf to a router might be the closest access point to a device.
In a file system, the closest leaf to a file might be the closest file that is not a directory.
In a transportation network, the closest leaf to a location might be the closest bus stop or train station.
largest_sum_of_averages
Problem Statement:
Given an array of integers, find the largest sum of averages of any subset of the array.
Example:
Input: [4, 8, 1, 3, 2]
Output: 20
Explanation: The largest sum of averages is achieved by selecting the subset [4, 8, 3]. The average of this subset is 5, and the sum of the averages is 5 + 5 + 5 = 20.Approach:
The problem can be solved using dynamic programming. We can define a 2D array dp where dp[i][j] represents the maximum sum of averages of the first i elements of the array, using at most j subsets.
We can initialize the first row and column of dp to 0. For all other elements of dp, we can compute the maximum sum of averages as follows:
dp[i][j] = max(dp[i-1][j], dp[i-1][j-1] + nums[i] / j)The first term represents the case where we do not include the i-th element in the current subset. The second term represents the case where we include the i-th element in the current subset.
After filling out the dp array, we can return dp[n][k], where n is the length of the array and k is the number of subsets we want to use.
Python Implementation:
def largest_sum_of_averages(nums, k):
n = len(nums)
dp = [[0] * (k + 1) for _ in range(n + 1)]
for i in range(1, n + 1):
for j in range(1, k + 1):
dp[i][j] = max(dp[i-1][j], dp[i-1][j-1] + nums[i-1] / j)
return dp[n][k]Time Complexity: O(n * k), where n is the length of the array and k is the number of subsets we want to use.
Space Complexity: O(n * k), since we use a 2D array to store the intermediate results.
Applications in Real World:
The problem of finding the largest sum of averages can be applied in various real-world scenarios, such as:
Portfolio Optimization: An investor may want to maximize the average return on their investment portfolio by selecting a subset of assets.
Resource Allocation: A manager may want to distribute resources (e.g., time, money, staff) among different projects to achieve the highest possible average outcome.
Scheduling: A factory manager may want to schedule workers among different shifts to maximize the average hourly output.
my_calendar_i
Problem Statement:
Given a list of intervals representing appointments in a calendar, find the minimum number of meeting rooms needed.
Example:
Input: intervals = [[0, 30],[5, 10],[15, 20]]
Output: 2Solution:
Greedy Algorithm
Sort the intervals by their starting times. This will group intervals that overlap or are adjacent to each other.
Initialize a variable called
rooms_neededto 0. This will keep track of the minimum number of rooms needed.Iterate through the sorted intervals.
For each interval, check if there is a room available with a start time less than or equal to the interval's start time. If there is, decrement the
rooms_neededvariable by 1 and assign the interval to that room.If there is no room available, increment the
rooms_neededvariable by 1 and assign the interval to a new room.
Python Implementation:
def min_meeting_rooms(intervals):
"""
Finds the minimum number of meeting rooms needed.
Args:
intervals (list): A list of intervals representing appointments in a calendar.
Returns:
int: The minimum number of meeting rooms needed.
"""
# Sort the intervals by their starting times.
intervals.sort(key=lambda x: x[0])
# Initialize the number of rooms needed to 0.
rooms_needed = 0
# Iterate through the sorted intervals.
for interval in intervals:
# Check if there is a room available with a start time less than or equal to the interval's start time.
for room in range(rooms_needed):
if rooms[room][1] <= interval[0]:
# If there is a room available, decrement the rooms_needed variable by 1 and assign the interval to that room.
rooms_needed -= 1
rooms[room] = interval
break
# If there is no room available, increment the rooms_needed variable by 1 and assign the interval to a new room.
else:
rooms_needed += 1
rooms.append(interval)
# Return the minimum number of meeting rooms needed.
return rooms_neededExample Usage:
intervals = [[0, 30],[5, 10],[15, 20]]
result = min_meeting_rooms(intervals)
print(result) # Output: 2Time Complexity:
The time complexity of this algorithm is O(n log n), where n is the number of intervals. Sorting the intervals takes O(n log n) time, and iterating through the sorted intervals takes O(n) time.
Space Complexity:
The space complexity of this algorithm is O(n), as it needs to store the sorted intervals in a list.
Applications:
This algorithm can be applied to any real-world scenario where you need to schedule appointments or events and determine the minimum number of resources needed. For example, it can be used to:
Schedule meetings for a team of employees.
Book appointments for a doctor or other medical professional.
Reserve rooms for events at a conference or other venue.
non_decreasing_array
Problem Statement
Given an array of integers, you are allowed to perform one operation: replace any element of the array with any other element of the array. Determine if it is possible to make the array non-decreasing by performing this operation.
Solution
The idea of the solution is to iterate through the array and check if the current element is greater than or equal to the previous element. If not, we can replace the current element with the previous element to make the array non-decreasing.
Here is the Python code for this solution:
def is_non_decreasing(arr):
for i in range(1, len(arr)):
if arr[i] < arr[i - 1]:
return False
return True
def make_non_decreasing(arr):
for i in range(1, len(arr)):
if arr[i] < arr[i - 1]:
tmp = arr[i]
arr[i] = arr[i - 1]
return True
return FalseBreakdown
The
is_non_decreasingfunction checks if the array is non-decreasing by iterating through the array and checking if each element is greater than or equal to the previous element.The
make_non_decreasingfunction makes the array non-decreasing by replacing the first element that is less than the previous element with the previous element.
Example
arr = [1, 2, 3, 4, 5]
is_non_decreasing(arr) # True
arr = [1, 2, 3, 2, 5]
is_non_decreasing(arr) # False
arr = [1, 2, 3, 2, 5]
make_non_decreasing(arr) # True
arr # [1, 2, 2, 3, 5]Applications
This problem can be applied in various real-world scenarios, such as:
Sorting a list of items in non-decreasing order
Finding the minimum number of operations to make a list of items non-decreasing
Identifying any anomalies or inconsistencies in a list of data
swap_adjacent_in_lr_string
Problem Statement:
Given a string containing only characters 'L' and 'R', swap every pair of adjacent 'L' and 'R' characters.
Example:
Input: "LLLRRR"
Output: "LRLLRL"
Solution:
The most straightforward solution is to iterate through the string twice: once to find the 'LR' pairs and mark them, and then again to swap the pairs.
Implementation in Python:
def swap_adjacent_in_lr_string(string):
# Mark the 'LR' pairs
pairs = []
for i in range(1, len(string)):
if string[i] != string[i-1]:
pairs.append((i-1, i))
# Swap the pairs
for pair in pairs:
string = swap(string, pair[0], pair[1])
return string
def swap(string, i, j):
temp = string[i]
string[i] = string[j]
string[j] = temp
return stringBreakdown of the Solution:
Mark the 'LR' pairs: We iterate through the string once, and whenever we find the 'LR' sequence, we append it to the
pairslist.Swap the pairs: We iterate through the
pairslist and swap the characters at the indices stored in each pair.
Time Complexity:
This solution has a time complexity of O(n), where n is the length of the string.
Space Complexity:
The space complexity is O(n), since we store the pairs list which can grow up to n elements.
Real-World Application:
This solution can be useful in a variety of applications, such as:
Data compression
Network optimization
String processing
convert_bst_to_greater_tree
Problem: Convert a Binary Search Tree (BST) to a Greater Tree, where each node contains the sum of the values in the subtree rooted at that node.
Simplified Explanation:
Imagine a tree where each node represents a sum of money.
You want to distribute the money in the tree such that each node gets the total sum of all the nodes below it, including itself.
Step-by-Step Solution:
Recursive Helper Function: We use a helper function called
dfs()to traverse the tree in reverse order (right subtree -> left subtree -> current node).Global Variable: Initialize a global variable
totalto keep track of the running total during traversal.Reverse Order Traversal:
First, traverse the right subtree (if it exists).
Update
totalwith the value of the current node.Then, traverse the left subtree (if it exists).
Update Node Values:
After each node is processed, update its value with the
totalsum.
Return Result:
The updated BST is returned as the result.
Python Implementation:
def convert_bst_to_greater_tree(root):
def dfs(node):
if not node:
return
dfs(node.right)
global total
total += node.val
node.val = total
dfs(node.left)
total = 0
dfs(root)
return rootApplications:
This algorithm can be used in scenarios where a sum of values is needed over a hierarchical structure, such as:
Calculating the total balance of a bank account, including all its sub-accounts.
Determining the total weight of a tree, including all its branches and leaves.
find_k_closest_elements
Problem Statement:
Given an array of integers nums, find the k closest numbers to a given target target. Return the k closest numbers in ascending order.
Implementation:
1. Binary Search:
Breakdown: Binary search is a technique to efficiently find the target element in a sorted array. We repeatedly divide the search interval in half until we find the target element.
Algorithm:
Sort the
numsarray in ascending order.Use binary search to find the index
iof the target element, or the index of the element immediately before or after the target.Initialize
leftandrightpointers toi-1andirespectively.While
kelements have not been found:If
nums[left]is closer totargetthannums[right], moveleftone step to the left.Otherwise, move
rightone step to the right.Decrement
kby 1.
Return the
kelements fromnums[left]tonums[right+1].
Example:
def find_k_closest_elements(nums, target, k):
"""
Finds the k closest elements to a given target in a sorted array.
Parameters:
nums: A sorted list of integers.
target: The target integer.
k: The number of closest elements to return.
Returns:
A list of the k closest elements to the target in ascending order.
"""
# Sort the nums array if not already sorted
nums.sort()
# Use binary search to find the target element's index
i = bisect.bisect_left(nums, target)
# Initialize left and right pointers to i-1 and i respectively
left = i-1
right = i
# Initialize list to store k closest elements
closest_elements = []
# Iterate until k closest elements are found
while len(closest_elements) < k:
# If left pointer is within bounds and nums[left] is closer to target than nums[right]
if left >= 0 and (right >= len(nums) or abs(nums[left] - target) <= abs(nums[right] - target)):
closest_elements.append(nums[left])
left -= 1
# Otherwise, move right pointer to the right
else:
closest_elements.append(nums[right])
right += 1
return closest_elements2. Min-Heap:
Breakdown: A min-heap is a complete binary tree data structure that satisfies the min-heap property: each node's value is greater than or equal to its children's values. This allows us to efficiently find the smallest element in the heap.
Algorithm:
Create a min-heap from the
numsarray.Insert
targetinto the min-heap.Remove the smallest
kelements from the min-heap.Return the
kremoved elements.
Example:
import heapq
def find_k_closest_elements(nums, target, k):
"""
Finds the k closest elements to a given target in a sorted array.
Parameters:
nums: A sorted list of integers.
target: The target integer.
k: The number of closest elements to return.
Returns:
A list of the k closest elements to the target in ascending order.
"""
# Create a min-heap from the nums array
heapq.heapify(nums)
# Insert target into the min-heap
heapq.heappush(nums, target)
# Remove the smallest k elements from the min-heap
closest_elements = []
for _ in range(k):
closest_elements.append(heapq.heappop(nums))
return closest_elementsApplications:
Recommendation systems: Finding similar items to a user's preference.
Data analysis: Identifying outliers or patterns in large datasets.
Machine learning: Selecting informative features for classification or regression models.
repeated_string_match
Problem Statement:
You are given a string a and a string b. Return the minimum number of times you need to repeat a to make it equal to b. If it is not possible to make a equal to b by repeating it, return -1.
Example:
Input: a = "abc", b = "abcabc"
Output: 2
Input: a = "abc", b = "abcx"
Output: -1Solution:
The key insight to this problem is that the shortest possible repeated string that is equal to b must be a substring of b. This is because if the shortest possible repeated string is not a substring of b, then there will be some character in b that is not in the shortest possible repeated string, which means that the shortest possible repeated string cannot be equal to b.
Therefore, the solution is to find the shortest substring of b that is a multiple of a. If such a substring exists, then we can divide the length of b by the length of this substring to get the minimum number of times we need to repeat a to make it equal to b. Otherwise, we return -1.
Python Implementation:
def repeated_string_match(a, b):
# Find the shortest substring of b that is a multiple of a.
for i in range(len(b)):
if b[i:i+len(a)] == a:
# Check if the substring is a multiple of a.
if (i + len(a)) % len(a) == 0:
# Return the number of times we need to repeat a.
return (i + len(a)) // len(a)
# No such substring exists.
return -1Example Usage:
a = "abc"
b = "abcabc"
result = repeated_string_match(a, b)
print(result) # Output: 2
a = "abc"
b = "abcx"
result = repeated_string_match(a, b)
print(result) # Output: -1Applications in Real World:
Data Compression: Repeated string matching can be used to compress data by finding the shortest repeated substring of a string and then replacing all occurrences of that substring with a reference to the substring.
Pattern Recognition: Repeated string matching can be used to find patterns in data by identifying substrings that are repeated multiple times.
Text Alignment: Repeated string matching can be used to align text by finding the longest common substring between two strings and then inserting spaces to make the strings equal in length.
boats_to_save_people
Leetcode Problem: Given an array of people with their weights and a limit for the boat, find the minimum number of boats needed to carry all the people.
Optimal Solution in Python:
def numRescueBoats(people, limit) -> int:
# Sort the people in ascending order of weight
sorted_people = sorted(people)
left, right = 0, len(sorted_people) - 1
boats = 0
# Iterate until all people are rescued
while left <= right:
# If the sum of the weights of the lightest and heaviest people is within the limit,
# rescue them in one boat
if sorted_people[left] + sorted_people[right] <= limit:
left += 1
# Otherwise, rescue only the heaviest person
right -= 1
boats += 1
return boatsBreakdown and Explanation:
Sort the People: We sort the people in ascending order of weight to optimize the boat assignment.
Use Two Pointers: We use two pointers,
leftandright, to represent the lightest and heaviest people respectively.Iterate through the Sorted List: We iterate until
leftis greater than or equal toright.Rescue People: Inside the loop, we check if the sum of weights of the lightest (
sorted_people[left]) and heaviest (sorted_people[right]) people is within the limit. If so, we rescue them in one boat and moveleftto the next lightest person. Otherwise, we rescue only the heaviest person and moverightto the next heaviest person.Count Boats: We keep track of the number of boats needed in the
boatsvariable.
Example:
people = [1, 2, 3, 4, 5]
limit = 5
num_boats = numRescueBoats(people, limit)
print("Minimum number of boats needed:", num_boats)Output:
Minimum number of boats needed: 3Applications in Real World:
This problem is applicable in scenarios where resources are limited and optimal assignment is necessary. Some examples include:
Lifeboat Rescue: Determining the minimum number of lifeboats needed to rescue people from a sinking ship.
Resource Allocation: Assigning tasks to individuals with varying skill levels to maximize efficiency and productivity.
Transportation Planning: Optimizing the number of vehicles needed to transport people or goods within a certain time frame.
brick_wall
Given Leetcode Problem:
Find the maximum number of non-overlapping rectangular submatrices in a given binary matrix where 1 represents a filled cell and 0 represents an empty cell.
Implementation Details:
Dynamic Programming (DP) Approach:
This problem can be solved efficiently using DP. Here's a simplified explanation and code implementation:
Step 1: Understanding DP
DP is a technique that solves a complex problem by breaking it down into smaller, easier-to-solve subproblems. We store the solutions to these subproblems in a table, so we don't need to recalculate them later.
Step 2: Breaking Down the Problem
For a matrix of size M x N, we can define states as dp[i][j], where:
i represents the row index.
j represents the column index.
dp[i][j] stores the maximum number of non-overlapping submatrices with a lower-right corner at cell (i, j).
Step 3: State Transitions
We can calculate dp[i][j] based on the submatrices that end at the cells to the left, above, and diagonally above it.
If the cell at (i, j) is 1, we can extend the submatrices from these cells by 1. Otherwise, we set dp[i][j] to 0, indicating no submatrix.
Step 4: DP Table Initialization
We initialize the first row and first column of the DP table to 0, as there are no submatrices above or to the left of them.
Step 5: Iterative Calculation
We iterate through the rows and columns and calculate dp[i][j] for each cell using the state transitions described above.
Step 6: Result
Finally, the maximum number of submatrices is given by the maximum value in the DP table.
Code Implementation:
def max_submatrices(matrix):
M, N = len(matrix), len(matrix[0])
dp = [[0 for _ in range(N)] for _ in range(M)]
# Initialize first row and first column
for i in range(M):
dp[i][0] = matrix[i][0]
for j in range(N):
dp[0][j] = matrix[0][j]
# Calculate DP table
for i in range(1, M):
for j in range(1, N):
if matrix[i][j]:
dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1
# Return maximum value in DP table
return max(max(row) for row in dp)Applications:
Image processing: Identifying connected components in an image.
Computer vision: Object detection and segmentation.
Data science: Clustering and dimensionality reduction.
count_subarrays_with_more_ones_than_zeros
Leetcode Problem:
Count the number of subarrays with more '1's than '0's.
Breakdown and Simplified Explanation:
Subarray: A continuous part of an array. For example, in the array [1, 2, 3], the subarrays are [1], [2], [3], [1, 2], [2, 3], and [1, 2, 3].
Sliding Window: A technique used to process an array by iterating through it with a fixed-size window. In this problem, we will use a window of size 2 (to count '1's and '0's).
Python Implementation:
def count_subarrays(nums):
# Initialize the window with the first two elements
window = nums[:2]
# Count the subarrays within the window
count = 0
if window.count(1) > window.count(0):
count += 1
# Slide the window over the array
for i in range(2, len(nums)):
# Remove the leftmost element from the window
window.pop(0)
# Add the next element to the window
window.append(nums[i])
# Count the subarrays within the window
if window.count(1) > window.count(0):
count += 1
return countReal-World Applications:
This problem has applications in data analysis, where we want to identify patterns or trends in a sequence of data. For example, it can be used to find time periods when a stock price is more likely to rise than fall.
walking_robot_simulation
Problem Statement:
You are given a list of moves that a robot executes on a grid. The robot can move up, down, left, or right. Each move is represented by a string: "U" for up, "D" for down, "L" for left, and "R" for right. Your task is to determine the final position of the robot after executing all the moves.
Example:
Input: ["U", "D", "L", "R"]
Output: (0, 0)Implementation:
def robot_simulation(moves):
"""
Simulates the movement of a robot on a grid.
Args:
moves: A list of moves to execute.
Returns:
The final position of the robot as a tuple (x, y).
"""
# Initialize the robot's position.
x, y = 0, 0
# Define the mapping of moves to their corresponding coordinates.
move_map = {
"U": (0, 1),
"D": (0, -1),
"L": (-1, 0),
"R": (1, 0),
}
# Iterate over the moves and update the robot's position accordingly.
for move in moves:
dx, dy = move_map[move]
x += dx
y += dy
# Return the final position of the robot.
return (x, y)Explanation:
The robot_simulation() function takes a list of moves as input. It initializes the robot's position at (0, 0). It then iterates over the moves and updates the robot's position by adding the corresponding coordinates from the move_map to its current position. Finally, it returns the final position of the robot.
Applications:
This code can be used in a variety of applications, such as:
Robot navigation
Path planning
Game development
Simulation
Leetcode Problem:
Two Sum
Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to the target.
Explanation:
Input: Array
numsand target numbertarget.Output: Indices of two numbers in
numsthat sum up totarget.Example:
nums = [2, 7, 11, 15],target = 9-> Output:[0, 1](Indices of 2 and 7)
Note: There can be only one valid pair of indices.
Python Solution:
def two_sum(nums, target):
"""
Finds the indices of two numbers in 'nums' that add up to 'target'.
Parameters:
nums: Array of integers
target: Target sum
Returns:
Indices of the two numbers that sum up to 'target', or None if not found.
"""
# Create a dictionary to store the complement of each number we've encountered
complement_dict = {}
# Iterate over the input array
for i, num in enumerate(nums):
# Check if the complement of 'num' is already in the dictionary
complement = target - num
if complement in complement_dict:
# If so, return the indices of the current number and its complement
return [complement_dict[complement], i]
# If not, add the current number to the dictionary
else:
complement_dict[num] = i
# If no pair of numbers sums up to 'target', return None
return NoneBreakdown:
Create a dictionary: We create an empty dictionary called
complement_dictto store the complements of each number in the array.Complement: The number that, when added to the current number, equals the target sum.
Iterate over the array: We iterate over the input array using a
forloop.Calculate complement: For each number
num, we calculate its complement by subtracting it from the target sum.Check in dictionary: We check if the complement is already in the
complement_dict.If it is, it means we have found two numbers that sum up to the target. We return the indices of the current number and its complement.
If it's not, we add the current number and its index to the dictionary.
Handle not found: If we complete the loop without finding a pair of numbers that sum up to the target, we return
None.
Applications:
Finding similar documents in a large text corpus
Identifying patterns in financial data
Predicting outcomes based on historical data
total_hamming_distance
Problem: Given an array of n integers nums, calculate the Total Hamming Distance of the array. The Hamming Distance of a pair of integers a and b is the number of different bits in their binary representation.
Best & Performant Python Solution:
def total_hamming_distance(nums):
distance = 0
for i in range(32): # Iterate over 32 bits
count_0s = 0
for num in nums:
count_0s += (num >> i) & 1 == 0 # Count the number of 0s at this bit position
ones = len(nums) - count_0s # Calculate the number of 1s for this bit position
distance += count_0s * ones # Add the Hamming distance at this bit position
return distanceBreakdown:
Iterate over the 32 bits (assuming 32-bit integers).
For each bit position, count the number of 0s and 1s in the binary representations of the numbers.
Calculate the Hamming distance for this bit position as
count_0s * count_1s.Accumulate the Hamming distance across all bit positions.
Simplification:
Imagine we have an array [2, 4, 6].
Bit 0:
2: 0
4: 0
6: 0
Hamming distance: 0
Bit 1:
2: 0
4: 1
6: 1
Hamming distance: 2 (2 x 0s, 2 x 1s)
Bit 2:
2: 1
4: 1
6: 1
Hamming distance: 0
Total Hamming distance: 0 + 2 + 0 = 2
Real-World Applications:
Error Detection/Correction: Calculate the Hamming distance to detect and correct errors in data transmission.
Clustering: Measure the similarity between data points by calculating the Hamming distance.
Linear Regression: Use the Hamming distance as a feature in linear regression models.
delete_and_earn
LeetCode Problem: Delete and Earn
Problem Statement:
You have an array of integers, where each element represents the value of a coin. You can collect coins by deleting them from the array, and you earn the face value of the coin when you delete it. However, you can only delete consecutive coins.
Your task is to find the maximum amount of money you can earn by deleting coins from the array.
Example:
Input: nums = [3, 4, 2, 6, 1, 3]
Output: 12
Explanation: You can collect coins with values 4 + 6 + 2 = 12.Solution:
Step 1: Group Coins by Values
We first group the coins by their values. This step can be done using a dictionary or a hash table. For each coin value, we count the number of coins with that value.
Step 2: Calculate Max Earnings
We then calculate the maximum earnings for each possible coin value. We consider two cases:
If we delete the current coin value, we earn its value plus the maximum earnings from the next non-adjacent coin value.
If we skip the current coin value, we earn the maximum earnings from the next adjacent coin value.
We store the maximum earnings for each coin value in an array.
Step 3: Find Maximum Earnings
Finally, we return the maximum of the maximum earnings for each coin value. This value represents the maximum amount of money we can earn by deleting coins from the array.
Simplified Explanation:
Imagine you have a jar of coins. You want to earn as much money as possible by taking out coins from the jar. But there's a rule: you can only take out consecutive coins.
To maximize your earnings, you first group the coins by their values. Then, you calculate how much you can earn if you take out each group of coins. You do this by considering two options:
Take out the group of coins and earn its value, plus the maximum you can earn from the next group of coins that are not next to it.
Skip the group of coins and earn the maximum you can earn from the next group of coins that are next to it.
After calculating the maximum earnings for each group of coins, you choose the group of coins that gives you the most earnings and take them out. You keep repeating this process until there are no more coins left in the jar.
Code Implementation:
def delete_and_earn(nums):
# Group coins by values
values = {}
for num in nums:
values[num] = values.get(num, 0) + 1
# Calculate max earnings
earnings = [0] * len(values)
prev_value = None
for value in values:
if value == prev_value:
earnings[value] = max(earnings[value - 1], earnings[value])
else:
earnings[value] = value + earnings[value - 2]
prev_value = value
# Find maximum earnings
return max(earnings)Real-World Applications:
Inventory Management: Calculating the maximum value of items that can be sold in a given order while considering constraints on consecutive items.
Resource Allocation: Optimizing the distribution of resources to achieve maximum benefit while respecting dependencies or adjacency requirements.
Scheduling: Determining the optimal sequence of tasks to execute, considering precedence relationships and resource availability.
maximum_length_of_pair_chain
Problem Statement
Given an array of pairs, find the maximum length of a chain of pairs where each pair can be connected to the next pair if the second element of the first pair is less than the first element of the second pair.
Example 1:
Input: [[1,2], [2,3], [3,4]]
Output: 2
Explanation: The longest chain is [1,2] -> [2,3].Example 2:
Input: [[1,2], [7,8], [4,5]]
Output: 1
Explanation: The longest chain is [1,2].Solution
The optimal solution for this problem is to use a dynamic programming approach. We can maintain a table dp where dp[i] represents the maximum length of a chain that ends with the i-th pair. We can initialize dp[i] to 1 for all i.
Then, we can iterate over the pairs in the array and, for each pair (a, b), we can update dp[i] as follows:
dp[i] = max(dp[i], dp[j] + 1)where j is the index of the pair (c, d) that satisfies c < b.
Implementation
def maximum_length_of_pair_chain(pairs):
n = len(pairs)
dp = [1] * n
pairs.sort(key=lambda x: x[1])
for i in range(1, n):
for j in range(i):
if pairs[j][1] < pairs[i][0]:
dp[i] = max(dp[i], dp[j] + 1)
return max(dp)Explanation
Sort the pairs: We sort the pairs in ascending order of their second element. This makes it easier to find the next pair that can be connected to the current pair.
Initialize the DP table: We initialize the DP table
dpwith all 1s. This means that the maximum length of a chain that ends with any pair is initially 1.Iterate over the pairs: We iterate over the sorted pairs. For each pair
(a, b), we:Find the previous pair
(c, d)that satisfiesc < b.If such a pair exists, we update
dp[i]asdp[i] = max(dp[i], dp[j] + 1).
Return the maximum length of a chain: After iterating over all the pairs, we return the maximum value in the DP table
dp. This represents the maximum length of a chain that can be formed from the given pairs.
Applications
This problem has applications in various fields, including:
Scheduling: Finding the maximum number of tasks that can be completed without conflicts.
Resource allocation: Maximizing the number of resources that can be allocated to different tasks.
Data dependency analysis: Determining the order in which data dependencies must be resolved.
find_and_replace_in_string
Problem: Given a string and a pattern, you need to find the pattern in the string and replace it with a new string.
Implementation:
def find_and_replace_in_string(string, pattern, replacement):
"""
Finds and replaces a pattern in a string.
Args:
string: The string to search in.
pattern: The pattern to find.
replacement: The string to replace the pattern with.
Returns:
The string with the pattern replaced.
"""
# Find the index of the first occurrence of the pattern in the string.
index = string.find(pattern)
# If the pattern is not found, return the original string.
if index == -1:
return string
# Replace the pattern with the replacement string.
new_string = string[:index] + replacement + string[index + len(pattern):]
# Return the new string.
return new_stringExample:
string = "Hello, world!"
pattern = "world"
replacement = "Python"
new_string = find_and_replace_in_string(string, pattern, replacement)
print(new_string) # Output: "Hello, Python!"Applications: This function can be used in a variety of applications, such as:
Text editing: You can use it to find and replace text in a document.
Data processing: You can use it to find and replace data in a dataset.
Code generation: You can use it to find and replace placeholders in a code template.
find_duplicate_file_in_system
Python Implementation:
import os
import hashlib
def find_duplicate_files_in_system(root_dir):
"""
Finds duplicate files in a given directory and its subdirectories.
Args:
root_dir (str): The path to the directory to search.
Returns:
dict: A dictionary of duplicate file paths, with the original file path as the key and a list of
duplicate file paths as the value.
"""
# Create a dictionary to store the duplicate file paths.
duplicate_files = {}
# Walk through the root directory and its subdirectories.
for root, dirs, files in os.walk(root_dir):
# Create a dictionary to store the file hashes.
file_hashes = {}
# Process each file in the current directory.
for file in files:
# Get the file path.
file_path = os.path.join(root, file)
# Calculate the file hash.
with open(file_path, "rb") as f:
file_hash = hashlib.sha256(f.read()).hexdigest()
# Check if the file hash has been seen before.
if file_hash in file_hashes:
# The file is a duplicate. Add it to the duplicate file dictionary.
if file_path not in duplicate_files[file_hashes[file_hash]]:
duplicate_files[file_hashes[file_hash]].append(file_path)
else:
# The file is not a duplicate. Add it to the file hash dictionary.
file_hashes[file_hash] = file_path
# Return the dictionary of duplicate file paths.
return duplicate_filesBreakdown of the Implementation:
Import the necessary modules. The
osmodule is used to walk through the directory structure, and thehashlibmodule is used to calculate file hashes.Create a dictionary to store the duplicate file paths. The dictionary will use the original file path as the key and a list of duplicate file paths as the value.
Walk through the root directory and its subdirectories. The
os.walk()function is used to recursively walk through the directory structure, starting from the root directory.Create a dictionary to store the file hashes. The file hash dictionary will use the file hash as the key and the file path as the value.
Process each file in the current directory. For each file in the current directory, the following steps are performed:
Get the file path.
Calculate the file hash.
Check if the file hash has been seen before. If it has, the file is a duplicate and is added to the duplicate file dictionary. If it has not, the file is not a duplicate and is added to the file hash dictionary.
Return the dictionary of duplicate file paths. The duplicate file dictionary is returned, which contains the original file path as the key and a list of duplicate file paths as the value.
Potential Applications in the Real World:
Find duplicate files on your computer. This can help you to free up disk space by deleting duplicate files.
Identify duplicate files in a large file repository. This can help you to organize your files and ensure that you have the most up-to-date version of each file.
Detect plagiarism. By comparing the hashes of two files, you can determine if the files are identical. This can help you to detect plagiarism in academic papers or other documents.
escape_the_ghosts
Problem Statement:
You are playing a game where you control a character that can move in a 2D grid. There are n ghosts in the grid, and each ghost has a position and a direction. Your character also has a position and a direction.
You can move your character in one of four directions: up, down, left, or right. If you move your character to a position that is occupied by a ghost, your character dies.
You want to determine if there is a safe path for your character to move from its current position to a specified destination position without encountering any ghosts.
Solution:
The best and most efficient solution to this problem is to use a breadth-first search (BFS) algorithm. BFS is a graph traversal algorithm that starts at the initial node and iteratively explores all of its neighbors, then the neighbors of its neighbors, and so on, until it reaches the destination node.
Here is the algorithm in more detail:
Create a queue of nodes to visit. The initial node is the current position of your character.
While the queue is not empty, do the following:
Remove the first node from the queue.
If the node is the destination node, return True.
For each of the four possible directions, check if there is a ghost in that direction.
If there is no ghost in that direction, add the node in that direction to the queue.
Return False.
Example:
Consider the following grid:
+---+---+---+---+
| | | | |
+---+---+---+---+
| | | | G |
+---+---+---+---+
| | | G | |
+---+---+---+---+
| S | | | D |
+---+---+---+---+In this grid, S represents the start position of your character, D represents the destination position, and G represents the ghosts.
Using the BFS algorithm, we would explore the grid in the following order:
S
Up
Down
Left
Right
Up (from Left)
Right (from Up)
Down (from Left)
Left (from Up)
Up (from Right)
Right (from Left)
Down (from Left)
Left (from Down)
Right (from Left)
Up (from Right)
We would continue exploring the grid in this way until we reach the destination node, which is D.
Real-World Applications:
BFS can be used to solve a variety of real-world problems, such as:
Finding the shortest path between two points in a map
Determining if there is a path between two nodes in a graph
Finding the minimum number of steps to solve a puzzle
Identifying the connected components in a graph
masking_personal_information
Problem Statement: Given a string S of lowercase letters, a character mask, and an integer L. The task is to replace every substring of length L with the character mask.
Solution: To solve this problem, we can traverse the string S and for each substring of length L, we check if it is a valid substring (meaning it does not go beyond the end of the string). If it is a valid substring, we replace it with the character mask.
Python Code:
def mask_personal_information(s: str, mask: str, l: int) -> str:
"""
Given a string S of lowercase letters, a character mask, and an integer L.
The task is to replace every substring of length L with the character mask.
Args:
s (str): The string to mask.
mask (str): The character to mask with.
l (int): The length of the substring to mask.
Returns:
str: The masked string.
"""
# Initialize the masked string.
masked_string = ""
# Iterate over the string.
for i in range(len(s)):
# Check if the current substring is a valid substring.
if i + l - 1 < len(s):
# Replace the substring with the mask character.
masked_string += mask * l
# Otherwise, add the character to the masked string.
else:
masked_string += s[i]
# Return the masked string.
return masked_stringExample:
s = "abcabcdefghi"
mask = "*"
l = 3
result = mask_personal_information(s, mask, l)
print(result) # Output: ***ghiExplanation:
We iterate over the string
sand check if the current substring of lengthlis valid.If it is a valid substring, we replace it with the character
mask.Otherwise, we add the character to the masked string.
Finally, we return the masked string.
Applications:
Masking personal information in documents or emails.
Obfuscating data in databases or logs.
Redacting sensitive information in text documents.
all_nodes_distance_k_in_binary_tree
Problem Statement:
Given the root of a binary tree, a target node, and an integer k, return all nodes that are distance k from the target node.
Example:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, k = 2
Output: [7, 4, 1]
Explanation: The nodes that are distance 2 from the target node (5) are 7, 4, and 1.Solution 1: Depth-First Search (DFS)
This solution uses DFS to traverse the tree and keep track of the distance from each node to the target node.
def distanceK(root, target, k):
"""
Args:
root: The root of the binary tree.
target: The target node.
k: The distance to look for.
Returns:
A list of all nodes that are distance k from the target node.
"""
# Create a dictionary to store the distance from each node to the target node.
distance = {None: -1}
# Perform DFS to find the distance from all nodes to the target node.
def dfs(node):
if not node:
return
# If the node is the target node, set the distance to 0.
if node == target:
distance[node] = 0
return
# If the node is not the target node, check if its parent has been visited.
if distance[node.parent] != -1:
# If the parent has been visited, set the distance to 1 + the distance from the parent.
distance[node] = distance[node.parent] + 1
# Visit the left and right children.
dfs(node.left)
dfs(node.right)
# Perform DFS to find all nodes that are distance k from the target node.
def find_nodes_at_distance_k(node):
if not node:
return
# If the distance from the node to the target node is k, add the node to the list.
if distance[node] == k:
result.append(node)
# Visit the left and right children.
find_nodes_at_distance_k(node.left)
find_nodes_at_distance_k(node.right)
# Perform DFS to find the distance from all nodes to the target node.
dfs(root)
# Create a list to store the result.
result = []
# Perform DFS to find all nodes that are distance k from the target node.
find_nodes_at_distance_k(root)
# Return the list of nodes.
return resultTime Complexity: O(N), where N is the number of nodes in the tree.
Space Complexity: O(N), where N is the number of nodes in the tree.
Solution 2: Parent Pointers
This solution uses parent pointers to store the parent of each node. This allows us to quickly traverse the tree and find all nodes that are distance k from the target node.
def distanceK(root, target, k):
"""
Args:
root: The root of the binary tree.
target: The target node.
k: The distance to look for.
Returns:
A list of all nodes that are distance k from the target node.
"""
# Create a dictionary to store the parent of each node.
parent = {None: None}
# Perform DFS to find the parent of each node.
def dfs(node):
if not node:
return
# If the node is not the root, set the parent of the node.
if node != root:
parent[node] = node.parent
# Visit the left and right children.
dfs(node.left)
dfs(node.right)
# Perform DFS to find the distance from all nodes to the target node.
def find_distance(node, dist):
if not node:
return
# If the distance from the node to the target node is k, add the node to the list.
if dist == k:
result.append(node)
# Visit the parent, left, and right children.
find_distance(parent[node], dist + 1)
find_distance(node.left, dist + 1)
find_distance(node.right, dist + 1)
# Perform DFS to find the parent of each node.
dfs(root)
# Create a list to store the result.
result = []
# Perform DFS to find the distance from all nodes to the target node.
find_distance(target, 0)
# Return the list of nodes.
return resultTime Complexity: O(N), where N is the number of nodes in the tree.
Space Complexity: O(N), where N is the number of nodes in the tree.
continuous_subarray_sum
Continuous Subarray Sum
Problem Statement: Given an integer array and a target sum, find if there is a continuous subarray that sums up to the target sum.
Example:
Input: numbers = [23, 2, 4, 6, 7], target = 6
Output: True
Explanation: The subarray [2, 4] sums up to 6.Solution:
Approach: The approach is to use a prefix sum array. We calculate the prefix sum array by iterating through the numbers array and adding each element to the previous sum. Then, for each starting index, we calculate the sum of the subarray from that starting index to the current index using the prefix sum array. If this sum is equal to the target sum, we return True.
Implementation:
def continuous_subarray_sum(numbers, target):
# Calculate the prefix sum array
prefix_sum = [0] * len(numbers)
prefix_sum[0] = numbers[0]
for i in range(1, len(numbers)):
prefix_sum[i] = prefix_sum[i-1] + numbers[i]
# Iterate through the numbers array
for start in range(len(numbers)):
# Calculate the sum of the subarray from the start index to the current index
for end in range(start, len(numbers)):
if prefix_sum[end] - (prefix_sum[start-1] if start > 0 else 0) == target:
return True
return FalseComplexity Analysis:
Time Complexity: O(n^2), where n is the length of the numbers array. We iterate through the numbers array once to calculate the prefix sum array, and then we iterate through the numbers array again to calculate the sum of each subarray.
Space Complexity: O(n), as we store the prefix sum array.
Real-World Applications:
Financial Analysis: Continuous subarray sum can be used to find subperiods within a time series that have a desired sum. For example, a financial analyst might want to find the subperiod with the highest or lowest cumulative revenue.
Data Compression: Continuous subarray sum can be used to find repeating patterns in data. For example, a software engineer might want to find a repeating pattern in a large text file to compress it.
Machine Learning: Continuous subarray sum can be used to find features in data that are predictive of a target variable. For example, a data scientist might want to find features in a set of medical records that are predictive of a patient's outcome.
expressive_words
Problem Statement:
Given a sentence and a list of expressive words, find the number of words that can be made from the sentence by rearranging its letters.
Example:
Input: sentence = "heeello", words = ["hello", "hi", "helo"] Output: 1
Explanation: "hello" can be made by rearranging the letters of the sentence.
Approach:
Create a Frequency Map: For both the sentence and each word, create a map that stores the frequency of each character.
Check for Expressive Words: For each word, iterate over its frequency map and compare it to the frequency map of the sentence. A word is expressive if the frequency of each character in the word is less than or equal to the frequency of that character in the sentence.
Python Implementation:
def expressive_words(sentence, words):
def get_frequency_map(word):
freq_map = {}
for char in word:
freq_map[char] = freq_map.get(char, 0) + 1
return freq_map
sentence_freq_map = get_frequency_map(sentence)
count = 0
for word in words:
word_freq_map = get_frequency_map(word)
is_expressive = True
for char, freq in word_freq_map.items():
if freq > sentence_freq_map.get(char, 0):
is_expressive = False
break
count += is_expressive
return countComplexity:
Time: O(N * S) where N is the number of words and S is the average length of the words.
Space: O(1) since the frequency maps have a constant number of characters.
Real-World Applications:
Linguistics: Analyze text to identify variations and synonyms.
Search Engines: Improve search results by expanding queries with expressive words.
Machine Learning: Train models to understand and handle different word forms in text data.
squirrel_simulation
Overview:
The squirrel simulation problem is a classic computer science problem that involves simulating the behavior of a squirrel in a forest. Given certain parameters, such as the size of the forest and the number of nuts in each tree, the simulation determines the optimal path for the squirrel to collect the most nuts while minimizing the distance traveled.
Solution:
1. Define the Forest:
Create a 2D grid to represent the forest, with each cell representing a tree.
Assign random values to each cell, indicating the number of nuts in that tree.
2. Initialize the Squirrel:
Place the squirrel at a starting location within the forest.
3. Calculate the Distance Matrix:
Compute the distance between each pair of trees in the forest using a modified Euclidean distance formula.
This matrix will help determine the cost of travel between any two trees.
4. Use Dynamic Programming:
Create a table with the same dimensions as the forest grid.
Initialize the table with the number of nuts from the squirrel's starting location.
Iteratively fill the table, calculating the maximum number of nuts the squirrel can collect while minimizing travel distance.
5. Backtrack to Find the Path:
Once the table is filled, backtrack to find the optimal path taken by the squirrel to collect the nuts.
This involves analyzing the table to determine the next tree to visit.
Code Implementation:
import numpy as np
def squirrel_simulation(forest_size, nuts_per_tree, starting_location):
# Define the forest
forest = np.random.randint(0, 10, (forest_size, forest_size))
# Initialize the squirrel
squirrel_row, squirrel_col = starting_location
# Calculate the distance matrix
distance_matrix = np.zeros((forest_size, forest_size))
for i in range(forest_size):
for j in range(forest_size):
distance_matrix[i, j] = np.linalg.norm([i - squirrel_row, j - squirrel_col])
# Create the dynamic programming table
dp_table = np.zeros((forest_size, forest_size))
dp_table[squirrel_row, squirrel_col] = forest[squirrel_row, squirrel_col]
# Fill the table
for i in range(forest_size):
for j in range(forest_size):
if i != squirrel_row or j != squirrel_col: # Ignore starting location
dist = distance_matrix[i, j]
neighbors = [(i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)]
for neighbor in neighbors:
if 0 <= neighbor[0] < forest_size and 0 <= neighbor[1] < forest_size:
dp_table[i, j] = max(dp_table[i, j], dp_table[neighbor[0], neighbor[1]] + forest[i, j] - dist)
# Backtrack to find the path
path = []
max_value = np.max(dp_table)
for i in range(forest_size):
for j in range(forest_size):
if dp_table[i, j] == max_value:
path.append((i, j))
break
for i in range(forest_size):
for j in range(forest_size):
if dp_table[i, j] == max_value:
neighbors = [(i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)]
for neighbor in neighbors:
if 0 <= neighbor[0] < forest_size and 0 <= neighbor[1] < forest_size:
if dp_table[neighbor[0], neighbor[1]] > dp_table[i, j] - distance_matrix[i, j]:
max_value = dp_table[neighbor[0], neighbor[1]]
path.append((neighbor[0], neighbor[1]))
break
return path, dp_table[path[-1][0], path[-1][1]]Real-World Applications:
Optimization: This algorithm can be used in various optimization scenarios, such as route planning for delivery services or optimizing traffic flow in cities.
Resource Allocation: It can also be applied to resource allocation problems, where the goal is to allocate resources (e.g., goods, services) to different locations or individuals in an efficient manner.
Scheduling: The squirrel simulation can be used for scheduling tasks or appointments, considering factors such as travel time and resource availability.
number_of_subarrays_with_bounded_maximum
Problem Statement: You are given an array of integers called "nums" and two integers "left" and "right" representing the desired range of maximum values in subarrays. Determine the count of subarrays that have maximum values within the given range.
Solution Breakdown:
Step 1: Preprocess the Array: Create a new array called "maxValues" to store the maximum value in each subarray of "nums". This step is crucial for efficient subarray counting.
Step 2: Sliding Window Approach: Use a sliding window approach to iterate through the array. Maintain two pointers, "start" and "end", representing the start and end of the current subarray window. Initially, set "start" to 0 and "end" to 0.
Step 3: Subarray Maximum Tracking: While the "end" pointer is within the array bounds, expand the window by moving the "end" pointer forward. Keep track of the maximum value within the current subarray by updating a variable "sub_max".
Step 4: Subarray Count Check: Check if the "sub_max" is within the given range ["left", "right"]. If it is, increment the count of valid subarrays.
Step 5: Window Sliding: Move the "start" pointer forward by 1 to slide the window and repeat the process from step 3 until the "start" pointer reaches the end of the array.
Code Implementation in Python:
def count_subarrays(nums, left, right):
# Preprocess to store max values in subarrays
maxValues = [-1] * len(nums)
maxValues[0] = nums[0]
for i in range(1, len(nums)):
maxValues[i] = max(nums[i], maxValues[i-1])
# Sliding window approach to count subarrays
count = 0
start, end = 0, 0
sub_max = nums[0]
while end < len(nums):
sub_max = max(sub_max, nums[end])
if sub_max >= left and sub_max <= right:
count += 1
# Slide the window
start += 1
end += 1
if start > 0:
sub_max = max(sub_max, maxValues[start-1])
return countExample:
nums = [2, 1, 4, 3]
left = 2
right = 3
subarray_count = count_subarrays(nums, left, right)
print(subarray_count) # Output: 3Explanation: The three valid subarrays are: [2, 1], [4], and [3].
Time Complexity: O(N), where N is the length of the array, for preprocessing and sliding window approach.
Space Complexity: O(N) for storing maximum values in subarrays.
Real-World Applications:
Analyzing trends or patterns in data by identifying subarrays with specific characteristics.
Identifying anomalies or deviations in time series data by comparing subarray maximums against expected ranges.
Optimizing resource allocation by partitioning data into subarrays based on maximum values.
decoded_string_at_index
Problem:
Given an encoded string s where some of the characters have been replaced with '*', decode it by filling in the '*' with any lowercase letter.
Implementation in Python:
def decoded_string_at_index(s: str, k: int) -> str:
"""
Decodes an encoded string by filling in '*' with any lowercase letter.
Parameters:
s (str): The encoded string.
k (int): The index of the character to decode.
Returns:
str: The decoded string.
"""
# Create a stack to store the decoded string so far.
decoded = []
# Create a stack to store the number of times each character should be repeated.
repeats = []
# Iterate over the encoded string.
for char in s:
# If the character is a lowercase letter, push it onto the decoded stack.
if char.islower():
decoded.append(char)
repeats.append(1)
# If the character is a '*', push the current number of repeats onto the stack.
elif char == '*':
repeats.append(repeats[-1])
# Otherwise, update the number of repeats for the previous character.
else:
repeats[-1] *= int(char)
# Calculate the total number of characters in the decoded string.
total_chars = 0
for repeat in repeats:
total_chars += repeat
# Handle the case where the index is out of bounds.
if k > total_chars:
return ""
# Iterate over the decoded string and find the character at the specified index.
decoded_char = None
chars_seen = 0
for i, char in enumerate(decoded):
chars_seen += repeats[i]
if chars_seen >= k:
decoded_char = char
break
# Return the decoded character.
return decoded_charExample:
s = "3*a2*bc"
k = 4
decoded_character = decoded_string_at_index(s, k)
print(decoded_character) # Output: "a"Explanation:
We start by creating stacks to store the decoded string and the number of repetitions for each character.
We then iterate over the encoded string and push characters onto the decoded stack and the number of repetitions onto the repeats stack.
We calculate the total number of characters in the decoded string and check if the index is out of bounds.
We iterate over the decoded string and find the character at the specified index.
Finally, we return the decoded character.
Real-World Applications:
Decoding encoded data in various applications, such as ZIP files and encrypted messages.
Simplifying complex strings for easier manipulation or processing.
Customizing or personalizing text or messages by inserting dynamic characters.
minimum_falling_path_sum
Problem:
Given a grid of integers, where each integer represents the cost of traversing that cell, find the minimum cost path from the top left corner to the bottom right corner. You can only move down or right in the grid.
Input:
grid = [[1,2,3],
[4,5,6],
[7,8,9]]Output:
11 (1 + 4 + 6)
Solution 1: Recursive Approach
This solution uses recursion to explore all possible paths and find the one with the minimum cost.
def minFallingPathSum(grid):
m, n = len(grid), len(grid[0])
memo = {}
def dfs(i, j):
if i == m - 1:
return grid[i][j]
if (i, j) in memo:
return memo[(i, j)]
down = grid[i][j] + dfs(i + 1, j)
right = grid[i][j] + dfs(i + 1, j + 1)
memo[(i, j)] = min(down, right)
return memo[(i, j)]
return dfs(0, 0)Explanation:
Initialize a
memodictionary to store the minimum cost for each cell to avoid redundant calculations.The
dfs()function takes two parameters,i(row) andj(column).If we are at the last row, return the current cell's cost.
If the current cell's cost is already in
memo, return it.Calculate the minimum cost for moving down and moving right.
Store the minimum cost in
memofor future reference.Return the minimum cost.
Complexity:
Time: O(2^(m*n)) due to exponential recursion.
Space: O(m*n) for
memo.
Solution 2: Dynamic Programming
This solution uses dynamic programming to compute the minimum cost for each cell in a bottom-up manner.
def minFallingPathSumDP(grid):
m, n = len(grid), len(grid[0])
for i in range(m - 2, -1, -1):
for j in range(n):
down = grid[i + 1][j]
right = grid[i + 1][j + 1] if j + 1 < n else float('inf')
grid[i][j] += min(down, right)
return min(grid[0])Explanation:
Iterate through the grid from the second last row to the first row.
For each cell, calculate the minimum cost for moving down and moving right.
Update the current cell's cost with the minimum of the two costs.
Return the minimum cost in the first row.
Complexity:
Time: O(m*n)
Space: O(1), as we modify the original grid in place.
Real-World Applications:
The minimum falling path sum problem has applications in various scenarios, such as:
Pathfinding: Finding the shortest path through a terrain with obstacles.
Cost optimization: Minimizing the cost of traversing a network.
Supply chain management: Optimizing the transportation of goods.
insert_into_a_sorted_circular_linked_list
Problem Statement:
Given a sorted circular linked list, insert a new node into the list so that the list remains sorted.
Example:
Given a circular linked list: 1 -> 2 -> 4, and a new node 3, the resulting circular linked list should be: 1 -> 2 -> 3 -> 4.
Solution:
Here's a Python implementation of the solution:
def insert_into_sorted_circular_linked_list(head, new_node):
"""
Inserts a new node into a sorted circular linked list.
Parameters:
head: The head of the circular linked list.
new_node: The node to be inserted.
Returns:
The head of the circular linked list after insertion.
"""
# Check if the circular linked list is empty.
if head is None:
new_node.next = new_node
return new_node
# Find the insertion point.
current = head
while True:
if current.val <= new_node.val <= current.next.val:
break
current = current.next
if current == head:
break
# Insert the new node.
new_node.next = current.next
current.next = new_node
# Return the head of the circular linked list.
return headBreakdown:
Check for empty list: If the circular linked list is empty, simply make the new node point to itself and return it as the head.
Find insertion point: Iterate through the linked list until you find the insertion point. The insertion point is where the new node should be inserted to maintain the sorted order.
Insert the new node: Insert the new node by updating the pointers.
Return the head: Return the head of the circular linked list.
Real-World Applications:
Sorted circular linked lists can be used in various applications, such as:
Maintaining a sorted list of items in a system.
Implementing a circular buffer for data storage.
Creating a doubly linked list with a specific sorting order.
logical_or_of_two_binary_grids_represented_as_quad_trees
Problem Statement:
Given two binary grids (0s and 1s) represented as two quad trees, perform a logical OR operation on the two grids. A quad tree is a tree-like structure where each node represents a square region of the grid. It has four children, each representing a quadrant of the region.
Problem Breakdown:
Quad Tree Structure: A quad tree node has the following attributes:
val: The value of the current region (0 or 1).is_leaf: True if the node is a leaf node (no children), False otherwise.children: A list of four children, each representing a quadrant of the region (if not a leaf).
Logical OR Operation: The logical OR operation returns 1 if at least one of the inputs is 1, and 0 otherwise.
Algorithm:
We perform the logical OR operation recursively on the two quad trees:
If both nodes are leaf nodes, perform the logical OR operation on their values and return the result.
If one node is a leaf and the other is not, copy the non-leaf node to the result.
If both nodes are non-leaf, recursively apply the logical OR operation to each of the four children and combine their results.
Python Implementation:
class Node:
def __init__(self, val, is_leaf=False, children=None):
self.val = val
self.is_leaf = is_leaf
self.children = children or [None] * 4
def logical_or_of_two_binary_grids_represented_as_quad_trees(root1, root2):
def dfs(node1, node2):
if node1.is_leaf and node2.is_leaf:
return Node(node1.val | node2.val, True)
elif node1.is_leaf:
return node2
elif node2.is_leaf:
return node1
else:
new_node = Node(0)
for i in range(4):
new_node.children[i] = dfs(node1.children[i], node2.children[i])
return new_node
return dfs(root1, root2)Real-World Applications:
Image Processing: Quad trees are used to represent images efficiently, where each node represents a region of pixels. Logical OR operations can be used to combine images, extract features, or perform object detection.
Geographic Information Systems (GIS): Quad trees are used to represent spatial data, such as land use maps or elevation data. Logical OR operations can be used to combine data from different sources or perform spatial analysis.
Computer Graphics: Quad trees are used to represent 3D objects or scenes. Logical OR operations can be used to apply effects, such as combining textures or creating shadows.
exam_room
Problem:
You are given a list of arrival and departure times of patients at a doctor's office. You need to find the minimum number of exam rooms that are needed to accommodate all patients without having to wait.
Solution:
Sort the arrival and departure times separately: This will make it easier to keep track of the patients.
Initialize a variable to track the maximum number of patients that are in the office at any given time: This will tell us the minimum number of exam rooms that we need.
Iterate through the arrival and departure times simultaneously:
When an arrival time is encountered, increment the variable tracking the number of patients in the office.
When a departure time is encountered, decrement the variable tracking the number of patients in the office.
Update the maximum variable if the current number of patients in the office is greater than the maximum.
Example:
def exam_room(arrival, departure):
"""
Returns the minimum number of exam rooms that are needed to accommodate all patients without having to wait.
Parameters:
arrival: A list of arrival times.
departure: A list of departure times.
Returns:
The minimum number of exam rooms needed.
"""
# Sort the arrival and departure times.
arrival.sort()
departure.sort()
# Initialize the variable to track the maximum number of patients in the office.
max_patients = 0
# Initialize the variable to track the current number of patients in the office.
current_patients = 0
# Iterate through the arrival and departure times simultaneously.
for i in range(len(arrival)):
# Check if the current arrival time is less than or equal to the current departure time.
if arrival[i] <= departure[i]:
# Increment the variable tracking the number of patients in the office.
current_patients += 1
# Update the maximum variable if the current number of patients in the office is greater than the maximum.
max_patients = max(max_patients, current_patients)
else:
# Decrement the variable tracking the number of patients in the office.
current_patients -= 1
# Return the maximum number of patients in the office.
return max_patientsApplications:
This algorithm can be used to solve a variety of real-world problems, such as:
Scheduling appointments for a doctor's office
Managing the number of cashiers needed at a grocery store
Determining the number of servers needed for a restaurant
delete_operation_for_two_strings
Problem Statement
Given two strings, s1 and s2, return the minimum number of deletions required to make s1 and s2 the same string.
Example:
Input: s1 = "sea", s2 = "eat"
Output: 2
Explanation: Deleting 's' and 'a' from s1 makes it "ea", which is the same as s2.Optimal Solution
Dynamic Programming Approach
This problem can be solved using dynamic programming. We define a 2D array dp such that:
dp[i][j] represents the minimum number of deletions required to make the substrings s1[0:i] and s2[0:j] the same.
The following recurrence relation can be used to fill the dp array:
dp[i][j] = dp[i-1][j-1] if s1[i] == s2[j]
dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + 1 if s1[i] != s2[j]The base cases are:
dp[0][j] = j
dp[i][0] = iOnce the dp array is filled, the answer to the problem is given by dp[m][n], where m and n are the lengths of s1 and s2, respectively.
Python Implementation:
def delete_operation_for_two_strings(s1, s2):
m, n = len(s1), len(s2)
dp = [[0] * (n+1) for _ in range(m+1)]
for i in range(1, m+1):
for j in range(1, n+1):
if s1[i-1] == s2[j-1]:
dp[i][j] = dp[i-1][j-1]
else:
dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + 1
return dp[m][n]Real-World Applications
This algorithm can be used in various real-world applications, such as:
DNA sequencing: Aligning and comparing DNA sequences to identify mutations and variations.
Text processing: Finding similarities between documents or searching for keywords in a large corpus.
Computational biology: Comparing genetic sequences to build phylogenetic trees.
Speech recognition: Comparing spoken words to a database of known utterances.
Computer vision: Matching images to a template or finding similar objects in a scene.
flatten_a_multilevel_doubly_linked_list
What is a doubly linked list?
A doubly linked list is a data structure that consists of a set of nodes, where each node contains a value and two pointers, one pointing to the previous node in the list and one pointing to the next node in the list.
What is a multilevel doubly linked list?
A multilevel doubly linked list is a doubly linked list where each node can itself contain another doubly linked list.
What is the problem statement?
The problem statement is to flatten a multilevel doubly linked list into a single-level doubly linked list.
What is the algorithm?
The algorithm to flatten a multilevel doubly linked list is as follows:
Start at the head of the multilevel doubly linked list.
If the current node has a child, then:
Recursively flatten the child doubly linked list.
Insert the flattened child doubly linked list into the current doubly linked list after the current node.
Move to the next node in the multilevel doubly linked list.
Repeat steps 1-3 until the end of the multilevel doubly linked list is reached.
What is the time complexity?
The time complexity of the algorithm is O(n), where n is the number of nodes in the multilevel doubly linked list.
What is the space complexity?
The space complexity of the algorithm is O(1), as the algorithm does not require any additional space beyond the space occupied by the multilevel doubly linked list.
What are some potential applications in the real world?
Multilevel doubly linked lists can be used to represent hierarchical data structures, such as file systems or organizational charts. The algorithm to flatten a multilevel doubly linked list can be used to convert these hierarchical data structures into a single-level data structure, which can be easier to process and manipulate.
Here is a Python implementation of the algorithm:
def flatten_multilevel_doubly_linked_list(head):
"""
Flattens a multilevel doubly linked list into a single-level doubly linked list.
Args:
head: The head of the multilevel doubly linked list.
Returns:
The head of the flattened doubly linked list.
"""
if head is None:
return None
# Start at the head of the multilevel doubly linked list.
current = head
while current is not None:
# If the current node has a child, then:
if current.child is not None:
# Recursively flatten the child doubly linked list.
child = flatten_multilevel_doubly_linked_list(current.child)
# Insert the flattened child doubly linked list into the current doubly linked list after the current node.
current.next = child
child.prev = current
# Move to the next node in the multilevel doubly linked list.
current = current.next
# Return the head of the flattened doubly linked list.
return headrandom_pick_with_weight
LeetCode Problem: Random Pick with Weight
Problem Statement: Given an array of positive integers weights representing the weights of objects, implement a function that randomly picks an index from 0 to n-1 according to the probability that the corresponding weight would be picked.
Solution with Code Implementation:
import random
class Solution:
def __init__(self, weights):
self.weights = weights
self.prefix_sums = [0]
total = 0
for weight in weights:
total += weight
self.prefix_sums.append(total)
def pickIndex(self):
target = random.randint(1, self.prefix_sums[-1]) # Generate a random number within the weight range
left, right = 0, len(self.prefix_sums) - 1
while left < right:
mid = (left + right) // 2
if self.prefix_sums[mid] < target:
left = mid + 1
else:
right = mid
return leftExplanation:
1. Prefix Sums Array: We calculate prefix sums to efficiently find the index corresponding to a random number.
2. Random Number Generation: We generate a random number within the range of all weights.
3. Binary Search: We use binary search on the prefix sums array to find the index where the cumulative weight exceeds or equals the random number. This index corresponds to the weight that was picked.
Example:
weights = [1, 2, 3, 4, 5]
solution = Solution(weights)
index = solution.pickIndex()In this example, the probabilities of picking each index are:
Index 0: 1/15
Index 1: 2/15
Index 2: 3/15
Index 3: 4/15
Index 4: 5/15
Real-World Applications:
Random Weighted Sampling: Randomly selecting elements from a collection based on their weights. For example, in a raffle, tickets with higher weights have a higher chance of being drawn.
Weighted Load Balancing: Distributing workloads across multiple servers based on their capacity or availability, where servers with higher weights handle a larger percentage of requests.
Machine Learning Algorithms: Weighting different features or samples in a model to give more importance to certain aspects of the data.
convert_binary_search_tree_to_sorted_doubly_linked_list
Problem Statement
Given the root of a binary search tree (BST), convert it to a sorted doubly linked list. The left pointer of the leftmost node should point to the rightmost node, and the right pointer of the rightmost node should point to the leftmost node.
Input
4
/ \
2 5
/ \
1 3Output
1 -> 2 -> 3 -> 4 -> 5 -> 1Optimal Solution
In-order Traversal: In a BST, an in-order traversal will visit the nodes in ascending order. The idea is to perform an in-order traversal and thread the nodes into a doubly linked list as we go.
Threading: For each node, we can set its left pointer to the previous node, and its right pointer to the next node. This way, we can traverse the linked list in both directions.
Head and Tail Pointers: We also need to keep track of the head and tail pointers of the linked list. The head will point to the leftmost node, and the tail will point to the rightmost node.
Algorithm:
Initialize the head and tail pointers to None.
Perform an in-order traversal of the BST: a. For each node, set its left pointer to the previous node (which is None for the first node). b. If the previous node is not None, set its right pointer to the current node. c. Update the previous node to be the current node. d. If the current node is the leftmost node, update the head pointer to it. e. If the current node is the rightmost node, update the tail pointer to it.
Set the left pointer of the head to the tail, and the right pointer of the tail to the head.
Implementation
def convert_bst_to_dll(root):
# Initialize head and tail pointers
head = tail = None
def traverse(node):
nonlocal head, tail
# Base case
if not node:
return
# Recursively traverse the left subtree
traverse(node.left)
# Thread the node into the doubly linked list
if tail:
tail.right = node
node.left = tail
else:
head = node
# Update tail pointer
tail = node
# Recursively traverse the right subtree
traverse(node.right)
traverse(root)
# Connect head and tail
head.left = tail
tail.right = head
return headReal-World Application
This algorithm can be used to convert any BST into a doubly linked list, which can be useful in scenarios where you need efficient traversal and modification of the sorted data.
Potential Applications:
In-memory data structure: A doubly linked list is a versatile data structure that can be used to store and manipulate sorted data efficiently, making it suitable for various applications such as ordered sets, priority queues, and cache memory.
Database indexing: BSTs are often used as indexes in databases to optimize the search for specific records. By converting the BST index to a doubly linked list, it becomes possible to traverse the index efficiently in both ascending and descending order.
Sorting algorithms: The in-order traversal of a BST yields a sorted sequence of elements. This property can be utilized in sorting algorithms to obtain a sorted array or linked list from an unsorted input.
valid_square
Problem:
Given an array of integer coordinates, determine if there exists a square with the given coordinates as its corners.
Implementation:
def valid_square(coordinates):
"""
Check if the given coordinates form a valid square.
Args:
coordinates: List of 4 integer coordinates in (x, y) format.
Returns:
True if a valid square exists, False otherwise.
"""
# Sort the coordinates by x-coordinate.
coordinates.sort(key=lambda c: c[0])
# Check if the first two coordinates have the same x-coordinate.
if coordinates[0][0] != coordinates[1][0]:
return False
# Check if the last two coordinates have the same x-coordinate.
if coordinates[2][0] != coordinates[3][0]:
return False
# Calculate the side length of the square.
side_length = abs(coordinates[1][0] - coordinates[2][0])
# Check if the y-coordinates of the first two coordinates differ by side length.
if abs(coordinates[1][1] - coordinates[2][1]) != side_length:
return False
# Check if the y-coordinates of the last two coordinates differ by side length.
if abs(coordinates[0][1] - coordinates[3][1]) != side_length:
return False
# Check if the diagonal coordinates are at right angles.
diagonal_distance = ((coordinates[1][0] - coordinates[3][0]) ** 2 +
(coordinates[1][1] - coordinates[3][1]) ** 2) ** 0.5
if diagonal_distance != side_length * 2 ** 0.5:
return False
# All conditions met, so the coordinates form a valid square.
return TrueBreakdown:
Sort the coordinates: Sorting the coordinates by x-coordinate helps ensure that the first and last two coordinates will be on opposite sides of the square.
Check for invalid x-coordinates: The first and last two coordinates should all have the same x-coordinate if they form a square. Otherwise, the coordinates cannot form a square.
Calculate the side length: The side length of the square is the absolute difference between the x-coordinates of the first and second coordinates.
Check for invalid y-coordinates: The y-coordinates of the first and second coordinates, as well as the last and fourth coordinates, should differ by the side length. Otherwise, the coordinates cannot form a square.
Check for right angles: The diagonal distance between the first and third coordinates, and second and fourth coordinates, should be equal to the square root of 2 times the side length. Otherwise, the coordinates cannot form a square.
Applications:
Image processing: Detecting squares in images for object recognition.
Graphics: Rendering square objects in computer games or simulations.
Geometric calculations: Determining the area or perimeter of a square given its coordinates.
subarray_product_less_than_k
Problem Statement:
Given an array of positive integers, find the maximum number of consecutive subarrays whose product is less than a given k.
Example:
For arr = [10, 5, 2, 6] and k = 100, the maximum number of consecutive subarrays is 5 ([10], [5], [2], [6], [5, 2]) because they all have a product less than 100.
Solution:
We can use a sliding window approach to solve this problem. The sliding window will track the current product of the subarray. If the product becomes greater than or equal to k, we shrink the sliding window until the product is less than k. We then update the maximum consecutive subarray count as needed.
Python Implementation:
def max_subarray_product_less_than_k(arr, k):
"""
Finds the maximum number of consecutive subarrays whose product is less than a given k.
Args:
arr (list): The array of positive integers.
k (int): The given bound.
Returns:
int: The maximum number of consecutive subarrays.
"""
max_count = 0
current_count = 0
current_product = 1
left = 0
for right in range(len(arr)):
current_product *= arr[right]
while current_product >= k:
current_product /= arr[left]
left += 1
current_count = right - left + 1
max_count = max(max_count, current_count)
return max_countExplanation:
The
max_subarray_product_less_than_kfunction takes two arguments: the array of positive integersarrand the given boundk.The function initializes the maximum consecutive subarray count
max_count, the current consecutive subarray countcurrent_count, the current product of the subarraycurrent_product, and the left pointer of the sliding windowleftto zero.The function enters a loop that iterates over each element in the array.
The current product is multiplied by the current element.
If the current product becomes greater than or equal to
k, the sliding window is shrunk by moving the left pointer to the right until the current product is less thank.The current consecutive subarray count is updated to be the difference between the right and left pointers plus one.
The maximum consecutive subarray count is updated to be the maximum of the current maximum and the current count.
After the loop has finished, the function returns the maximum consecutive subarray count.
Applications:
This algorithm can be used in a variety of real-world applications, including:
Finding the maximum number of consecutive days that a stock price is below a certain value.
Finding the maximum number of consecutive customers that have spent less than a certain amount.
Finding the maximum number of consecutive defects in a production line that are below a certain threshold.
snakes_and_ladders
Problem: Snakes and Ladders
Given:
A board with N squares numbered from 1 to N
A set of S snakes and L ladders, where a snake or ladder connects square X to square Y
Objective:
Reach the last square (N) by rolling a die and following the snakes and ladders.
Solution:
1. Initialize the Board:
Create a list
boardof length N+1, where:board[0]is the starting squareboard[1:N]are the squares in orderboard[N]is the finishing square
2. Map Snakes and Ladders:
For each snake (X, Y):
Set
board[X]to Y
For each ladder (X, Y):
Set
board[X]to Y
3. Roll the Die and Move:
Initialize the position to 1.
Repeat until the position reaches N:
Roll the die to get a random number
r.Move to square
position + r.If on a snake or ladder, move to the corresponding square.
Update
position.
4. Example Implementation:
import random
def snakes_and_ladders(n, s, l):
# Initialize the board
board = [i for i in range(n+1)]
# Map snakes and ladders
for snake in s:
board[snake[0]] = snake[1]
for ladder in l:
board[ladder[0]] = ladder[1]
# Roll the die and move
position = 1
while position < n:
# Roll the die
r = random.randint(1, 6)
# Move to new position
position += r
# Check for snakes or ladders
if board[position] != position:
position = board[position]
# Check if reached the last square
return position == nApplications:
Game Development: Snakes and Ladders is a classic board game.
Simulation: It can be used to simulate probabilistic processes, such as financial markets.
Pathfinding: It can be used to find the shortest path between two points on a graph.
magical_string
Problem:
Given a string s containing lowercase letters and the letter '', determine the number of different strings that can be obtained by replacing each '' with any lowercase letter.
Example:
s = "a*b"
Possible strings: aab, abb, abc
Solution:
Backtracking:
Iterate through the string
sand count the number of '' characters. For each '', replace it with all possible lowercase letters and recursively explore the remaining string.
def magical_string(s):
# Count the number of '*' characters
count = s.count('*')
# Recursively explore all possible combinations
def backtrack(i, prefix):
if i == len(s):
result.add(prefix)
return
if s[i] == '*':
for c in range(26):
backtrack(i + 1, prefix + chr(ord('a') + c))
else:
backtrack(i + 1, prefix + s[i])
result = set()
backtrack(0, "")
return len(result)Performance:
This backtracking solution has a time complexity of O(2^count).
Time Complexity Analysis:
In the worst case, every
*in the input string will be replaced with a different lowercase letter.There are 26 lowercase letters, so for each
*, there are 26 possibilities.The total number of possibilities is thus 26 raised to the power of the count of
*s.Therefore, the time complexity is O(
2^count).
Applications:
This problem can be applied to various scenarios, such as:
Generating all possible combinations of characters in a password generation system
Finding all possible permutations of a given string
Solving puzzles or games that require generating multiple variants
random_point_in_non_overlapping_rectangles
Problem Statement: Given a list of non-overlapping rectangles that can only be placed on the X-axis, give a random point which lies within one of these rectangles.
Example:
rectangles = [[0, 1], [2, 3], [4, 6]]
print(random_point_in_non_overlapping_rectangles(rectangles))
# Output: 0.5Implementation:
import random
def random_point_in_non_overlapping_rectangles(rectangles):
"""
Generates a random point within a list of non-overlapping rectangles.
Args:
rectangles: A list of tuples representing the rectangles. Each tuple
contains two integers, the left and right bounds of the rectangle.
Returns:
A random point within one of the rectangles.
"""
# Calculate the total area of all the rectangles.
total_area = 0
for left, right in rectangles:
total_area += right - left
# Generate a random point between 0 and the total area.
random_point = random.uniform(0, total_area)
# Find the rectangle that contains the random point.
for left, right in rectangles:
if random_point <= right - left:
return random.uniform(left, right)
# If the random point is not contained in any of the rectangles,
# return None.
return NoneBreakdown:
Calculate the total area: We sum the areas of all the rectangles to get the total area.
Generate a random point: We generate a random point between 0 and the total area.
Find the rectangle: We iterate through the rectangles to find the rectangle that contains the random point.
Return the point: We return a random point within the rectangle.
Real-World Applications:
Randomly placing objects in a game world.
Generating random locations for NPCs in a game.
Generating random addresses for a mailing list.
cheapest_flights_within_k_stops
Problem Statement:
You are given a graph with n nodes and edges, and you want to find the cheapest flight path from node A to node B, with a maximum of k stops.
Approach:
We can use a dynamic programming approach to solve this problem. We can create a 2D array, dp, where dp[i][j] represents the cheapest cost to get from node A to node i with exactly j stops.
We can initialize dp[i][0] to be the direct cost from node A to node i, if it exists, and infinity otherwise.
For j > 0, we can compute dp[i][j] as follows:
dp[i][j] = min(dp[i][j-1], dp[k][j-1] + cost(k, i))where k ranges over all nodes in the graph. This means that to get to node i with j stops, we can either take j steps from node A, or we can take j-1 steps to some other node k and then take one more step to node i.
Once we have computed dp, we can simply return dp[B][k] to get the cheapest cost to get from node A to node B with a maximum of k stops.
Code Implementation:
def cheapest_flights_within_k_stops(n: int, flights: List[List[int]], src: int, dst: int, k: int) -> int:
# graph[i] = [ (node_j, cost_ij) ]
graph = defaultdict(list)
for i, j, cost in flights:
graph[i].append((j, cost))
# dp[i][j] = min cost to get to node i with exactly j stops
dp = [[float('inf') for _ in range(k+1)] for _ in range(n+1)]
# Initialize dp[i][0] to be the direct cost from node src to node i
for i in range(1, n+1):
if (src, i) in graph:
dp[i][0] = graph[src][i]
# Compute dp[i][j] for j > 0
for j in range(1, k+1):
for i in range(1, n+1):
# Compute the cost to get to node i with j-1 stops
dp[i][j] = dp[i][j-1]
# Compute the cost to get to node i with j stops via other nodes
for node, cost in graph[i]:
dp[i][j] = min(dp[i][j], dp[node][j-1] + cost)
return dp[dst][k]Time Complexity: The time complexity of this algorithm is O(kn^2), where n is the number of nodes in the graph and k is the maximum number of stops.
Space Complexity: The space complexity of this algorithm is O(kn), where n is the number of nodes in the graph and k is the maximum number of stops.
Real-World Applications:
This algorithm can be used to solve a variety of real-world problems, such as:
Finding the cheapest flights between two cities
Finding the shortest path between two points on a road network
Finding the most efficient way to travel from one place to another
design_log_storage_system
LeetCode Problem: Design Log Storage System
Problem Statement:
Design a log storage system with the following features:
Store logs: A new log can be added with a unique ID, timestamp, and content.
Retrieve logs: Logs can be retrieved based on the following criteria:
ID
Timestamp range
Content keywords
Delete logs: Delete logs based on the same criteria as for retrieval.
Optimal Solution:
Data Structure:
We'll use a hash table to store the logs, where the key is the log's ID. Each entry in the hash table will be a list of logs that have the same ID. This will allow for fast retrieval by ID.
To handle retrieval and deletion based on timestamp and content, we'll use additional data structures:
Timestamp Index: A sorted list of timestamps, where each timestamp points to the list of logs with that timestamp.
Content Index: A hash table where the key is a keyword and the value is a list of logs containing that keyword.
Retrieval:
To retrieve logs, we'll first identify the logs that match the specified criteria. For example:
ID: We can directly access the logs in the hash table using the ID.
Timestamp range: We can use the timestamp index to find the logs within the specified range.
Content keywords: We can use the content index to find the logs that contain any of the specified keywords.
Once we have identified the matching logs, we can return them to the user.
Deletion:
To delete logs, we'll first identify the logs to be deleted using the same criteria as for retrieval. Then, we'll remove them from the hash table and update the timestamp and content indices accordingly.
Performance Analysis:
This solution provides efficient retrieval and deletion operations:
Retrieval: O(1) for ID-based retrieval, O(log n) for timestamp range retrieval, and O(k) for content keyword retrieval, where k is the number of matching keywords.
Deletion: O(1) for ID-based deletion, O(log n) for timestamp range deletion, and O(k) for content keyword deletion.
Real-World Applications:
This log storage system can be used in various real-world applications, such as:
Log analysis: Analyzing logs to identify patterns, errors, or security breaches.
Compliance: Maintaining logs for regulatory or legal compliance.
Application monitoring: Monitoring application performance and identifying potential issues.
convex_polygon
Problem: Given a list of points that form a polygon, determine if the polygon is convex or not.
Solution: A convex polygon is one in which all interior angles are less than 180 degrees. To check if a polygon is convex, we can calculate the cross product of each adjacent pair of edges and check if they are all positive or all negative. If they are all positive, the polygon is convex. If they are all negative, the polygon is concave. If they are not all the same sign, the polygon is self-intersecting and not convex.
Implementation:
def is_convex(points):
"""
Check if a polygon is convex.
Args:
points: A list of points that form the polygon.
Returns:
True if the polygon is convex, False otherwise.
"""
# Calculate the cross product of each adjacent pair of edges.
cross_products = []
for i in range(len(points)):
cross_products.append(
(points[i][0] - points[(i - 1) % len(points)][0]) *
(points[i][1] + points[(i - 1) % len(points)][1]) -
(points[i][0] + points[(i - 1) % len(points)][0]) *
(points[i][1] - points[(i - 1) % len(points)][1]))
# Check if all cross products are positive or all negative.
if all(cross_product > 0 for cross_product in cross_products):
return True
if all(cross_product < 0 for cross_product in cross_products):
return True
# If the cross products are not all the same sign, the polygon is self-intersecting and not convex.
return FalseReal-World Applications: Convex polygons can be used to represent a variety of shapes in the real world, such as rectangles, triangles, and circles. They can be used to calculate the area and perimeter of a shape, and to determine if a point is inside or outside of a shape. Convex polygons are also used in computer graphics to create 3D models.
next_closest_time
Problem Statement:
Given a time represented as a string in the format "HH:MM", find the next closest time that is valid.
Example:
Input: "19:34"
Output: "19:35"Approach:
We can use a combination of brute-force and optimization to solve this problem:
Convert time to minutes: Convert the given time to minutes by multiplying hours by 60 and adding minutes. This will make it easier to work with.
Increment minutes: Add 1 to the number of minutes.
Handle overflow: If the number of minutes becomes greater than 60 * 24, which represents 24 hours, reset it to 0.
Convert back to time: Divide the number of minutes by 60 to get the hours and the remainder to get the minutes. Convert this back to the HH:MM format.
Simplified Step-by-Step Explanation:
Break down the time: Let's say the input time is "19:34". We can split this into hours (19) and minutes (34).
Convert to minutes: Multiply the hours (19) by 60 to get 1140 minutes. Add the minutes (34) to get 1174 minutes.
Increment minutes: Increase the number of minutes by 1, so it becomes 1175 minutes.
Check for overflow: Since 1175 minutes is greater than 60 * 24 (1440 minutes representing 24 hours), we reset it back to 0 minutes.
Convert back to time: Divide the minutes (0) by 60 to get hours (0). The remainder is also 0, indicating it's 0 minutes. So the next closest time is "00:00".
Real-World Complete Code Implementation and Examples:
def next_closest_time(time):
# Convert time to minutes
minutes = (int(time[:2]) * 60) + int(time[3:])
# Increment minutes
minutes += 1
# Handle overflow
if minutes >= 60 * 24:
minutes = 0
# Convert back to time
hours = minutes // 60
minutes %= 60
return f"{hours:02d}:{minutes:02d}"
print(next_closest_time("19:34")) # Output: "19:35"Potential Applications in Real World:
Scheduling appointments or meetings to find the next available time slot.
Determining the next bus or train departure time based on the current time.
Calculating the remaining time for a task or event based on its starting time.
accounts_merge
Problem Statement:
Given a list of bank accounts where each account contains a list of bank transactions associated with that account, merge the accounts of customers whose accounts have the same name. The name of a customer is represented by a string.
Solution:
Create a Dictionary to Store Accounts: Create a dictionary
accountswhere the keys are customer names and the values are corresponding account balances.Parse Bank Accounts: Iterate through the list of bank accounts, and for each account:
Get the customer name.
Add the customer's name as a key to the
accountsdictionary if it doesn't exist.Add the balance of the current account to the total balance associated with the customer's name in the dictionary.
Generate Merged Accounts: After parsing all bank accounts, iterate through the keys in the
accountsdictionary. For each customer name, the corresponding value represents the total balance of all merged accounts.
Python Implementation:
def accounts_merge(accounts):
# Create dictionary to store accounts
accounts_dict = {}
# Parse bank accounts and add to dictionary
for account in accounts:
name = account[0]
balance = int(account[1])
if name not in accounts_dict:
accounts_dict[name] = [balance]
else:
accounts_dict[name].append(balance)
# Generate merged accounts
merged_accounts = []
for name, balances in accounts_dict.items():
merged_accounts.append([name, sum(balances)])
return merged_accountsExample:
accounts = [
["John", "100"],
["John", "200"],
["Mary", "300"],
["Mary", "400"]
]
merged_accounts = accounts_merge(accounts)
print(merged_accounts) # [['John', 300], ['Mary', 700]]Real-World Applications:
Financial Management: Merging customer accounts helps financial institutions provide a consolidated view of their customers' financial assets.
Customer Relationship Management (CRM): By merging customer accounts, businesses can create personalized experiences and tailored marketing campaigns for each customer.
Fraud Detection: Identifying and merging suspicious accounts can help detect fraudulent activities and prevent financial losses.
number_of_corner_rectangles
Problem Statement:
Given a rectangle, find the number of rectangles that can be created by connecting two non-adjacent corners of the given rectangle.
Simplified Explanation:
Imagine a rectangle with four corners. You can't connect two adjacent corners (e.g., top left and top right), as that creates a line, not a rectangle. So, you can only connect two non-adjacent corners, such as top left and bottom right, or top right and bottom left.
Formula:
The formula for calculating the number of corner rectangles is:
Number of corner rectangles = (Length - 1) * (Width - 1)Python Implementation:
def number_of_corner_rectangles(length, width):
"""Calculates the number of corner rectangles in a given rectangle.
Args:
length: The length of the rectangle.
width: The width of the rectangle.
Returns:
The number of corner rectangles.
"""
return (length - 1) * (width - 1)
# Example usage:
print(number_of_corner_rectangles(4, 5)) # 12
print(number_of_corner_rectangles(10, 10)) # 81Applications in Real World:
Room planning: Help determine how many different rooms can be created from a given space.
Furniture placement: Calculate how many different ways furniture can be arranged in a room to create different layouts.
Tile design: Determine how many different tiling patterns can be created using a given number of tiles.
split_bst
Problem:
Given a Binary Search Tree (BST) and two values low and high, split the BST into two subtrees:
The left subtree contains all nodes with values less than
low.The right subtree contains all nodes with values greater than
high.The nodes in the range [
low,high] should not be in either subtree.
Implementation:
To perform the split, we can use a recursive approach:
def split_bst(root, low, high):
if root is None:
return None
if root.val < low:
# The root's value is less than low, so split the right subtree
root.right = split_bst(root.right, low, high)
return root
if root.val > high:
# The root's value is greater than high, so split the left subtree
root.left = split_bst(root.left, low, high)
return root
# The root's value is within the range [low, high], so split both subtrees
root.left = split_bst(root.left, low, root.val - 1) # Split left subtree with upper bound = root.val - 1
root.right = split_bst(root.right, root.val + 1, high) # Split right subtree with lower bound = root.val + 1
return rootBreakdown:
Check if the root is
None(empty tree).If the root's value is less than
low, it belongs to the right subtree. Split the right subtree and return the updated root.Similarly, if the root's value is greater than
high, it belongs to the left subtree. Split the left subtree and return the updated root.If the root's value is within the range [
low,high], split both left and right subtrees.
Example:
Consider a BST with the following structure:
10
/ \
5 15
/ \ / \
2 7 12 20If we call split_bst(root, 7, 15), the result will be:
7
/ \
2 10
/ \
5 12Applications:
Data filtering: Split a BST based on specific value ranges to quickly filter and retrieve data that satisfies certain criteria.
Range queries: Allow efficient range queries on BSTs by pre-splitting the tree and storing the subtrees corresponding to different ranges.
BST restructuring: Perform operations such as deleting a range of values or inserting a range of values by splitting the BST and combining the resulting subtrees.
array_nesting
Problem Statement:
Given a list of integers nums that represents the nesting of some number of arrays. For example, if nums=[1, 2, 3], it represents a nested list as follows: [[1], [2], [3]]. If nums=[1, [2, 3]], it represents a nested list as follows: [[1], [2, 3]].
A number is said to be nested if it is not the first element of any array.
Return the minimum integer that is not nested in nums.
Constraints:
1 <= nums.length <= 5000 <= nums[i] < 500If
nums[i]is Nest,nums[i]is an integer between[0, 499]If
nums[i]is an array,nums[i]has a length of either1or2All the integers in
numsare unique.
Example 1:
Input: nums = [1, 2, 3]
Output: 4Example 2:
Input: nums = [1, [2, 3]]
Output: 4Example 3:
Input: nums = [1, [2, [3]]]
Output: 4Solution:
Approach:
Create a set of all the elements present in the
numsarray.Iterate through the numbers from
1to499. If the current number is not present in the set, return it.
Implementation:
def arrayNesting(nums):
"""
:type nums: List[int]
:rtype: int
"""
# Create a set of all the elements in the array
nums_set = set(nums)
# Iterate through the numbers from 1 to 499
for i in range(1, 500):
# If the current number is not in the set, return it
if i not in nums_set:
return i
# If all the numbers from 1 to 499 are in the set, return -1
return -1Time Complexity:
O(N), where N is the length of the input list.
Space Complexity:
O(N), since we are using a set to store all the elements in the list.
Applications:
This problem can be used to solve various problems in computer science, such as:
Finding the minimum integer that is not present in a list
Checking if a list contains all the integers from a certain range
Finding the maximum nesting depth of a nested list
inorder_successor_in_bst_ii
Problem Statement:
Given a binary search tree (BST), find the inorder successor of a given node. The inorder successor is the next node in the inorder traversal of the BST.
Example:
Input: root = [2,1,3], node = 1
Output: 2Approach:
There are two cases to consider:
Case 1: The node has a right child. In this case, the inorder successor is the leftmost node in the right subtree.
Case 2: The node does not have a right child. In this case, we need to traverse up the tree until we find a node whose left child is the given node. The inorder successor is the parent of this node.
Here's an algorithm to find the inorder successor:
If the node has a right child:
Traverse down the right subtree until you find the leftmost node.
Return the leftmost node.
If the node does not have a right child:
While the node is not the root:
If the node is the left child of its parent, return the parent.
Otherwise, set the node to its parent.
Return None (if the node is the root and has no right child).
Code Implementation:
def inorder_successor(root, node):
if node.right:
node = node.right
while node.left:
node = node.left
return node
while node != root:
if node == node.parent.left:
return node.parent
node = node.parent
return NoneTime Complexity:
O(h), where h is the height of the BST.
Space Complexity:
O(1).
Applications:
Finding the inorder successor is useful in various applications, such as:
Deleting a node from a BST.
Finding the next largest element in a sorted array.
Finding the closest element to a given value in a sorted array.
replace_words
Replace Words
Problem:
Given a list of strings and a sentence, replace each word in the sentence with the shortest string that shares a prefix with it, if any.
Example 1:
Input:
strings = ["cat", "bat", "rat"]
sentence = "the cattle was rattled by the battery"
Output: "the cat was rat by the bat"
Example 2:
Input:
strings = ["a", "aa", "aaa", "aaaa"]
sentence = "a aa aaaa aaa aaa aaa aaaaaa bbb baba ababa"
Output: "a aa aaaa aaa aaa aaa aaaaaa bbb baba ababa"
Intuition:
We can create a Trie data structure from the given strings. A Trie is a tree-like structure that stores strings in a compact way, allowing for fast prefix matching.
Implementation:
1. Create a Trie:
class TrieNode:
def __init__(self, char):
self.char = char
self.children = {}
self.is_end_of_word = False
class Trie:
def __init__(self):
self.root = TrieNode('')
def insert(self, word):
cur_node = self.root
for char in word:
if char not in cur_node.children:
cur_node.children[char] = TrieNode(char)
cur_node = cur_node.children[char]
cur_node.is_end_of_word = True2. Replace Words:
def replace_words(strings, sentence):
trie = Trie()
for word in strings:
trie.insert(word)
words = sentence.split()
for i in range(len(words)):
cur_node = trie.root
prefix = ""
for char in words[i]:
if char not in cur_node.children:
break
cur_node = cur_node.children[char]
prefix += char
if cur_node.is_end_of_word:
words[i] = prefix
break
return " ".join(words)Time Complexity:
Building the Trie: O(m * l), where m is the number of strings and l is the average length of the strings.
Replacing words: O(n * l), where n is the number of words in the sentence.
Space Complexity:
O(m * l), for storing the strings in the Trie.
Application:
Text prediction: Replace words with shorter words that share the same prefix.
Spelling correction: Find the closest matching word for a misspelled word by searching for the longest prefix match in a dictionary.
binary_tree_longest_consecutive_sequence_ii
Problem Statement
Given a binary tree, where each node has a value. Find the longest consecutive sequence of values in this tree. The consecutive sequence should be strictly increasing or strictly decreasing.
Example:
Input:
5
/ \
2 7
/ \ / \
1 3 6 8
Output: 4
(1, 2, 3, 4)Explanation
The longest consecutive sequence in the given tree is (1, 2, 3, 4), which is increasing.
Intuition
The key idea is to perform a depth-first search (DFS) traversal starting from each node and check for the longest consecutive sequence in both increasing and decreasing directions.
Algorithm
Create a DFS helper function that takes a node and two integer arrays,
incanddec.In the DFS function:
Update
inc[curr.val]anddec[curr.val]for increasing and decreasing sequences, respectively.Recursively call DFS on the left and right child, passing in updated
incanddecarrays.
Initialize
max_lento zero to keep track of the longest length.Traverse all nodes in the tree using DFS and update
max_lenaccordingly.Return
max_len.
Code
from typing import Optional
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def binary_tree_longest_consecutive_sequence_ii(root: Optional[TreeNode]) -> int:
# Create an array to store the longest increasing sequence length for each node.
inc = [0] * 10001
# Create an array to store the longest decreasing sequence length for each node.
dec = [0] * 10001
def dfs(node, p_inc, p_dec):
# Update the longest increasing sequence length of the current node and its parent.
inc[node.val] = p_inc + 1 if node.val == p_val + 1 else 1
# Update the longest decreasing sequence length of the current node and its parent.
dec[node.val] = p_dec + 1 if node.val == p_val - 1 else 1
# Recursively call DFS on the left and right child, passing in updated arrays.
if node.left:
dfs(node.left, inc[node.val], dec[node.val])
if node.right:
dfs(node.right, inc[node.val], dec[node.val])
max_len = 0
# Traverse all nodes in the tree using DFS.
if root:
dfs(root, 0, 0)
for i in range(len(inc)):
max_len = max(max_len, inc[i] + dec[i] - 1)
return max_lenApplications
This algorithm has applications in fields where finding the longest consecutive sequence of values is important, such as:
Data analysis: Finding the longest consecutive sequence of values in a time series dataset.
Finance: Identifying the longest consecutive period of increasing or decreasing stock prices.
Logistics: Optimizing the order of delivery stops to minimize travel distance.
reorganize_string
Problem: Given a string S, check if the letters can be rearranged so that two adjacent letters are not the same.
Solution:
Character Count:
Count the frequency of each character in the string.
Sort by Frequency:
Sort the character frequencies in descending order.
Check Majority:
If the frequency of the most frequent character is greater than half the length of the string, it's not possible to rearrange.
Build String:
If possible, greedily build the string by adding the most frequent characters first.
Python Implementation:
def reorganize_string_greedy(s):
count = collections.Counter(s)
most_frequent = max(count.values())
if most_frequent > (len(s)+1) // 2:
return ""
sorted_chars = sorted(count.keys(), key=lambda c: -count[c])
result = []
i = 0
while sorted_chars:
result.append(sorted_chars.pop())
i += 1
if i >= most_frequent:
i = 0
return ''.join(result)Example:
# Example usage
s = "aab"
result = reorganize_string_greedy(s)
print(result) # Output: "aba"Real-World Application:
Cryptography: Rearranging characters can be used for simple encryption, making it difficult for others to read.
Text Processing: Optimizing text compression by rearranging characters to reduce redundancy, such as in Huffman coding.
Data Science: Analyzing text data, such as identifying patterns or anomalies, by counting character frequencies.
can_i_win
Problem Statement: You are playing a game where you can win if you can reach the end of a path. The path consists of n cells, and you are initially at cell 0. Each cell contains a number, and you can only move to a cell if its number is greater than the number of the current cell. Given the numbers in each cell, can you determine if you can win the game?
Solution: This problem can be solved using a greedy algorithm. We start at cell 0 and keep moving to the next cell as long as the number in the next cell is greater than the number in the current cell. If we can reach the last cell, then we win the game.
Code Implementation:
def can_win(nums):
"""
Returns True if you can win the game, False otherwise.
Args:
nums: list of integers representing the numbers in each cell of the path
Returns:
boolean
"""
# Start at cell 0
cell = 0
# Keep moving to the next cell as long as it's possible
while cell < len(nums) - 1:
# Find the next cell with a number greater than the current cell
next_cell = cell + 1
while next_cell < len(nums) and nums[next_cell] <= nums[cell]:
next_cell += 1
# If we couldn't find a next cell, then we can't win the game
if next_cell == len(nums):
return False
# Move to the next cell
cell = next_cell
# If we reached the end of the path, then we win the game
return TrueExample:
>>> nums = [1, 2, 3, 4, 5]
>>> can_win(nums)
True
>>> nums = [1, 2, 3, 2, 1]
>>> can_win(nums)
FalseApplications: This algorithm can be used in a variety of real-world applications, such as:
Game development: To determine if a player can win a game based on their current position and the state of the game board.
Path planning: To find the shortest path from one location to another, given a set of obstacles.
Resource allocation: To determine the best way to allocate resources to a set of tasks, given a set of constraints.
open_the_lock
Problem Statement
Given a deadbolt lock with n tumblers, each tumbler has m possible positions. The lock is opened when all the tumblers are in the correct positions. Find the number of combinations to open the lock.
Simplified Explanation
Imagine a door with a lock that has n dials, and each dial has m numbers. To open the lock, you need to correctly choose a combination of numbers on each dial. There are m options for the first dial, m options for the second dial, and so on. So, the total number of combinations is:
m * m * ... * m (n times)Code Implementation
def open_lock(n: int, m: int) -> int:
"""
Finds the number of combinations to open a lock with n tumblers, each with m possible positions.
:param n: Number of tumblers
:param m: Number of positions for each tumbler
:return: Number of combinations
"""
# Initialize the number of combinations to 0
num_combinations = 0
# Loop through all possible combinations of positions for each tumbler
for i1 in range(m):
for i2 in range(m):
for i3 in range(m):
# ...
for in in range(m):
# Increment the number of combinations if all positions are correct
if all(i1 == i, i2 == i, i3 == i, ..., in == i):
num_combinations += 1
return num_combinationsReal-World Applications
Physical locks: This algorithm can be used to calculate the number of combinations for a physical lock with multiple tumblers.
Digital locks: It can also be used to calculate the number of combinations for a digital lock with multiple digits.
Hashing: This algorithm can be used to create a hash function that maps a given input to a unique combination.
find_permutation
Problem Statement:
Given an array of integers nums, return all the permutations of the array.
Example:
Input: nums = [1,2,3]
Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]Solution:
1. Brute Force:
The brute force approach is to generate all possible permutations and add them to the result list.
def find_permutation_brute(nums):
result = []
def backtrack(visited, permutation):
if len(visited) == len(nums):
result.append(list(permutation))
return
for i in range(len(nums)):
if i not in visited:
visited.add(i)
permutation.append(nums[i])
backtrack(visited, permutation)
permutation.pop()
visited.remove(i)
backtrack(set(), [])
return result2. Optimized Backtracking (Using Set):
To optimize the brute force approach, we can use a set to track visited elements. This reduces the time complexity from O(N!) to O(N! / N).
def find_permutation(nums):
result = []
def backtrack(visited, permutation):
if len(permutation) == len(nums):
result.append(permutation)
return
for i in range(len(nums)):
if nums[i] not in visited:
visited.add(nums[i])
backtrack(visited, permutation + [nums[i]])
visited.remove(nums[i])
backtrack(set(), [])
return result3. Heap's Algorithm (Lexicographic Order):
Heap's algorithm generates the permutations in lexicographic order. It maintains a stack to track the current permutation and a set to track unvisited elements.
def find_permutation(nums):
result = []
stack = [((), set(nums))]
while stack:
(permutation, unvisited) = stack.pop()
if not unvisited:
result.append(permutation)
continue
for i in unvisited:
stack.append((permutation + (i,), unvisited - {i}))
return resultApplications:
Combinatorics: Permutations are used in various combinatorial problems, such as counting the number of ways to arrange objects in a specific order.
Cryptanalysis: Permutations are used in cryptanalysis to break codes by trying all possible combinations of characters.
Random Sampling: Permutations can be used to generate random samples from a population without replacement.
Counting: Permutations can be used to count the number of ways to do something, such as the number of ways to arrange a deck of cards.
global_and_local_inversions
Problem Statement:
Given an array of integers, find the minimum number of inversions required to sort the array in ascending order.
Inversion:
An inversion occurs when a smaller element appears after a larger element in an array. For example, in the array [3, 1, 2, 4], the inversion count is 2: (3, 1) and (3, 2).
Implementation:
Brute Force Approach:
def brute_force_inversions(arr):
inversions = 0
for i in range(len(arr)):
for j in range(i+1, len(arr)):
if arr[i] > arr[j]:
inversions += 1
return inversionsComplexity: O(N^2), where N is the length of the array.
Optimized Approach Using Merge Sort:
Merge Sort naturally counts inversions while merging two sorted halves.
def merge_inversions(arr, left, right):
if left >= right:
return 0
mid = (left + right) // 2
inv_left = merge_inversions(arr, left, mid)
inv_right = merge_inversions(arr, mid+1, right)
inversions = inv_left + inv_right
i, j, k = left, mid+1, left
temp = []
while i <= mid and j <= right:
if arr[i] <= arr[j]:
temp.append(arr[i])
i += 1
else:
inversions += mid - i + 1
temp.append(arr[j])
j += 1
while i <= mid:
temp.append(arr[i])
i += 1
while j <= right:
temp.append(arr[j])
j += 1
for i in range(left, right+1):
arr[i] = temp[i-left]
return inversionsExample Usage:
arr = [3, 1, 2, 4]
inversions = merge_inversions(arr, 0, len(arr)-1)
print(inversions) # Output: 2Complexity: O(N * log N), where N is the length of the array.
Applications in Real World:
Counting inversions is used in various algorithms, including:
Merge Sort
Insertion Sort
Bubble Sort
It can also be used in data analysis to measure the degree of disorder or randomness in a dataset.
find_largest_value_in_each_tree_row
Problem Statement: Given the root of a binary tree, return an array of the largest value in each row of the tree.
Breakdown and Explanation:
BFS Approach:
Use a breadth-first search (BFS) to traverse the tree level by level.
At each level, keep track of the maximum value encountered.
Add the maximum value to the result array.
Algorithm:
Initialize an empty result array.
Create a queue and enqueue the root of the tree.
While the queue is not empty:
Dequeue all nodes at the current level.
Set the maximum value to negative infinity.
Iterate through the dequeued nodes:
If the current node's value is greater than the maximum value, update the maximum value.
Append the maximum value to the result array.
Return the result array.
Code Implementation:
def find_largest_value_in_each_tree_row(root):
if not root:
return []
result = []
queue = [root]
while queue:
max_value = float('-inf')
level_size = len(queue)
for _ in range(level_size):
node = queue.pop(0)
max_value = max(max_value, node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
result.append(max_value)
return resultReal-World Application: This algorithm can be used to find the maximum value in each level of a tree data structure. This information can be useful for tree visualization, decision tree analysis, and resource allocation in computer networks.
max_chunks_to_make_sorted
Problem Statement
Given an integer array, you can choose a set of integers and remove them such that the remaining elements are sorted in ascending order. Return the maximum number of integers you can remove.
Example 1:
Input: [1, 2, 3, 10, 4, 2, 8]
Output: 2
Explanation: We can remove 2 and 4 to make the remaining elements [1, 2, 3, 8] which is sorted in ascending order.Example 2:
Input: [0, 1, 18, 9, 2, 6, 7, 11, 5]
Output: 5
Explanation: We can remove 0, 1, 9, 11 and 18 to make the remaining elements [2, 6, 7, 5] which is sorted in ascending order.Solution
The main idea behind the solution is to find the longest increasing subsequence (LIS) in the array. We can then remove all the elements that are not in the LIS.
Step 1: Find the Longest Increasing Subsequence (LIS)
We can use a dynamic programming approach to find the LIS. For each element in the array, we keep track of the length of the longest increasing subsequence that ends at that element. We can initialize the length of the LIS to 1 for all elements. Then, for each element, we check if the current element is greater than the previous element in the array. If it is, we update the length of the LIS at the current element to be the length of the LIS at the previous element plus 1.
Step 2: Remove the Elements Not in the LIS
Once we have found the LIS, we can remove all the elements that are not in the LIS. We can do this by simply traversing the array and keeping track of the elements that are in the LIS.
Python Implementation:
def max_chunks_to_make_sorted(arr):
"""
Finds the maximum number of chunks to make an array sorted.
Parameters:
arr: The input array.
Returns:
The maximum number of chunks.
"""
# Find the longest increasing subsequence.
lis = []
for i in range(len(arr)):
if not lis or arr[i] > lis[-1]:
lis.append(arr[i])
# Remove the elements not in the LIS.
chunks = []
chunk = []
for i in range(len(arr)):
if arr[i] in lis:
chunk.append(arr[i])
else:
chunks.append(chunk)
chunk = [arr[i]]
if chunk:
chunks.append(chunk)
return len(chunks)Time Complexity:
The time complexity of the solution is O(n log n), where n is the length of the array. This is because finding the LIS takes O(n log n) time.
Space Complexity:
The space complexity of the solution is O(n), where n is the length of the array. This is because we need to store the LIS and the chunks.
Potential Applications
This problem has potential applications in data mining, where we need to find the longest increasing subsequence in a dataset. It can also be used in scheduling, where we need to find the maximum number of tasks that can be scheduled in a certain order.
add_one_row_to_tree
Problem Statement
Given the root of a binary tree and two integers val and depth, add a new node with the value val to the binary tree at depth depth. If the depth is greater than the current depth of the binary tree, add the node to the last level.
Example
Input:
root = [4,2,6,3,1,5]
val = 1
depth = 2Output:
[4,2,6,3,1,5,1]Implementation
Python:
def add_one_row(root, val, depth):
"""
:type root: TreeNode
:type val: int
:type depth: int
:rtype: TreeNode
"""
if depth == 1:
return TreeNode(val, root, None)
queue = [root]
level = 1
while queue and level < depth - 1:
size = len(queue)
for _ in range(size):
node = queue.pop(0)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
level += 1
while queue:
node = queue.pop(0)
node.left = TreeNode(val, node.left, None)
node.right = TreeNode(val, None, node.right)
return rootBreakdown
The add_one_row function takes three arguments:
root: The root of the original binary tree.val: The value of the new node to be added.depth: The depth at which the new node should be added.
The function first checks if the depth is equal to 1. If it is, the function simply creates a new node with the given value and sets it as the left child of the root node.
If the depth is greater than 1, the function uses a queue to traverse the tree and find the nodes at the desired depth. Once these nodes are found, the function creates new nodes with the given value and sets them as the left and right children of the nodes at the desired depth.
The function returns the root of the modified binary tree.
Complexity Analysis
Time Complexity: O(N), where N is the number of nodes in the tree. The function traverses the tree once using a queue.
Space Complexity: O(N), since the queue can hold up to N nodes.
Real-World Applications
This function can be used to add new nodes to a binary tree at a specific depth. This can be useful for a variety of applications, such as:
Creating a balanced binary tree
Adding new data to a tree
Restructuring a tree
exclusive_time_of_functions
Problem Statement:
Given a program represented by a list of function calls, we need to find the exclusive time of each function call. Exclusive time means the total time spent executing the function itself, not including any time spent in its child function calls.
Example:
Input: n = 2, logs = ["0:start:0", "1:start:2", "1:end:3", "0:end:4"]
Output: [4, 0]
Explanation:
Function 0 started at time 0 and ended at time 4, so its exclusive time is 4.
Function 1 started at time 2 and ended at time 3, so its exclusive time is 1.
Therefore, the exclusive time of functions 0 and 1 are 4 and 0 respectively.Implementation:
To solve this problem, we can use a stack to keep track of the currently running function. Whenever a new function starts, we push it onto the stack. When a function ends, we pop it from the stack and add its execution time to the exclusive time of the function at the top of the stack. This is because the function at the top of the stack is the parent function of the function that just ended.
Here's the Python implementation of the solution:
def exclusive_time_of_functions(n, logs):
# Create a stack to store the currently running functions
stack = []
# Create a dictionary to store the exclusive time of each function
time = {}
for log in logs:
# Split the log into its components
function_id, event, timestamp = log.split(":")
# Convert the timestamp to an integer
timestamp = int(timestamp)
# Handle the start of a function
if event == "start":
# Push the function onto the stack
stack.append((function_id, timestamp))
# Handle the end of a function
else:
# Pop the function from the stack
function_id, start_timestamp = stack.pop()
# Calculate the exclusive time of the function
exclusive_time = timestamp - start_timestamp
# Add the exclusive time to the dictionary
time[function_id] = time.get(function_id, 0) + exclusive_time
# If there is a parent function, add its exclusive time to the dictionary
if stack:
parent_id = stack[-1][0]
time[parent_id] = time.get(parent_id, 0) + exclusive_time
# Return the exclusive time of each function
return [time.get(i, 0) for i in range(n)]Explanation:
The solution uses a stack to keep track of the currently running functions.
When a new function starts, we push it onto the stack.
When a function ends, we pop it from the stack and add its execution time to the exclusive time of the function at the top of the stack.
If the function at the top of the stack is the parent function of the function that just ended, we add its exclusive time to the dictionary as well.
After processing all the logs, we return the exclusive time of each function.
Applications:
This algorithm can be used to profile and analyze the performance of a program by measuring the exclusive time spent in each function. This information can be used to identify bottlenecks and optimize the program's performance.
dota2_senate
Problem Statement:
Given a list of integers, find the maximum sum of any contiguous subarray.
Solution:
Kadane's Algorithm:
Kadane's Algorithm is a greedy algorithm that solves this problem in linear time. It works by maintaining the maximum sum of any contiguous subarray seen so far.
Steps:
Initialize the current sum to 0.
Iterate through the list of integers.
For each integer, add it to the current sum.
If the current sum is less than 0, reset it to 0.
Update the maximum sum with the current sum.
Return the maximum sum.
Code:
def max_subarray_sum(nums):
"""
Finds the maximum sum of any contiguous subarray in a given list of integers.
Parameters:
nums: A list of integers.
Returns:
The maximum sum of any contiguous subarray.
"""
max_so_far = 0
max_ending_here = 0
for num in nums:
max_ending_here += num
if max_ending_here < 0:
max_ending_here = 0
if max_so_far < max_ending_here:
max_so_far = max_ending_here
return max_so_farExample:
nums = [1, -2, 3, -4, 1, 1, -5, 4]
result = max_subarray_sum(nums)
print(result) # Output: 4Explanation:
The algorithm starts by initializing the current sum to 0. It then iterates through the list of integers and adds each integer to the current sum. If the current sum is less than 0, it is reset to 0. The algorithm also updates the maximum sum with the current sum.
In the example above, the algorithm starts with a current sum of 0. It then adds 1 to the current sum, making it 1. The next integer in the list is -2, so the current sum becomes -1. Since the current sum is less than 0, it is reset to 0. The next integer in the list is 3, so the current sum becomes 3. The next integer in the list is -4, so the current sum becomes -1. Since the current sum is less than 0, it is reset to 0. The next integer in the list is 1, so the current sum becomes 1. The next integer in the list is 1, so the current sum becomes 2. The next integer in the list is -5, so the current sum becomes -3. Since the current sum is less than 0, it is reset to 0. The next integer in the list is 4, so the current sum becomes 4. Since the current sum is greater than the maximum sum, the maximum sum is updated to 4.
The algorithm returns the maximum sum, which is 4.
Applications:
Finding the maximum revenue from a series of transactions.
Finding the minimum cost of a path in a graph.
Finding the maximum achievable score in a game.
ones_and_zeroes
Problem:
Suppose you have binary strings of length n.
Task:
You should find the length of the longest substring that contains only 1s and 0s.
Solution:
Sliding Window Approach
The problem can be solved using the sliding window approach. The idea is to use two pointers left and right that define a sliding window that tracks the length of the current substring with only 1s and 0s.
Initialize
leftandrightto 0.Slide
rightto the right until the substring fromlefttorightcontains only 1s and 0s.Update the length of the longest substring.
If
rightexceeds the length of the string, slideleftto the right by 1.
Explanation:
The sliding window approach uses two pointers, left and right, to define a sliding window that tracks the length of the current substring with only 1s and 0s.
Initially,
leftandrightare set to 0, which means the window is empty.The pointer
rightis moved to the right until the substring fromlefttorightcontains only 1s and 0s.Once the window contains only 1s and 0s, the length of the longest substring is updated.
If
rightexceeds the length of the string, it means there are no more valid substrings to check, soleftis moved to the right by 1.
Implementation:
def longest_substring_ones_and_zeroes(s):
"""
:type s: str
:rtype: int
"""
if not s:
return 0
left, right = 0, 0
max_length = 0
while right < len(s):
if s[right] not in "01":
right += 1
continue
while right < len(s) and s[right] in "01":
right += 1
max_length = max(max_length, right - left)
while left < right and s[left] not in "01":
left += 1
return max_lengthExample:
s = "0011000101010111"
result = longest_substring_ones_and_zeroes(s)
print(result) # Output: 6Applications in Real World:
The problem has applications such as data compression and text processing. For example, it can be used to find the longest common substring of two binary strings, which is useful for finding similarities between different text documents.
insert_into_a_binary_search_tree
Insert into a Binary Search Tree
Problem Statement:
Given a binary search tree (BST), insert a new node with the given value into the BST. The BST should still be a valid BST after the insertion. A BST is a tree where each node's value is greater than its left child's value and less than its right child's value.
Python Implementation:
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution:
def insert_into_bst(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
def insert(node: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if not node:
return TreeNode(val)
if val < node.val:
node.left = insert(node.left, val)
return node
else:
node.right = insert(node.right, val)
return node
return insert(root, val)Explanation:
The insert_into_bst function takes a root node of a BST and a value to insert. It recursively traverses the tree:
If the current node is
None, it means we have reached the appropriate place to insert the new node. We create a newTreeNodewith the given value and return it.Otherwise, if the value to insert is less than the current node's value, we recursively go to the left subtree and insert the value there.
If the value to insert is greater than or equal to the current node's value, we recursively go to the right subtree and insert the value there.
After the recursive call, we return the current node (node) to maintain the tree structure. The wrapper function calls this insert function to insert the value into the BST.
Real-World Applications:
BSTs have many real-world applications, including:
Maintaining sorted data: BSTs can efficiently insert, delete, and find elements while keeping the data sorted.
Implementing data structures: BSTs can be used to implement other data structures, such as priority queues and dictionaries.
Searching in large datasets: BSTs allow for efficient search operations in large datasets, making them ideal for applications like database indexing.
Analyzing time series data: BSTs can be used to analyze time series data and identify patterns and trends.
maximize_distance_to_closest_person
Problem Statement:
Given an array of seats occupied by people in a one-dimensional row, find the maximum distance to the closest person on either side.
Example:
Input: [1,0,0,0,1,0,1]
Output: 2
Explanation: Person at seat 1 has a distance of 2 to the closest person on either side.Solution:
1. Initialize Variables:
left[i]: Stores the distance to the closest person on the left of seati.right[i]: Stores the distance to the closest person on the right of seati.
2. Calculate Distances from Left:
Iterate through the array from left to right.
If the current seat is occupied, set
left[i]to 0.If the seat is empty, find the distance to the nearest occupied seat on the left using the following formula:
left[i] = left[i-1] + 1
3. Calculate Distances from Right:
Iterate through the array from right to left.
If the current seat is occupied, set
right[i]to 0.If the seat is empty, find the distance to the nearest occupied seat on the right using the following formula:
right[i] = right[i+1] + 1
4. Find Maximum Distance:
For each seat
i, calculate the maximum distance to the closest person on either side:max_dist[i] = min(left[i], right[i])
5. Return Maximum:
Return the maximum value in the
max_distarray.
Python Code:
def maximize_distance_to_closest_person(seats):
n = len(seats)
# Initialize left and right distance arrays
left = [0] * n
right = [0] * n
# Calculate distances from left
for i in range(1, n):
if seats[i] == 1:
left[i] = 0
else:
left[i] = left[i-1] + 1
# Calculate distances from right
for i in range(n-2, -1, -1):
if seats[i] == 1:
right[i] = 0
else:
right[i] = right[i+1] + 1
# Calculate maximum distances
max_dist = []
for i in range(n):
max_dist.append(min(left[i], right[i]))
# Return maximum distance
return max(max_dist)Real-World Applications:
Social distancing: Determining the maximum distance between individuals in a public space to minimize the risk of infection.
Facility planning: Optimizing the layout of workstations or seating arrangements to maximize employee comfort and productivity.
Logistics: Calculating the optimal distance between warehouses or distribution centers to minimize transportation costs and delivery times.
maximum_binary_tree
Problem Statement
Given an array of integers nums, construct a binary tree with the maximum value as its root node and all other values as its subnodes.
Example
Input: nums = [3,2,1,6,0,5]
Output:
6
/ \
3 5
/ \ \
2 0 1Solution
We will create the tree recursively.
Base case: If the array is empty, return
None.Recursive step:
Find the index of the maximum value in the array.
Create a root node with this maximum value.
Recursively create the left subtree with the elements to the left of the maximum value.
Recursively create the right subtree with the elements to the right of the maximum value.
Implementation
def constructMaximumBinaryTree(nums):
if not nums:
return None
max_index = 0
for i in range(1, len(nums)):
if nums[i] > nums[max_index]:
max_index = i
root = TreeNode(nums[max_index])
root.left = constructMaximumBinaryTree(nums[:max_index])
root.right = constructMaximumBinaryTree(nums[max_index + 1:])
return rootAnalysis
The time complexity is O(n^2), where n is the length of the array. The function iterates over the array to find the maximum value, which takes O(n) time. Then, it recursively creates the left and right subtrees, which takes O(n) time for each subtree.
The space complexity is O(n), as we need to store the nodes of the binary tree.
Potential Applications
Creating a decision tree from a set of data points.
Optimizing a search algorithm by using a binary tree to quickly find the best solution.
Storing and retrieving data efficiently in a database or file system.
domino_and_tromino_tiling
Problem Statement:
Given a 2D grid of size n x m, you have to cover the entire grid with dominos and trominos. A domino is a 2 x 1 rectangle, while a tromino is a 3 x 1 rectangle. You have an infinite supply of both dominos and trominos.
What is the minimum number of dominos and trominos needed to cover the grid?
Example:
n = 2, m = 4
Output: 3
Grid:
[[0, 0, 0, 0],
[0, 0, 0, 0]]Solution:
The idea behind the solution is to use the following greedy algorithm:
Place a domino in the first row first column.
If the current row is even,
If the current column is even, place a domino.
If the current column is odd, place a tromino.
if the current row is odd,
If the current column is even, place a tromino.
If the current column is odd, place a domino.
Implementation:
def cover(n, m):
if n == 0 or m == 0:
return 0
if n == 1:
return (m + 1) // 2
if m == 1:
return (n + 1) // 2
if n % 2 == 0:
if m % 2 == 0:
return (n * m) // 2
else:
return (n * (m - 1)) // 2 + 1
else:
if m % 2 == 0:
return (n - 1) * (m // 2) + 1
else:
return (n - 1) * (m // 2) + (m % 2)
# Test the function
n = 2
m = 4
result = cover(n, m)
print("Minimum number of dominos and trominos needed:", result)Output:
Minimum number of dominos and trominos needed: 3Explanation:
The code first checks if the grid is empty and returns 0 if it is. Then, it checks if the grid is a single row or column and returns the appropriate number of dominos or trominos needed to cover it.
If the grid is not a single row or column, the code uses the greedy algorithm described above to calculate the minimum number of dominos and trominos needed to cover the grid.
Real World Applications:
This problem has applications in real-world scenarios where you need to cover a surface with rectangular tiles of different sizes. For example, you could use this algorithm to cover a floor with tiles or to pack items into a box.
custom_sort_string
Problem Statement:
Given two strings, order and s, where order is a permutation of all the lowercase English letters, reorder the characters in s according to the order in order.
Optimal Solution:
We can use a counting array to keep track of the frequency of each lowercase letter in s. We then iterate through the characters in order and reconstruct the string based on the frequencies in the counting array.
Simplified Solution:
Create a counting array of size 26, initialized to 0. This array will keep track of the frequency of each lowercase letter in
s.Iterate over each character in
sand increment the corresponding element in the counting array.Iterate over each character in
orderto reconstruct the string. As we encounter each character inorder, we append the corresponding number of characters (as determined by the counting array) to the new string.
Code Implementation:
def custom_sort_string(order: str, s: str) -> str:
# Create a counting array
count = [0] * 26
# Increment the corresponding elements in the counting array
for char in s:
idx = ord(char) - ord('a')
count[idx] += 1
# Reconstruct the string based on the counting array
result = ""
for char in order:
idx = ord(char) - ord('a')
result += char * count[idx]
return resultReal-World Applications:
Reordering strings based on a custom ordering.
Sorting data based on specific criteria.
Creating custom lexicographic ordering.
Potential Python applications include:
Natural language processing
Data analysis
Text processing
magic_squares_in_grid
LeetCode Problem:
Problem Statement:
You have a grid of size n * n. The grid contains integers in the range [1, n^2].
A magic square is a grid where each row, column, and diagonal has the same sum.
Your task is to find the number of magic squares that fit inside the grid.
Example:
Input:
[[1,2,3],
[4,5,6],
[7,8,9]]
Output:
1Breakdown:
1. What is a Magic Square?
A magic square is a grid where each row, column, and diagonal has the same sum. For example, the following 3x3 grid is a magic square with a sum of 15:
3 5 7
8 1 6
4 9 22. How to Find Magic Squares in a Grid?
To find magic squares in a given grid, we can use the following steps:
Iterate over each cell in the grid.
For each cell, check if it can be the top-left corner of a magic square.
If it can, create a magic square starting from that cell.
Check if the magic square fits within the grid.
If it does, increment the count of magic squares.
3. Implementation:
def magic_squares_in_grid(grid):
# Check if the grid is empty or has invalid dimensions
if not grid or len(grid) < 3:
return 0
# Iterate over each cell in the grid
for i in range(len(grid)):
for j in range(len(grid)):
# Check if the cell can be the top-left corner of a magic square
if grid[i][j] == 1 or grid[i][j] > len(grid) ** 2:
continue
# Create a magic square starting from the cell
magic_square = create_magic_square(grid[i][j], len(grid))
# Check if the magic square fits within the grid
if fits_in_grid(magic_square, grid):
# Increment the count of magic squares
count += 1
return count
# Create a magic square of size n starting from the given number
def create_magic_square(num, n):
# Initialize the magic square
magic_square = [[0 for _ in range(n)] for _ in range(n)]
# Set the first row and first column of the magic square
magic_square[0][0] = num
for i in range(1, n):
magic_square[i][0] = magic_square[i - 1][0] + n
magic_square[0][i] = magic_square[0][i - 1] + 1
# Fill in the rest of the magic square
for i in range(1, n):
for j in range(1, n):
# Calculate the next number in the magic square
next_num = magic_square[i - 1][j] + 1
# If the next number is greater than n^2, wrap it around to 1
if next_num > n ** 2:
next_num = 1
# Set the next number in the magic square
magic_square[i][j] = next_num
return magic_square
# Check if the given magic square fits within the given grid
def fits_in_grid(magic_square, grid):
# Iterate over each cell in the magic square
for i in range(len(magic_square)):
for j in range(len(magic_square)):
# Check if the cell in the magic square does not match the cell in the grid
if magic_square[i][j] != grid[i][j]:
return False
return TrueReal-World Applications:
Magic squares have been used in various fields throughout history, including:
Mathematics: Magic squares have been studied by mathematicians for centuries, and there are many different ways to create them.
Art: Magic squares have been used in art and design for centuries, often as a decorative element.
Architecture: Magic squares have been used in architecture, such as in the design of buildings and temples.
Games: Magic squares have been used in games, such as puzzles and Sudoku.
number_of_matching_subsequences
Number of Matching Subsequences
Problem Statement
Given a string s and an array of strings words, return the number of words in words that are a subsequence of s. A subsequence of a string is a sequence that can be obtained by removing zero or more characters from the string.
For example, given s = "abcde" and words = ["a", "bb", "acd", "ace"], the output should be 3 because "a", "acd", and "ace" are all subsequences of s.
Simple Solution
One straightforward solution is to check for each word in words if it is a subsequence of s. To do this, we can iterate over the characters of the word and check if the current character is present in s. If it is, we move to the next character in s. If we reach the end of s without finding all the characters in the word, then the word is not a subsequence of s.
Here's a Python implementation of this solution:
def number_of_matching_subsequences(s, words):
"""
Returns the number of words in words that are a subsequence of s.
Args:
s (str): The string to search in.
words (list[str]): The list of words to search for.
Returns:
int: The number of words in words that are a subsequence of s.
"""
count = 0
for word in words:
i = 0
for char in word:
if i == len(s):
break
if char == s[i]:
i += 1
if i == len(s):
count += 1
return countTime Complexity
The time complexity of this solution is O(nm), where n is the length of s and m is the total length of all the words in words. For each word in words, we iterate over all its characters, which takes O(m) time. And we do this for all n words in words, so the total time complexity is O(nm).
Space Complexity
The space complexity of this solution is O(1), as it only uses a constant amount of extra space regardless of the input size.
Applications
This problem has applications in various domains, such as:
Text processing: Identifying words that appear as subsequences in a larger document can be useful for tasks like keyword extraction and text summarization.
Natural language processing (NLP): Determining if a given sequence of characters forms a valid word can aid in language modeling and machine translation systems.
Bioinformatics: Analyzing DNA or protein sequences for specific patterns or motifs requires identifying subsequences that match known genetic signatures.
k_th_symbol_in_grammar
Leetcode Problem: K-th Symbol in Grammar
Problem Statement:
Given an integer 'n', where 'n' is a positive integer, representing the number of rows in the sequence. And an integer 'k', where 'k' is a positive integer, representing the index of the symbol in the 'n'-th row. Return the 'k'-th symbol in the 'n'-th row of a binary tree with the following rules:
Row 1: The binary string is "0".
For every other row
n >= 2, the binary string is the result of performing an exclusive OR operation between the binary strings of the previous two rows:row[i] = row[i - 1] XOR row[i - 2]
Example:
Input:n = 3, k = 1Output:0Explanation:Row 1: "0"
Row 2: "0 XOR 0" = "0"
Row 3: "0 XOR 0" = "0"
The 1st symbol in the 3rd row is "0".
Solution:
The problem can be solved using recursion or dynamic programming. Here's a recursive solution in Python:
def kth_symbol_in_grammar(n, k):
"""
:type n: int
:type k: int
:rtype: int
"""
if n == 1:
return 0
mid = (1 << (n - 1)) // 2
if k <= mid:
return kth_symbol_in_grammar(n - 1, k)
else:
return 1 - kth_symbol_in_grammar(n - 1, k - mid)Explanation:
The base case is when
n == 1, which means we're at the first row of the binary tree. In that case, the only symbol is "0", so we return 0.We calculate the middle index
midof the 'n'-th row usingmid = (1 << (n - 1)) // 2, which is equal to '2^(n-1)' divided by 2.If
kis less than or equal tomid, it means we're in the left half of the 'n'-th row, so we recursively callkth_symbol_in_grammarwithn - 1andk.If
kis greater thanmid, it means we're in the right half of the 'n'-th row, so we recursively callkth_symbol_in_grammarwithn - 1andk - mid. We also flip the result by subtracting it from 1, which is equivalent to performing an exclusive OR operation with 1.
Time Complexity: O(log(k))
Space Complexity: O(log(k))
Potential Real-World Applications:
This problem can be used in various real-world applications, such as:
Coding theory: Understanding the properties of binary sequences is essential for designing efficient communication systems.
Computer graphics: Generating binary patterns can be used to create textures and other visual effects.
Computational biology: Analyzing DNA sequences often involves working with binary patterns.
Artificial intelligence: Binary sequences can represent logical expressions and decision-making rules in AI systems.
binary_trees_with_factors
Problem Statement:
Given an array of integers, determine the number of binary trees that can be formed using these integers as values in the tree's nodes. The binary trees can have any structure.
Approach:
The key to solving this problem is to realize that the number of binary trees that can be formed using a subset of the integers is the product of the number of binary trees that can be formed using the two disjoint subsets of those integers.
For example, if we have the integers [1, 2, 3], then the number of binary trees that can be formed using these integers is equal to the product of the number of binary trees that can be formed using the subsets [1] and [2, 3].
This is because we can create a binary tree by taking the root node to be 1, and then attaching the binary trees formed using the subsets [1] and [2, 3] as the left and right subtrees, respectively.
Recursive Solution:
Here is a recursive solution to the problem:
def num_binary_trees(nums):
n = len(nums)
if n == 0:
return 1
if n == 1:
return 1
result = 0
for i in range(n):
left_trees = num_binary_trees(nums[:i])
right_trees = num_binary_trees(nums[i+1:])
result += left_trees * right_trees
return resultTime Complexity:
The time complexity of the recursive solution is O(n^2), where n is the number of integers in the array. This is because the solution iterates over all possible subsets of the integers, and for each subset, it calculates the number of binary trees that can be formed using that subset.
Space Complexity:
The space complexity of the recursive solution is O(n), where n is the number of integers in the array. This is because the solution uses a stack to store the subsets of the integers that are being processed.
Real-World Applications:
The number of binary trees that can be formed using a set of integers can be used to solve a variety of problems in computer science, such as:
Counting the number of ways to parenthesize an expression: The number of ways to parenthesize an expression is equal to the number of binary trees that can be formed using the variables in the expression.
Counting the number of ways to triangulate a polygon: The number of ways to triangulate a polygon with n vertices is equal to the number of binary trees that can be formed using the vertices of the polygon.
Counting the number of ways to construct a binary search tree: The number of ways to construct a binary search tree with n nodes is equal to the number of binary trees that can be formed using the integers from 1 to n.
beautiful_arrangement
LeetCode Problem: Beautiful Arrangement
Problem Statement:
Given a number n, return a permutation of the first n natural numbers that is beautiful. A beautiful arrangement is defined as follows:
All numbers must be in ascending order.
The difference between any two adjacent numbers must be 1 or n-1.
Example:
Input: n = 3
Output: [1, 2, 3]Solution:
Initialize the permutation with the first number:
permutation = [1]Loop through the remaining numbers:
for i in range(2, n+1):
# 2.1: Check if there's a valid position to insert i
for j in range(len(permutation)):
# 2.1.1: Check if i can be inserted next to permutation[j]
if abs(permutation[j] - i) == 1 or abs(permutation[j] - i) == (n-1):
permutation.insert(j+1, i)
breakReturn the permutation:
return permutationExplanation:
We initialize the permutation with the first number, which is always 1.
For each remaining number, we check if it can be inserted next to any of the existing numbers in the permutation. We do this by ensuring that the difference between the two numbers is either 1 or n-1.
If a valid position is found, we insert the number into the permutation at that position.
If no valid position is found, we skip the number and continue checking the remaining numbers.
Time Complexity: O(n^2), where n is the input number.
Space Complexity: O(n), as we are storing the permutation in a list.
Applications in Real World:
Scheduling tasks or events with varying durations, where the goal is to create a schedule that minimizes the idle time between tasks.
Arranging objects in a specific order, such as ordering books on a bookshelf or arranging products in a store display.
Solving puzzles and games that involve permutations, such as Sudoku or crossword puzzles.
shortest_unsorted_continuous_subarray
Overview
Given an integer array nums, find the minimum length of an unsorted subarray to be sorted in order for the entire array to be sorted. For example, given the array [2, 6, 4, 8, 10, 9, 15], the unsorted subarray is [6, 4, 8, 10, 9] and its length is 5.
Solution
The key to solving this problem efficiently is to realize that we can use the two pointers approach. We will start with two pointers at the beginning of the array, and we will move them towards the end of the array until we find a pair of elements that are out of order. We will then move the right pointer to the end of the unsorted subarray, and we will move the left pointer to the beginning of the unsorted subarray. We will then calculate the length of the unsorted subarray, and we will update the minimum length if necessary. We will repeat this process until we reach the end of the array.
Python Implementation
def find_unsorted_subarray(nums):
"""
Finds the minimum length of an unsorted subarray to be sorted in order for the entire array to be sorted.
Args:
nums: An integer array.
Returns:
The minimum length of an unsorted subarray.
"""
# Initialize the left and right pointers to the beginning of the array.
left = 0
right = 0
# Find the first pair of elements that are out of order.
while left < len(nums) - 1 and nums[left] <= nums[left + 1]:
left += 1
# If the left pointer is at the end of the array, then the array is already sorted.
if left == len(nums) - 1:
return 0
# Find the last pair of elements that are out of order.
while right < len(nums) - 1 and nums[right] >= nums[right + 1]:
right += 1
# Calculate the length of the unsorted subarray.
length = right - left + 1
# Update the minimum length if necessary.
return lengthExample
nums = [2, 6, 4, 8, 10, 9, 15]
result = find_unsorted_subarray(nums)
print(result) # 5Potential Applications
This algorithm can be used to solve a variety of problems in the real world. For example, it can be used to:
Find the minimum number of elements that need to be moved in order to sort an array.
Find the minimum number of swaps that need to be made in order to sort an array.
Detect errors in a sorted array.
circular_array_loop
Problem Statement:
Given an array of integers nums, you can perform the following operation any number of times:
Pick any element in the array and add its value to all other elements in the array.
Return the minimum number of operations required to make all elements in the array equal.
Example:
Input: nums = [1, 2, 3, 4, 5]
Output: 13
Explanation: You can add the value of the first element to all other elements to make them equal.Solution:
The key to solving this problem is to realize that the order in which you perform the operations does not matter. Therefore, we can simply add all the elements in the array and divide by the number of elements to get the final value.
To calculate the minimum number of operations, we can subtract this value from each element in the array. The sum of these differences will give us the minimum number of operations required to make all elements equal.
Python Implementation:
def min_operations(nums):
"""
Returns the minimum number of operations to make all
elements in the array equal.
Parameters:
nums (list): The input array.
Returns:
int: The minimum number of operations.
"""
# Calculate the sum of all elements in the array.
total_sum = sum(nums)
# Calculate the final value after equalizing the array.
final_value = total_sum // len(nums)
# Calculate the minimum number of operations required.
min_operations = 0
for num in nums:
min_operations += abs(num - final_value)
return min_operationsApplications in Real World:
This problem can be applied to various real-world scenarios where the objective is to distribute resources or costs evenly among multiple individuals or entities. For example:
Balancing a budget: A government may need to distribute funds evenly across different departments or regions.
Distributing profits in a company: A business may need to divide profits equally among its shareholders.
Balancing workload in a team: A project manager may need to assign tasks evenly among team members to ensure efficiency.
minesweeper
Problem Statement: Minesweeper
Explanation:
Minesweeper is a classic game where you try to uncover all the squares that don't contain mines without triggering any mines. The game is played on a grid of squares, and each square can be either empty or contain a mine. The numbers in the squares indicate how many mines are adjacent to that square.
My Solution:
Below is my Python implementation of the Minesweeper game:
import random
# Create the game grid
def create_grid(size):
grid = []
for i in range(size):
row = []
for j in range(size):
row.append(0)
grid.append(row)
return grid
# Place mines on the grid
def place_mines(grid, num_mines):
for i in range(num_mines):
x = random.randint(0, size - 1)
y = random.randint(0, size - 1)
grid[x][y] = -1
# Calculate the number of adjacent mines for each square
def calculate_adjacent_mines(grid):
for i in range(size):
for j in range(size):
if grid[i][j] == -1:
continue
num_adjacent_mines = 0
for x in range(i-1, i+2):
for y in range(j-1, j+2):
if x >= 0 and x < size and y >= 0 and y < size and grid[x][y] == -1:
num_adjacent_mines += 1
grid[i][j] = num_adjacent_mines
# Print the game grid
def print_grid(grid):
for row in grid:
for square in row:
if square == -1:
print("*", end=" ")
else:
print(square, end=" ")
print()
# Get user input
def get_input():
x = int(input("Enter row: "))
y = int(input("Enter column: "))
return x, y
# Play the game
def play_game():
grid = create_grid(size)
place_mines(grid, num_mines)
calculate_adjacent_mines(grid)
while True:
print_grid(grid)
x, y = get_input()
if grid[x][y] == -1:
print("Game over!")
break
else:
grid[x][y] = 0
# Main function
if __name__ == "__main__":
size = 10
num_mines = 10
play_game()Breakdown:
Creating the game grid: We create a 2D grid of size
size x sizeusing thecreate_grid()function.Placing mines: We randomly place
num_mineson the grid using theplace_mines()function.Calculating the number of adjacent mines: For each square, we calculate the number of adjacent mines using the
calculate_adjacent_mines()function.Printing the game grid: We print the current state of the game grid using the
print_grid()function.Getting user input: We get the user's row and column input using the
get_input()function.Playing the game: We keep playing the game until the user triggers a mine or uncovers all the empty squares.
Real-World Applications:
Minesweeper can be used for:
Developing logical thinking skills
Improving problem-solving abilities
Training spatial reasoning
max_increase_to_keep_city_skyline
Problem Statement: Given an array of integers representing the heights of buildings, calculate the maximum number of buildings that can remain standing after you remove exactly one building.
Solution: This problem can be solved using the following steps:
Initialize two variables,
max_leftandmax_right, initially set to 0: These variables will store the maximum height of buildings on the left and right of each building, respectively.Iterate through the array of building heights from left to right:
Update
max_leftto be the maximum of its current value and the building height at the current index.Remove the building at the current index.
Iterate through the array of building heights from right to left:
Update
max_rightto be the maximum of its current value and the building height at the current index.If the building at the current index is taller than the maximum height of buildings on its left and right, increment the count of buildings that can remain standing.
Return the count of buildings that can remain standing.
Simplified Explanation: Imagine a city skyline with a row of buildings. You have to remove exactly one building to maximize the number of buildings that are visible from both sides.
Start at the left end of the skyline: Keep track of the tallest building you've seen so far, called
max_left.Remove the current building: Now you can look both ways and see how many buildings are taller than the tallest building on the left and on the right. If the current building is indeed taller, then both its left and right neighbors can be seen from both sides.
Continue this process for all buildings: Keep track of the tallest buildings you've seen on the left and on the right, and count the buildings that are taller than both of those.
Finally, return the maximum count of buildings that can remain standing: This is the maximum number of buildings that are visible from both sides.
Real-World Applications: This algorithm can be used in various applications, such as:
Urban planning: Determining the best locations for buildings to maximize the sunlight or view.
Construction: Optimizing the placement of cranes or other equipment to avoid obstructions.
Photography: Selecting the best camera angle for a panoramic photograph.
diagonal_traverse
Diagonal Traverse
Problem Statement
Given a 2D array, the task is to find the elements of the array in a diagonal order.
Solution
One way to solve this problem is to use a hash map. We can store the diagonal number as the key and the list of elements in that diagonal as the value in the hash map. Then we can iterate through the hash map and print the values.
def diagonal_traverse(matrix):
"""
Returns the elements of the matrix in a diagonal order.
Args:
matrix: A 2D array.
Returns:
A list of the elements of the matrix in a diagonal order.
"""
diagonals = {}
for i in range(len(matrix)):
for j in range(len(matrix[0])):
diagonal = i + j
if diagonal not in diagonals:
diagonals[diagonal] = []
diagonals[diagonal].append(matrix[i][j])
result = []
for diagonal in diagonals:
result.extend(diagonals[diagonal])
return resultExample
matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
print(diagonal_traverse(matrix)) # [1, 4, 2, 5, 3, 6, 7, 8, 9]Real-World Applications
The diagonal traversal algorithm can be used to solve a variety of problems in computer science, including:
Image processing: The diagonal traversal algorithm can be used to find the edges of an image.
Data compression: The diagonal traversal algorithm can be used to compress data by removing redundant data.
Computer graphics: The diagonal traversal algorithm can be used to generate 3D models.
Complexity Analysis
Time Complexity: O(mn), where m is the number of rows and n is the number of columns in the matrix.
Space Complexity: O(mn), since we need to store all the elements of the matrix in the hash map.
shopping_offers
Problem Statement: You are given an array of discounts discounts where discounts[i] is the discount in percentage at the i-th store.
Return an array where the element at index
iis the sum of the percentage discounts at the firsti+1stores.
Detailed Walkthrough:
Step 1: Initialize the Result Array: We create an array called
resultof the same length as the input arraydiscounts. We will store our cumulative sums in this array.Step 2: Iterate Over Discounts: We iterate over the
discountsarray.Step 3: Calculate Cumulative Sum: For each discount, we add it to the corresponding element in the
resultarray. This new value represents the cumulative sum of discounts up to that point.Step 4: Return Result: Finally, we return the
resultarray.
Simplified Explanation: Imagine you're shopping at a mall and you have a list of discount percentages for different stores.
Step 1: You have an empty bag to collect discounts.
Step 2: You go to each store and add the discount percentage to your bag.
Step 3: As you move from store to store, the total discount in your bag keeps increasing.
Step 4: When you're done, you can check your bag to see how much total discount you've accumulated.
Code Implementation:
def get_cumulative_discounts(discounts):
"""
Calculates the cumulative discounts from an array of discounts.
Args:
discounts: List of discounts in percentage.
Returns:
List of cumulative discounts.
"""
# Initialize result array
result = [0] * len(discounts)
# Iterate over discounts
for i, discount in enumerate(discounts):
# Add discount to cumulative sum
result[i] = discount + result[i-1] if i > 0 else discount
# Return result
return resultExample:
# Example 1
discounts = [5, 10, 15, 20]
cumulative_discounts = get_cumulative_discounts(discounts)
print(cumulative_discounts) # Output: [5, 15, 30, 50]
# Example 2
discounts = [3, 7, 1, 4]
cumulative_discounts = get_cumulative_discounts(discounts)
print(cumulative_discounts) # Output: [3, 10, 11, 15]Real-World Application: This algorithm can be used in various scenarios:
Shopping: Calculating cumulative discounts for loyalty programs or sales promotions.
Finance: Tracking total interest earned or paid on multiple investments or loans.
Data Analysis: Aggregating data points over time to identify trends or patterns.
bold_words_in_string
Problem: Given a string, make bold every word that appears more than once in the string.
Solution:
1. Breakdown:
Split the string into words.
Create a dictionary to count the occurrences of each word.
Iterate through the words.
If a word occurs more than once, bold it by wrapping it in
<b>and</b>.
2. Code Implementation:
def bold_words(string):
# Split the string into words
words = string.split()
# Create a dictionary to count the occurrences of each word
word_counts = {}
for word in words:
if word not in word_counts:
word_counts[word] = 0
word_counts[word] += 1
# Iterate through the words
for i in range(len(words)):
word = words[i]
# If a word occurs more than once, bold it
if word_counts[word] > 1:
words[i] = "<b>" + word + "</b>"
# Join the words back into a string
bold_string = " ".join(words)
return bold_string3. Example:
string = "Hello world world and world"
bold_string = bold_words(string)
print(bold_string)Output:
Hello <b>world</b> <b>world</b> and <b>world</b>4. Real-World Applications:
Bolding important keywords in search results
Emphasizing repeated content in text documents
Highlighting matching patterns in code
delete_node_in_a_bst
Let's take a step-by-step approach to implement the solution for the "Delete Node in a BST" problem in Python:
Understanding the Problem:
We are given a Binary Search Tree (BST) and a target value. Our task is to delete the node with the given value from the BST. A BST is a tree data structure where the left subtree contains nodes with values less than the root, and the right subtree contains nodes with values greater than or equal to the root.
Implementation:
class Node:
def __init__(self, val):
self.val = val
self.left = None
self.right = None
def delete_node_in_bst(root, target):
if root is None:
return None
if target < root.val:
# Target is in the left subtree
root.left = delete_node_in_bst(root.left, target)
elif target > root.val:
# Target is in the right subtree
root.right = delete_node_in_bst(root.right, target)
else:
# Target is found
if root.left is None and root.right is None:
# Node has no children
return None
elif root.left is None:
# Node has only right child
return root.right
elif root.right is None:
# Node has only left child
return root.left
else:
# Node has both children
# Find the smallest node in the right subtree
min_node = find_min_node(root.right)
# Copy the value of the smallest node to the current node
root.val = min_node.val
# Delete the smallest node from the right subtree
root.right = delete_node_in_bst(root.right, min_node.val)
return root
def find_min_node(root):
if root is None:
return None
while root.left is not None:
root = root.left
return rootExplanation:
We start by defining a
Nodeclass that represents a node in the BST. Each node has aval(value), and pointers to its left and right child nodes.The
delete_node_in_bstfunction takes the root node of the BST and the target value as inputs. It recursively traverses the BST to find the target node.If the target value is less than the root's value, we know it must be in the left subtree, so we recursively call the function on the
root.leftnode. Similarly, if the target value is greater than the root's value, we recursively call the function on theroot.rightnode.When we find the target node, we have to consider three cases:
If the node has no children (both
root.leftandroot.rightareNone), we simply returnNone, which effectively deletes the node from the BST.If the node has only one child (either
root.leftorroot.right), we return the child node, which replaces the deleted node in the BST.If the node has both children, we find the smallest node in the right subtree (using the
find_min_nodehelper function) and replace the current node's value with the smallest node's value. Then, we recursively delete the smallest node from the right subtree.
The
find_min_nodefunction traverses the right subtree of a given node to find the smallest node (the leftmost node).
Example:
Suppose we have the following BST:
10
/ \
5 15
/ \ / \
2 7 12 20If we want to delete the node with value 15, we can call the delete_node_in_bst function as follows:
root = delete_node_in_bst(root, 15)The resulting BST will be:
10
/ \
5 12
/ \ / \
2 7 7 20Real-World Applications:
Deleting nodes from a BST is a common operation in various real-world applications, such as:
Maintaining sorted data in databases
Implementing priority queues
Storing and managing hierarchical data structures
Deleting items from a shopping cart or inventory system
optimal_division
Problem Statement:
You are given a number n and a set of positive integers cuts. Your goal is to divide n into the smallest possible number of integers using only the cuts in the set.
Best & Performant Solution (Dynamic Programming):
The best solution to this problem is a dynamic programming solution. It works by building up a table where each element represents the minimum number of cuts needed to divide a portion of n.
Here's how the algorithm works:
Initialize a table
dpof sizen+1, wheredp[0] = 0.For each element
iinn+1, iterate through the cuts:If
i - cut[j] >= 0, then calculatedp[i] = min(dp[i], dp[i - cut[j]] + 1)
Return
dp[n].
Breakdown:
Initialize the table: We start by initializing a table of size
n+1to store the minimum number of cuts needed to divide each portion ofn. We setdp[0]to 0 because no cuts are needed to divide 0.Iterate through the cuts: For each element
iin the table, we iterate through the set of cuts. If we can divideiusing a cutcut[j], we calculate the minimum number of cuts needed to divide the remaining portioni - cut[j], and add 1 to it (for the current cut). We store this value indp[i]if it is smaller than the current value.Return the result: Finally, we return
dp[n], which represents the minimum number of cuts needed to dividen.
Real-World Application:
This problem has applications in many real-world scenarios, such as:
Cutting a piece of wood into smaller pieces
Dividing a set of tasks among a team of workers
Optimizing the layout of a warehouse or factory
Python Implementation:
def optimal_division(n, cuts):
dp = [float('inf')] * (n + 1)
dp[0] = 0
for i in range(1, n + 1):
for cut in cuts:
if i - cut >= 0:
dp[i] = min(dp[i], dp[i - cut] + 1)
return dp[n]Example:
n = 7
cuts = [2, 3, 5]
result = optimal_division(n, cuts)
print(result) # Output: 2In this example, the optimal way to divide n into smaller integers is to make two cuts: one at cut[0] = 2 and another at cut[2] = 5. This results in three smaller integers: 2, 2, and 3.
beautiful_array
Beautiful Array
Given an integer n, return any array containing n unique integers such that they are in beautiful arrangement. That is, the array is beautiful if it is alternatingly strictly increasing and strictly decreasing.
Examples:
Input: n = 3
Output: [1, 2, 3]Input: n = 4
Output: [1, 3, 2, 4]Approach:
Create a base array of 1 to n:
arr = [1, 2, ..., n]Split the array into two:
left = arr[:n//2],right = arr[n//2:]Sort the left array in ascending order:
left.sort()Reverse the right array:
right.reverse()Merge the two arrays:
res = left + right
Python Implementation:
def beautifulArray(n: int) -> list[int]:
# Create a base array of 1 to n
arr = list(range(1, n+1))
while len(arr) > 1:
# Split the array into two
left = arr[:len(arr)//2]
right = arr[len(arr)//2:]
# Sort the left array in ascending order
left.sort()
# Reverse the right array
right.reverse()
# Merge the two arrays
arr = left + right
return arrReal-World Applications
Beautiful arrays can be used in various real-world applications, such as:
Permutation generation: Generating beautiful arrays is a way to generate permutations with alternating increases and decreases.
Optimization problems: Beautiful arrays can be used in optimization problems where alternating increases and decreases are desired.
Scheduling algorithms: Beautiful arrays can be used to schedule tasks in a way that ensures a balanced load.
card_flipping_game
Problem:
You have a deck of cards with N hidden faces. You can only flip one card at a time, and you can only flip the card directly to your left or right. What is the minimum number of flips required to reveal all the faces?
Solution:
The best and performant solution is to start from the center of the deck and flip the card to the right. Then, flip the card to the left. Repeat this process until all cards are revealed.
Breakdown:
Starting from the center: This ensures that you have the same number of cards to flip on both sides.
Flipping to the right: This exposes the card to the left.
Flipping to the left: This exposes the card to the right.
Example:
Let's say you have a deck of 5 cards:
[1, 2, 3, 4, 5]Flip card 3 to the right:
[1, 2, 3, 4, 5] -> [1, 2, 4, 3, 5]Flip card 3 to the left:
[1, 2, 4, 3, 5] -> [1, 2, 4, 5, 3]Flip card 2 to the right:
[1, 2, 4, 5, 3] -> [1, 2, 5, 4, 3]Flip card 2 to the left:
[1, 2, 5, 4, 3] -> [1, 2, 5, 3, 4]Flip card 1 to the right:
[1, 2, 5, 3, 4] -> [1, 2, 5, 4, 3]All cards are now revealed. It took 5 flips.
Real-World Applications:
This algorithm can be applied in real-world scenarios such as:
Optimizing card shuffling algorithms in games
Minimizing time spent flipping pancakes or burgers
Scheduling tasks to minimize execution time
valid_parenthesis_string
Problem Statement:
Determine if a given string of parentheses is valid. A valid string of parentheses follows these rules:
Open parentheses must have a corresponding closing parentheses.
Parentheses must appear in pairs with no unmatched parentheses.
The opening of a new pair of parentheses must always precede its corresponding closing parentheses.
Example:
Input: "()"
Output: True
Input: "()[]{}"
Output: True
Input: "(]"
Output: FalseSolution:
Approach:
We can use a stack to keep track of the opening parentheses. While iterating through the string, we push opening parentheses onto the stack and pop them off when we encounter a matching closing parentheses. If the stack is empty at the end of the string, the string is valid.
Algorithm:
Initialize an empty stack.
Iterate through the string:
If the current character is an opening parentheses, push it onto the stack.
If the current character is a closing parentheses, check if the stack is empty. If it is, return False. Otherwise, pop the top element from the stack.
After iterating through the string, check if the stack is empty. If it is, return True. Otherwise, return False.
Code Implementation:
def valid_parenthesis_string(s):
stack = []
for char in s:
if char == '(':
stack.append(char)
elif char == ')':
if stack:
stack.pop()
else:
return False
return not stackExplanation:
The valid_parenthesis_string function takes a string s as input. It initializes an empty stack to keep track of opening parentheses.
It then iterates through the string. If the current character is an opening parentheses, it is pushed onto the stack. If the current character is a closing parentheses, it checks if the stack is empty. If it is, the string is not valid and the function returns False. Otherwise, it pops the top element from the stack.
After iterating through the string, the function checks if the stack is empty. If it is, the string is valid and the function returns True. Otherwise, it returns False.
Applications:
Validating parentheses is a common task in programming, especially when working with expressions and data structures like stacks and queues. For example, it is used to ensure that HTML and XML documents are properly formatted, and to check the validity of mathematical expressions.
sentence_similarity_ii
Sentence Similarity II
Problem Statement:
Given two sentences, determine if they have similar meanings.
Example:
sentence1 = "I love dogs."
sentence2 = "Dogs are my favorite."Output:
True
Approach:
Tokenize the sentences: Split the sentences into individual words.
Stem the words: Remove the suffixes from the words to get their root form (e.g., "dogs" -> "dog").
Create bag-of-words representations: Count the occurrences of each stemmed word in both sentences.
Calculate cosine similarity: Use the cosine similarity formula to measure the similarity between the two bag-of-words representations.
Code Implementation:
import nltk
from nltk.stem import PorterStemmer
from sklearn.metrics.pairwise import cosine_similarity
def sentence_similarity(sentence1, sentence2):
# Tokenize the sentences
tokens1 = nltk.word_tokenize(sentence1)
tokens2 = nltk.word_tokenize(sentence2)
# Stem the words
stemmer = PorterStemmer()
stemmed_tokens1 = [stemmer.stem(token) for token in tokens1]
stemmed_tokens2 = [stemmer.stem(token) for token in tokens2]
# Create bag-of-words representations
bag_of_words1 = nltk.FreqDist(stemmed_tokens1)
bag_of_words2 = nltk.FreqDist(stemmed_tokens2)
# Calculate cosine similarity
similarity = cosine_similarity([list(bag_of_words1.values())], [list(bag_of_words2.values())])
# Return True if similarity is above a threshold (e.g., 0.5)
return similarity[0][0] > 0.5Explanation:
The
nltk.word_tokenize()function splits the sentences into individual words.The
PorterStemmer()class removes suffixes from the words to get their root form.The
nltk.FreqDist()function counts the occurrences of each word in the list.The
cosine_similarity()function calculates the cosine similarity between two vectors.If the cosine similarity is above a threshold, the sentences are considered similar.
Potential Applications:
Natural language processing (NLP) tasks like text classification, summarization, and question answering.
Chatbots and virtual assistants to understand user queries.
Search engines to find relevant documents.
Marketing and advertising to identify similar products or services.
2_keys_keyboard
2-Keys Keyboard
Problem Statement: You have a keyboard with only two keys: 'A' and 'B'. You want to type a target string, and you can type any key 'A' or 'B' at any time. Return the minimum number of times you have to press the keys to type the target string.
Understanding the Problem: Imagine you have a typewriter with only two keys, 'A' and 'B'. To type a given word, you need to press these keys strategically to create the desired sequence of letters. The goal is to find the minimum number of keystrokes required to type the word.
Solution: The key to solving this problem lies in identifying patterns within the target string. Here's a step-by-step solution to find the minimum number of keystrokes:
Split the String into Blocks:
Divide the target string into consecutive blocks of the same letter ('A' or 'B'). For example, "AABBB" can be split into ["AA", "BBB"].
Count Keystrokes for Each Block:
For each block, the number of keystrokes required to type it is simply its length. For instance, "AA" requires 2 keystrokes, while "BBB" requires 3.
Minimum Keystrokes:
To type the target string, you need to type each block in sequence. The minimum number of keystrokes is simply the sum of the keystrokes required for each block. For example, "AABBB" has blocks ["AA", "BBB"], which require 2 + 3 = 5 keystrokes in total.
Implementation:
def min_keystrokes(target):
# Split the string into blocks of same letter
blocks = []
curr_block = ""
for char in target:
if curr_block and char != curr_block[-1]:
blocks.append(curr_block)
curr_block = ""
curr_block += char
blocks.append(curr_block)
# Count keystrokes for each block
keystrokes = 0
for block in blocks:
keystrokes += len(block)
return keystrokes
target = "AABBB"
result = min_keystrokes(target)
print(result) # Output: 5Applications: This problem has applications in various fields, including:
Text Manipulation: Optimizing the efficiency of text processing algorithms.
Data Compression: Minimizing the size of data files by efficiently encoding strings.
Sequence Analysis: Identifying patterns and optimizing sequences in bioinformatics or computer science.
unique_substrings_in_wraparound_string
Problem Statement: Given a string s, return the number of unique substrings that can be formed by wrapping around the string.
Example:
Input: "pqr"
Output: 6
Explanation: The unique substrings that can be formed are: "p", "q", "r", "pq", "qr", "pr".Approach: We can solve this problem using a sliding window approach.
Algorithm:
Start with a window of size 1.
Use a hashmap to store the number of occurrences of each character in the window.
Slide the window by moving the left pointer to the right until the number of unique characters in the window is greater than or equal to k.
At each step, update the hashmap to reflect the change in the window.
If the number of unique characters in the window is greater than or equal to k, then the substring is a valid substring and we increment the count of unique substrings.
Repeat steps 2-5 until the left pointer reaches the end of the string.
Implementation:
def unique_substrings_in_wraparound_string(s: str) -> int:
"""
Returns the number of unique substrings that can be formed by wrapping around the string.
Parameters:
s: The string to check.
Returns:
The number of unique substrings.
"""
# Create a hashmap to store the number of occurrences of each character in the window.
char_counts = {}
# Initialize the count of unique substrings.
unique_substrings = 0
# Start with a window of size 1.
left = 0
right = 1
# Slide the window until the left pointer reaches the end of the string.
while left < right and right <= len(s):
# Update the hashmap to reflect the change in the window.
if s[right - 1] in char_counts:
char_counts[s[right - 1]] += 1
else:
char_counts[s[right - 1]] = 1
# Check if the number of unique characters in the window is greater than or equal to k.
if len(char_counts) >= right - left:
# If the number of unique characters in the window is greater than or equal to k, then the substring is a valid substring and we increment the count of unique substrings.
unique_substrings += 1
# Slide the window by moving the left pointer to the right.
while len(char_counts) > right - left:
char_counts[s[left]] -= 1
if char_counts[s[left]] == 0:
del char_counts[s[left]]
left += 1
# Move the right pointer to the right.
right += 1
# Return the count of unique substrings.
return unique_substringsReal-World Applications: This problem can be used to find the number of unique substrings in a DNA sequence, which is useful for DNA analysis. It can also be used to find the number of unique substrings in a protein sequence, which is useful for protein analysis.
reach_a_number
Problem Statement: Given a starting number num, calculate the number of steps it takes to reach target. In each step, you can either add or subtract 1 from num. Steps must be done in the correct order and cannot be skipped. Return the minimum number of steps to reach the target, or -1 if the target cannot be reached.
Simplified Explanation:
You have a starting point and a destination. To reach the destination, you can take small steps of +1 or -1. The goal is to reach the destination in the fewest possible steps.
Example:
Starting Number: 3
Destination: 9
Solution:
Step 1: Add 1 to 3, resulting in 4.
Step 2: Add 1 to 4, resulting in 5.
Step 3: Add 1 to 5, resulting in 6.
Step 4: Add 1 to 6, resulting in 7.
Step 5: Add 1 to 7, resulting in 8.
Step 6: Add 1 to 8, resulting in 9.
Total Steps: 6
Code Implementation:
def reach_a_number(num):
"""
:param num: The starting number.
:return: The minimum number of steps to reach the target, or -1 if the target cannot be reached.
"""
# Initialize the number of steps.
steps = 0
# While the target has not been reached.
while num != 0:
# If the target is positive.
if num > 0:
# Add 1 to the target.
num -= 1
# Otherwise, the target is negative.
else:
# Subtract 1 from the target.
num += 1
# Increment the number of steps.
steps += 1
# Return the number of steps.
return steps Applications:
In computer science, this problem is a classic example of a "greedy" algorithm, where you make the best choice at each step without considering the future.
It can also be used in real-world applications, such as calculating the minimum number of moves needed to solve a puzzle or reaching a destination in a game.
monotone_increasing_digits
Problem Statement
Given a non-negative integer n, return the largest number that can be formed by rearranging its digits in a non-decreasing order.
Solution
Step 1: Convert the integer to a list of digits
We can use the str() function to convert the integer to a string, and then use the list() function to convert the string to a list of characters. Finally, we can use the int() function to convert each character back to an integer.
def convert_to_digits(n):
digits = list(str(n))
digits = [int(digit) for digit in digits]
return digitsStep 2: Sort the digits in non-decreasing order
We can use the sorted() function to sort the digits in non-decreasing order.
def sort_digits(digits):
digits.sort()
return digitsStep 3: Convert the sorted list of digits back to an integer
We can use the join() function to concatenate the digits into a string, and then use the int() function to convert the string back to an integer.
def convert_to_integer(digits):
number = int(''.join(map(str, digits)))
return numberStep 4: Put it all together
We can now put all the steps together to create a function that returns the largest number that can be formed by rearranging the digits of a non-negative integer in a non-decreasing order.
def monotone_increasing_digits(n):
digits = convert_to_digits(n)
digits = sort_digits(digits)
number = convert_to_integer(digits)
return numberExample
n = 1234
result = monotone_increasing_digits(n)
print(result) # Output: 1234Applications
This algorithm can be used in a variety of real-world applications, such as:
Data sorting: The algorithm can be used to sort a list of numbers in non-decreasing order.
Financial analysis: The algorithm can be used to find the highest possible value of a stock portfolio.
Scheduling: The algorithm can be used to find the optimal schedule for a set of tasks.
sort_an_array
Problem Statement
Given an unsorted array of integers, sort the array in ascending order.
Solution
1. Built-in Sort Function
The simplest and most efficient way to sort an array in Python is to use the built-in sort() function:
nums = [5, 3, 1, 2, 4]
nums.sort()
print(nums) # Output: [1, 2, 3, 4, 5]2. Bubble Sort
Bubble sort repeatedly compares adjacent elements and swaps them if they are in the wrong order. It continues to iterate through the array until no more swaps are needed.
def bubble_sort(nums):
n = len(nums)
for i in range(n):
swapped = False
for j in range(1, n - i):
if nums[j] < nums[j - 1]:
nums[j], nums[j - 1] = nums[j - 1], nums[j]
swapped = True
if not swapped:
break3. Merge Sort
Merge sort follows the divide-and-conquer approach. It recursively divides the array into smaller chunks, sorts them, and then merges them back together.
def merge_sort(nums):
n = len(nums)
if n <= 1:
return nums
mid = n // 2
left = merge_sort(nums[:mid])
right = merge_sort(nums[mid:])
return merge(left, right)
def merge(left, right):
merged = []
l = 0
r = 0
while l < len(left) and r < len(right):
if left[l] < right[r]:
merged.append(left[l])
l += 1
else:
merged.append(right[r])
r += 1
merged.extend(left[l:])
merged.extend(right[r:])
return merged4. Quick Sort
Quick sort also follows the divide-and-conquer approach. It selects a pivot element, partitions the array into two subarrays, and recursively sorts each subarray.
def quick_sort(nums):
n = len(nums)
if n <= 1:
return nums
pivot = nums[0]
left = [x for x in nums[1:] if x < pivot]
right = [x for x in nums[1:] if x >= pivot]
return quick_sort(left) + [pivot] + quick_sort(right)Applications
Sorting algorithms are used in various real-world applications, including:
Data processing and analysis
Searching and retrieval in databases
Sorting files and directories
Optimizing search results
Machine learning and artificial intelligence
split_linked_list_in_parts
ERROR OCCURED split_linked_list_in_parts
Can you please implement the best & performant solution for the given leetcode problem in python, then simplify and explain the given content for competitive coding?
breakdown and explain each topic or step in detail and simplified manner (simplify in very plain english like explaining to a child).
give real world complete code implementations and examples for each. provide potential applications in real world.
The response was blocked.
maximum_sum_circular_subarray
Problem Statement
Given an array of n numbers, find the maximum sum of a circular subarray.
A circular subarray is a subarray that may wrap around the end of the array. For example, in the array [1, 2, 3, 4, 5], the subarray [2, 3, 4, 5, 1] is a circular subarray.
Solution
The problem can be solved using a dynamic programming approach. Let's define dp[i] as the maximum sum of a circular subarray ending at index i. Then, we can compute dp[i] as follows:
dp[i] = max(dp[i-1] + nums[i], nums[i])This means that dp[i] is either the maximum sum of a circular subarray ending at index i-1 plus the number at index i, or the number at index i itself.
To find the overall maximum sum of a circular subarray, we can simply take the maximum value of dp[i] for all i in the range [0, n-1].
Here is the Python code for the solution:
def maximum_sum_circular_subarray(nums):
n = len(nums)
dp = [0] * n
# Compute dp[i] for all i in the range [0, n-1]
for i in range(n):
dp[i] = max(dp[i-1] + nums[i], nums[i])
# Find the maximum value of dp[i] for all i in the range [0, n-1]
return max(dp)Example
Consider the array nums = [1, 2, 3, 4, 5]. The following table shows the values of dp[i] for all i in the range [0, 4]:
0
1
1
3
2
6
3
10
4
15
The maximum value of dp[i] is 15, which corresponds to the circular subarray [2, 3, 4, 5, 1].
Applications
The problem of finding the maximum sum of a circular subarray has applications in a variety of real-world problems. For example, it can be used to:
Find the most profitable subsegment of a stock price time series.
Find the best way to schedule a set of tasks to maximize their total profit.
Find the most efficient way to travel a circular route.
construct_binary_tree_from_string
Problem: Construct Binary Tree from String Statement: Given a string representing a binary tree in a pre-order traversal, construct the binary tree.
Solution:
Breakdown:
Pre-order traversal: A traversal where we visit the root, then the left subtree, then the right subtree.
Parse the string: We can split the string into a list of tokens, where each token represents a node in the tree.
Recursively build the tree:
If the token is "-", indicating an empty node, return None.
Otherwise, create a new node with the token as its value.
Recursively build the left and right subtrees by parsing the remaining tokens.
Return the root node.
Implementation:
def construct_binary_tree_from_string(s):
"""
:type s: str
:rtype: TreeNode
"""
# Split the string into tokens
tokens = s.split(" ")
# Parse the root node
root = parse_node(tokens)
return root
def parse_node(tokens):
if tokens[0] == "-":
return None
# Create a new node
node = TreeNode(int(tokens[0]))
# Recursively build the left and right subtrees
left = parse_node(tokens[1:])
right = parse_node(tokens[1:])
# Set the left and right children of the current node
node.left = left
node.right = right
# Return the current node
return nodeExample:
Input: "1 2 3 - - 4 - -"
Output:
1
/ \
2 3
/
4
Explanation:
The string represents the following pre-order traversal:
1
/ \
2 3
/
4
Therefore, the output is the binary tree shown above.Applications:
Parsing tree-like structures stored as strings.
Reconstructing trees from serialized data.
Implementing binary tree operations on trees parsed from strings.
output_contest_matches
Problem Statement:
Given a list of contest matches, where each match is represented by a pair of team indices, compute the winner for each match and the final winner.
Solution:
1. Define a Winner Function:
def winner(team1, team2):
# Check if team1 won
if team1 > team2:
return team1
# Otherwise, team2 won
else:
return team22. Create a Bracket Tree:
def create_bracket_tree(matches):
# Initialize the bracket as a dictionary
bracket = {}
# Iterate over the matches
for match in matches:
# Add the winner of the match to the bracket
winner = winner(match[0], match[1])
bracket[winner] = []3. Simulate Matches:
def simulate_matches(bracket):
# Create a stack to store the teams that have already played
played = []
# Iterate over the bracket
for team, opponents in bracket.items():
# If the team is in the stack, it has already played
if team in played:
continue
# Simulate the matches against all opponents
for opponent in opponents:
if opponent not in played:
winner = winner(team, opponent)
played.append(winner)
bracket[winner].append(opponent)4. Find Final Winner:
def find_final_winner(bracket):
# Get the list of teams that have not played yet
remaining_teams = [team for team in bracket if team not in played]
# The final winner is the only remaining team
return remaining_teams[0]Time Complexity: O(n * log n), where n is the number of teams.
Example:
matches = [(1, 2), (3, 4), (5, 6), (7, 8)]
bracket = create_bracket_tree(matches)
simulate_matches(bracket)
final_winner = find_final_winner(bracket)
print(f"The final winner is Team {final_winner}")Applications:
Computing the winner of a tournament or competition.
Generating a bracket for a sports competition.
Scheduling matches in a round-robin tournament.
longest_word_in_dictionary
Longest Word in Dictionary
Problem Statement:
You are given a dictionary of words and a string formed by concatenating these words. Find the longest word in the dictionary that is a subsequence of the given string.
Solution:
Define Subsequence: A subsequence is a sequence that can be obtained from another sequence by deleting some elements without changing the order of the remaining elements. For example, "abc" is a subsequence of "abcbd".
Create a Trie: A Trie is a data structure that organizes words based on their prefixes. Each node in the Trie corresponds to a letter in the alphabet. For this problem, create a Trie using the words in the dictionary.
Traverse the String: Start at the first character of the concatenated string. For each character, check if it matches the next character in the longest word in the dictionary that is a subsequence of the string so far. If so, move to the next character in the dictionary word. If not, backtrack to the previous node in the Trie and try a different path.
Update Longest Word: If you reach the end of a dictionary word that is a subsequence of the string, update the longest word variable to be the current dictionary word.
Python Implementation:
class TrieNode:
def __init__(self):
self.children = {}
self.is_word = False
class Trie:
def __init__(self):
self.root = TrieNode()
def insert(self, word):
node = self.root
for char in word:
if char not in node.children:
node.children[char] = TrieNode()
node = node.children[char]
node.is_word = True
def longest_word_in_dictionary(dictionary, string):
trie = Trie()
for word in dictionary:
trie.insert(word)
longest_word = ""
start = 0
while start < len(string):
current_word = ""
node = trie.root
for i in range(start, len(string)):
if string[i] in node.children:
node = node.children[string[i]]
current_word += string[i]
if node.is_word and len(current_word) > len(longest_word):
longest_word = current_word
else:
break
start += 1
return longest_wordExplanation:
The
TrieNodeclass represents a node in the Trie. It has a dictionary of child nodes and a boolean flag indicating if it represents the end of a word.The
Trieclass creates and manages the Trie. It inserts words into the Trie by iterating through the characters of each word and creating nodes for any missing characters.The
longest_word_in_dictionaryfunction takes a dictionary and a string as input. It creates a Trie from the dictionary, then iterates through the string, trying to find the longest word in the Trie that is a subsequence of the string. It does this by starting at the first character of the string and moving to the next character in the dictionary word if it matches the current character in the string. If the current character in the string does not match any character in the Trie, it backtracks to the previous node and tries a different path. If it reaches the end of a dictionary word that is a subsequence of the string, it updates the longest word variable.
Real-World Applications:
This algorithm can be used in various real-world applications, such as:
Finding the longest common subsequence of two strings
Spell checking
Autocomplete suggestions
Natural language processing
stone_game
Stone Game
Problem:
Two players are playing a game with a pile of stones. Each player takes turns removing either 1 or 2 stones from the pile. The player who takes the last stone wins.
Given an array of integers representing the number of stones in the pile, determine if the first player can win the game.
Best Solution:
Using Dynamic Programming
Breakdown:
Define a DP Array: Create a 2D array
dpof size(n+1) * (n+1), wherenis the number of stones in the pile.dp[i][j]will represent whether the first player can win if the number of stones in the pile is betweeniandj, inclusive.Base Cases: Fill in the base cases:
dp[i][i] = Truefor alli, as the first player can always win if there is only one stone left.dp[i][i+1] = Truefor alli, as the first player can also win if there are only two stones left.
Recursive Relation: For each range
(i, j)such thati+2 <= j, calculatedp[i][j]as follows:If
dp[i+1][j] = Falseanddp[i+2][j] = False, then the first player can force a win by taking 1 or 2 stones, sodp[i][j] = True.Otherwise,
dp[i][j] = False, meaning the first player cannot force a win.
Final Answer: Return
dp[0][n], which indicates whether the first player can win if the initial number of stones isn.
Time Complexity: O(n^3), where n is the number of stones.
Python Implementation:
def can_win_nim(stones):
n = len(stones)
dp = [[False] * (n+1) for _ in range(n+1)]
for i in range(n):
dp[i][i] = True
dp[i][i+1] = True
for i in range(n-1,-1,-1):
for j in range(i+2, n+1):
dp[i][j] = not (dp[i+1][j] and dp[i+2][j])
return dp[0][n]Real-World Applications:
This problem can be applied in game theory, where it is used to analyze optimal strategies for games with multiple players and incomplete information. It can also be used in computer science for resource allocation and scheduling problems.
fruit_into_baskets
Fruit Into Baskets
Problem Statement:
You have some fruit baskets of different capacities. You want to fill the baskets with fruits. You have an array of integers where each integer represents the capacity of a basket. You also have an array of integers where each integer represents the type of fruit.
The task is to find the maximum number of fruits that can be placed into the baskets without violating the following rule:
Each basket can have only two different types of fruits.
Solution:
The solution is based on the sliding window technique. We maintain a window of size 2 (two baskets) and slide it over the array of fruit capacities. We keep track of the maximum number of fruits that can be placed into the baskets within the window.
Python Code:
def max_fruits_into_baskets(baskets, fruits):
"""
Finds the maximum number of fruits that can be placed into the baskets
without violating the two types of fruits rule.
Parameters:
baskets (list): A list of integers representing the capacities of the baskets.
fruits (list): A list of integers representing the types of fruit.
Returns:
int: The maximum number of fruits that can be placed into the baskets.
"""
# Initialize the window start and end pointers.
start = 0
end = 0
# Initialize the count of the two different types of fruits in the window.
fruit_counts = {fruits[start]: 1, fruits[end]: 1}
# Initialize the maximum number of fruits in the window.
max_fruits = 0
# Slide the window over the array of fruit capacities.
while end < len(baskets):
# Increment the count of the current fruit type in the window.
fruit_counts[fruits[end]] += 1
# Update the maximum number of fruits in the window.
max_fruits = max(max_fruits, sum(fruit_counts.values()))
# If the number of different types of fruits in the window exceeds 2, move the start pointer forward until it reaches a basket with a different fruit type.
while len(fruit_counts) > 2:
# Decrement the count of the fruit type at the start of the window.
fruit_counts[fruits[start]] -= 1
# If the count of the fruit type at the start of the window becomes 0, remove it from the dictionary.
if fruit_counts[fruits[start]] == 0:
del fruit_counts[fruits[start]]
# Move the start pointer forward.
start += 1
# Move the end pointer forward.
end += 1
# Return the maximum number of fruits in the window.
return max_fruitsExample:
baskets = [2, 3, 4, 1, 5]
fruits = [1, 2, 1, 2, 2]
max_fruits = max_fruits_into_baskets(baskets, fruits)
print(max_fruits) # Output: 5Applications:
The algorithm can be applied in any scenario where you need to allocate resources (e.g., baskets) to items (e.g., fruits) with constraints (e.g., the two types of fruits rule). For example, the algorithm can be used to:
Allocate students to classrooms with limited capacity and subject constraints.
Assign tasks to workers with specific skills and workload limits.
Distribute products to warehouses with varying storage capacities and product compatibility constraints.
print_binary_tree
Problem Statement: LeetCode 589. N-ary Tree Preorder Traversal
Problem Breakdown:
What is an N-ary tree?
An N-ary tree is a tree where each node can have any number of children (as opposed to a binary tree which has at most two children).
What is Preorder Traversal?
Preorder traversal is a depth-first search algorithm that visits the root node, followed by its children (in left-to-right order), and then recursively visits the children of each child.
Python Solution:
def preorder(root):
if not root:
return []
result = [root.val]
for child in root.children:
result.extend(preorder(child))
return resultExplanation:
Check if the root node is None. If it is, return an empty list.
Create a result list and add the root node's value to it.
Iterate through the root node's children.
For each child, recursively call the preorder function and append the result to the result list.
Return the result list.
Real-World Application:
Preorder traversal can be used to print the nodes of an N-ary tree in a human-readable format. It can also be used to serialize a tree so that it can be stored or transmitted.
Example:
# Create an N-ary tree
tree = Node(1)
tree.children.append(Node(2))
tree.children.append(Node(3))
tree.children[0].children.append(Node(4))
tree.children[0].children.append(Node(5))
# Perform preorder traversal
result = preorder(tree)
# Print the result
print(result) # [1, 2, 3, 4, 5]Simplified Explanation for a Child:
Imagine a tree with branches and leaves. Preorder traversal is like walking along the branches of the tree, starting at the top and going down. As you go, you write down the value of each node (like a leaf or a branch) that you see. When you reach the end of a branch, you go back up to the last branch you came from and write down the values of the nodes on that branch. You keep doing this until you have written down all the values in the tree.
my_calendar_ii
Problem:
The MyCalendar class is a scheduling system that allows you to book events over a given period. Implement a function myCalendar.book, which takes two arguments, start and end, and returns True if the event can be booked, and False otherwise.
Simplified Explanation:
Imagine a calendar where each day is divided into slots. You want to book an event that starts at a particular slot and ends at another slot. The function myCalendar.book checks if the requested slots are available. If they are, it "reserves" them and returns True. If the slots are already booked or overlap with an existing event, it returns False.
Code Implementation:
class MyCalendar:
def __init__(self):
self.calendar = {} # Dictionary to store booked slots
def book(self, start, end):
# Check if the requested slots are available
for i in range(start, end):
if i in self.calendar:
return False # Slot already booked
# If available, reserve the slots
for i in range(start, end):
self.calendar[i] = True
return True # Event booked successfullyReal-World Applications:
Meeting scheduling: A meeting scheduling system uses a calendar to track available time slots and book meetings accordingly.
Resource management: Businesses use calendars to schedule resources such as equipment, vehicles, or rooms to avoid conflicts and optimize usage.
Event planning: Event planners use calendars to manage bookings for events such as conferences, weddings, or parties.
Time management: Individuals use calendars to keep track of appointments, deadlines, and other important events.
sort_characters_by_frequency
Sort Characters By Frequency
Given a string, sort its characters by their frequency in descending order.
Example 1:
Input: "tree"
Output: "eert"Example 2:
Input: "cccaaa"
Output: "aaaccc"Solution
The most straightforward way to solve this problem is to use a dictionary to count the frequency of each character. Then, sort the dictionary by the frequency in descending order and concatenate the characters accordingly.
from collections import Counter
def sort_characters_by_frequency(s: str) -> str:
# Create a dictionary to store the frequency of each character
char_count = Counter(s)
# Sort the dictionary by frequency in descending order
sorted_char_count = sorted(char_count.items(), key=lambda x: x[1], reverse=True)
# Concatenate the characters accordingly
return ''.join([char * count for char, count in sorted_char_count])Breakdown
Create a dictionary to store the frequency of each character. This can be done using the
Counterclass from thecollectionsmodule.Sort the dictionary by frequency in descending order. This can be done using the
sortedfunction and thekeyandreversearguments.Concatenate the characters accordingly. This can be done using a list comprehension and the
joinmethod.
Complexity Analysis
Time complexity: O(n log n), where n is the length of the string. This is because sorting the dictionary takes O(n log n) time.
Space complexity: O(n), since we need to store the frequency of each character.
Applications
This problem can be used to solve a variety of real-world problems, such as:
Text compression. By sorting the characters by frequency, we can create a Huffman code, which is a lossless data compression technique.
Natural language processing. By sorting the characters by frequency, we can identify the most common words in a text, which can be used for tasks such as text classification and language modeling.
Cryptography. By sorting the characters by frequency, we can create a frequency analysis attack, which is a technique for breaking simple ciphers.
contiguous_array
Problem Statement:
Given an array of integers nums, find the maximum sum of a contiguous subarray.
Example:
Input: nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4] Output: 6 Explanation: The contiguous subarray [4, -1, 2, 1] has the maximum sum of 6.
Approaching the Problem
Brute Force Approach
The simplest solution is to iteratively check all possible contiguous subarrays and calculate their sums. The subarray with the maximum sum is the answer. However, this approach has a time complexity of O(n^2), which is very inefficient for large arrays.
Efficient Solution: Kadane's Algorithm
Kadane's algorithm is a greedy algorithm that finds the maximum subarray sum in linear time (O(n)). The key idea is to maintain a running sum, curr_sum, which represents the current contiguous sum, and a max_sum to track the maximum sum encountered so far.
Algorithm:
Initialize
curr_sumandmax_sumto 0.Iterate over the array
nums.For each element
nums[i]:Update
curr_sumby addingnums[i].Update
max_sumifcurr_sumis greater thanmax_sum.If
curr_sumbecomes negative, reset it to 0.
Example:
For the input nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]:
Iteration
nums[i]
curr_sum
max_sum
1
-2
-2
0
2
1
-1
0
3
-3
-4
0
4
4
0
0
5
-1
-1
0
6
2
1
1
7
1
2
2
8
-5
-3
2
9
4
1
4
In the end, max_sum will contain the maximum subarray sum, which is 6 in this case.
Simplified Explanation
Imagine you have a pile of coins, and your goal is to find the heaviest group of coins that are touching each other (a contiguous subarray).
Brute Force: You would weigh every possible combination of coins, which takes a lot of time.
Kadane's Algorithm: You would start by weighing the first coin. Then, for each subsequent coin, you would add its weight to the current heaviest group of coins (the running sum). If the current heaviest group becomes lighter than your overall heaviest group, you would reset the current group to the current coin's weight.
Real-World Applications
Finance: Analyzing stock prices to identify trends and potential investment opportunities.
Weather Forecasting: Calculating average temperatures or precipitation over a period of time.
Data Analysis: Identifying patterns and anomalies in time series data, such as sales figures or website traffic.
rabbits_in_forest
Problem:
Given an array of integers representing the number of rabbits of each age in a forest, find the total number of rabbits in the forest.
Simplifying the Problem:
Imagine you have a forest with many rabbits. Suppose you know the number of rabbits at each age in the forest. Your task is to figure out the total number of rabbits in the forest.
Solution:
Understanding the Problem: Each age group has a fixed number of rabbits. We need to find the sum of all the rabbits in each age group.
Code Implementation:
def RabbitsInForest(ages):
"""
:param ages: List of integers representing the number of rabbits of each age
:return: Total number of rabbits
"""
# Initialize a dictionary to store the number of rabbits at each age
age_count = {}
# Count the number of rabbits at each age
for age in ages:
if age not in age_count:
age_count[age] = 0
age_count[age] += 1
# Sum up the number of rabbits at each age
total_rabbits = 0
for age in age_count:
total_rabbits += age_count[age]
return total_rabbitsExample:
If you have a forest with the following number of rabbits at each age:
ages = [1, 2, 3, 4, 5]Calling the RabbitsInForest function with this list will return the total number of rabbits in the forest:
total_rabbits = RabbitsInForest(ages)
print(total_rabbits) # Output: 15Real-World Application:
This problem can be applied in real-world scenarios like wildlife conservation, where researchers want to estimate the total population of a particular animal species in a specific area. By knowing the age distribution of the population, they can make accurate estimates about the overall population size.
smallest_range_ii
Problem Statement (Simplified):
Imagine you have a set of numbers and you want to make the difference between the largest and smallest numbers as small as possible. To do this, you can either increase the smallest number or decrease the largest number. You can only do one of these operations at a time.
Your task is to find the smallest difference you can achieve by performing this operation over and over until there's only one number left.
Breakdown:
Numbers: The starting set of numbers.
Difference: The difference between the largest and smallest numbers.
Operation: Increasing the smallest number or decreasing the largest number by 1.
Goal: Minimize the difference.
Solution (Simplified):
Sort the numbers: This helps us easily find the largest and smallest numbers.
Loop until there's only one number left:
Calculate the difference: Find the difference between the largest and smallest numbers.
Check the operation: Choose the operation that will result in a smaller difference.
If increasing the smallest number results in a smaller difference, do it.
If decreasing the largest number results in a smaller difference, do it.
Return the final difference: After the loop completes, there's only one number left. The difference is 0.
Implementation (Python):
def smallest_range_ii(nums):
nums.sort() # Sort the numbers
# Initialize the difference as the initial difference between the largest and smallest numbers
diff = nums[-1] - nums[0]
# Loop until there's only one number left
while len(nums) > 1:
# Calculate the difference if we increase the smallest number
diff1 = nums[1] - nums[0]
# Calculate the difference if we decrease the largest number
diff2 = nums[-1] - nums[-2]
# Choose the operation that results in a smaller difference
if diff1 < diff2:
nums[0] += 1 # Increase the smallest number
else:
nums[-1] -= 1 # Decrease the largest number
# Update the difference with the new value
diff = min(diff, nums[-1] - nums[0])
# Return the final difference
return diffExample Usage:
nums = [3, 5, 7, 1]
result = smallest_range_ii(nums)
print(result) # Output: 2Real-World Applications:
This problem can be useful in various applications, such as:
Balancing data: Distributing items evenly among different categories to minimize discrepancies.
Resource allocation: Optimizing resource allocation to ensure equal distribution and prevent shortages or surpluses.
Scheduling: Creating schedules that minimize the time difference between events or appointments.
friends_of_appropriate_ages
Problem Statement:
Given a list of people's ages, return a list of ages that are considered appropriate for friendship.
Input:
ages = [16, 22, 28, 34, 40, 46, 52, 58, 64]Output:
[16, 22, 28, 34, 40]Optimal Solution in Python:
def friends_of_appropriate_ages(ages):
# Initialize an empty list to store the appropriate ages
appropriate_ages = []
# Iterate over the ages
for age in ages:
# Check if the age is less than or equal to 40
if age <= 40:
# If the age is less than or equal to 40, add it to the list of appropriate ages
appropriate_ages.append(age)
# Return the list of appropriate ages
return appropriate_agesBreakdown and Explanation:
Step 1: Initialize an empty list to store the appropriate ages. This list will hold the ages that are considered appropriate for friendship.
Step 2: Iterate over the ages. This loop iterates over each age in the input list.
Step 3: Check if the age is less than or equal to 40. This condition checks if the age is within the acceptable range for friendship.
Step 4: If the age is less than or equal to 40, add it to the list of appropriate ages. If the age meets the criteria, it is added to the list.
Step 5: Return the list of appropriate ages. This returns the list of ages that are suitable for friendship.
Time Complexity:
O(n), where n is the number of ages in the input list. The loop iterates over each age in the list, so the time complexity is proportional to the length of the list.
Space Complexity:
O(n), where n is the number of ages in the input list. The appropriate_ages list can potentially store all of the ages in the input list, so the space complexity is proportional to the length of the list.
Real-World Applications:
This algorithm can be used in any application that needs to determine which ages are appropriate for friendship. For example, it could be used in a social networking site to recommend friends for a user based on their age.
top_k_frequent_words
Problem Statement:
Given a list of words, find the top k most frequent words.
Solution:
Create a dictionary to store word frequencies:
Iterate over the words in the list.
For each word, check if it already exists in the dictionary.
If it does, increment its frequency.
If it doesn't, add it to the dictionary with a frequency of 1.
Sort the dictionary by frequency:
Use the
sorted()function with thekey=lambda x: x[1]argument to sort the dictionary by the value (frequency) of each key (word).The resulting list will have the words sorted in descending order of frequency.
Find the top k most frequent words:
Slice the sorted list to get the top k words.
Code Implementation:
from collections import Counter
import heapq
def top_k_frequent_words(words, k):
# Create a dictionary to store word frequencies
word_freq = Counter(words)
# Sort the dictionary by frequency
sorted_words = sorted(word_freq.items(), key=lambda x: x[1], reverse=True)
# Find the top k most frequent words
top_k_words = [word for word, freq in sorted_words[:k]]
return top_k_wordsExample:
words = ["the", "quick", "brown", "fox", "jumped", "over", "the", "lazy", "dog"]
k = 3
top_k_words = top_k_frequent_words(words, k)
print(top_k_words)
# Output: ['the', 'quick', 'dog']Applications:
Text summarization
Search engine optimization
Language modeling
sum_of_square_numbers
Problem: Calculate the sum of the squares of the first N natural numbers.
Solution:
Initialize the sum to 0.
Loop through the first N natural numbers (1 to N).
For each number, calculate its square and add it to the sum.
Code Implementation:
def sum_of_square_numbers(n):
sum = 0
for i in range(1, n + 1):
sum += i ** 2
return sumExample:
n = 5
result = sum_of_square_numbers(n)
print(result) # Output: 55Explanation:
The function initializes the
sumto 0.It loops through the numbers 1 to 5.
For each number, it squares it (e.g., 1 squared is 1, 2 squared is 4, etc.) and adds it to the
sum.After the loop, the
sumis 15.
Real-World Applications:
Calculating the sum of squares is useful in various applications, such as:
Summing up the squares of deviations from the mean in statistics.
Calculating the variance and standard deviation of a dataset.
Modeling physical systems where the square of a quantity is proportional to another quantity (e.g., the kinetic energy of an object is proportional to the square of its velocity).
design_circular_deque
Problem Statement:
Design a circular deque, which is a double-ended queue (deque) where elements can be added and removed from either end. It should support the following operations:
insertFront(value): Inserts an element at the front of the deque.insertLast(value): Inserts an element at the end of the deque.deleteFront(): Deletes an element from the front of the deque.deleteLast(): Deletes an element from the end of the deque.getFront(): Returns the element at the front of the deque.getRear(): Returns the element at the end of the deque.isEmpty(): Returns true if the deque is empty, false otherwise.isFull(): Returns true if the deque is full, false otherwise.
Solution:
We can implement a circular deque using an array. The key idea is to use two pointers: front and rear. front always points to the front of the deque, and rear points to the end of the deque.
class CircularDeque:
def __init__(self, capacity):
"""
Initialize your data structure here. Set the size of the deque to be capacity.
"""
self.capacity = capacity
self.arr = [None] * capacity
self.size = 0
self.front = 0
self.rear = 0
def insertFront(self, value):
"""
Inserts an element at the front of the deque.
"""
if self.isFull():
raise Exception("Deque is full")
self.front = (self.front - 1) % self.capacity
self.arr[self.front] = value
self.size += 1
def insertLast(self, value):
"""
Inserts an element at the end of the deque.
"""
if self.isFull():
raise Exception("Deque is full")
self.arr[self.rear] = value
self.rear = (self.rear + 1) % self.capacity
self.size += 1
def deleteFront(self):
"""
Deletes an element from the front of the deque.
"""
if self.isEmpty():
raise Exception("Deque is empty")
self.arr[self.front] = None
self.front = (self.front + 1) % self.capacity
self.size -= 1
def deleteLast(self):
"""
Deletes an element from the end of the deque.
"""
if self.isEmpty():
raise Exception("Deque is empty")
self.rear = (self.rear - 1) % self.capacity
self.arr[self.rear] = None
self.size -= 1
def getFront(self):
"""
Returns the element at the front of the deque.
"""
if self.isEmpty():
raise Exception("Deque is empty")
return self.arr[self.front]
def getRear(self):
"""
Returns the element at the end of the deque.
"""
if self.isEmpty():
raise Exception("Deque is empty")
return self.arr[self.rear]
def isEmpty(self):
"""
Returns true if the deque is empty, false otherwise.
"""
return self.size == 0
def isFull(self):
"""
Returns true if the deque is full, false otherwise.
"""
return self.size == self.capacitySimplification:
The circular deque is a data structure that allows insertion and deletion of elements from both the front and the rear. We can implement it using an array and two pointers: front and rear. front always points to the front of the deque, and rear points to the end of the deque.
When we insert an element at the front, we decrement front by 1. If front becomes negative, we wrap it around to the end of the array. When we insert an element at the rear, we increment rear by 1. If rear reaches the end of the array, we wrap it around to the front.
When we delete an element from the front, we increment front by 1. If front reaches the end of the array, we wrap it around to the front. When we delete an element from the rear, we decrement rear by 1. If rear becomes negative, we wrap it around to the end of the array.
Real-world Applications:
Circular deques can be used in various real-world applications, such as:
Implementing a queue or a stack
Managing a buffer pool
Caching data
Scheduling tasks
Example:
# Create a circular deque with a capacity of 5
deque = CircularDeque(5)
# Insert elements into the deque
deque.insertFront(1)
deque.insertFront(2)
deque.insertLast(3)
deque.insertLast(4)
deque.insertLast(5)
# Print the elements of the deque
print("Elements of the deque:", deque.arr)
# Get the front and rear elements
print("Front element:", deque.getFront())
print("Rear element:", deque.getRear())
# Delete elements from the deque
deque.deleteFront()
deque.deleteLast()
# Print the elements of the deque
print("Elements of the deque:", deque.arr)Output:
Elements of the deque: [2, 1, 3, 4, 5]
Front element: 2
Rear element: 5
Elements of the deque: [None, 1, 3, 4, None]find_the_derangement_of_an_array
Leetcode Problem: Find the Derangement of an Array
Problem Statement: Given an array of n integers, find the number of ways to rearrange the array such that no element appears in its original position.
Solution: The derangement of an array is a permutation of the array in which no element appears in its original position. To count the number of derangements, we can use the following recursive relation:
D(n) = (n-1) * (D(n-1) + D(n-2))where D(n) is the number of derangements of an array of size n.
Python Implementation:
def find_derangement(n):
if n == 0:
return 1
elif n == 1:
return 0
else:
return (n - 1) * (find_derangement(n - 1) + find_derangement(n - 2))Breakdown:
If the array is empty, there is only one possible derangement: the empty array itself.
If the array contains only one element, there are no derangements.
Otherwise, we can choose an element to appear in its original position. This leaves n-1 elements to be permuted.
For each of the n-1 elements, we have two options:
Place it in its original position.
Place it in a different position.
If we place it in a different position, we have two subproblems to solve:
The derangement of the remaining n-2 elements
The derangement of the remaining n-2 elements where the element we chose to place in its original position is excluded
Example:
find_derangement(4)Output:
9Real-World Applications:
Derangements are used in a variety of applications, such as:
Scheduling: Assigning tasks to people such that no person is assigned to their own task.
Combinatorics: Counting the number of ways to rearrange objects such that no object appears in its original position.
Probability: Calculating the probability that a random permutation is a derangement.
4_keys_keyboard
Problem Statement:
You are given a string containing letters from 'a' to 'c'. You have a keyboard with 4 keys: 'a', 'b', 'c', and 'backspace'. The 'backspace' key can delete the last character typed.
Your goal is to find the minimum number of keystrokes needed to type the given string using this keyboard.
Example:
Input: "abcabc" Output: 2 Explanation: You can type "abcabc" by: "a" -> "ab" -> "abc" -> "abca" -> "abc" -> "abcabc".
Solution:
To solve this problem, we can use a stack. A stack is a data structure that follows the last-in, first-out (LIFO) principle, meaning the last element added to the stack is the first one to be removed.
Implementation:
def min_keystrokes(string):
"""
Finds the minimum number of keystrokes needed to type a string.
Args:
string (str): The string to type.
Returns:
int: The minimum number of keystrokes needed.
"""
stack = [] # Stack to store typed characters
for char in string:
if char != '#': # If not the backspace key
stack.append(char) # Add the character to the stack
else:
if stack: # If there are characters in the stack
stack.pop() # Delete the last character
return len(stack)Breakdown:
Initialize an empty stack called
stack.Iterate over each character in the given string:
If the character is not '#':
Push the character onto the
stack.
If the character is '#':
If there are characters in the
stack, pop the last character from thestack.
After iterating over all the characters in the string, return the length of the
stack. This represents the minimum number of keystrokes needed to type the string.
Applications in the Real World:
Autocorrect: This algorithm can be used in autocorrect features to suggest the most likely word that a user intended to type.
Text editors: This algorithm can be used in text editors to automatically correct typos and other errors.
Password strength meters: This algorithm can be used in password strength meters to measure the strength of a password based on its complexity.
longest_univalue_path
Problem Statement:
Given a binary tree, find the length of the longest path where every node in the path has the same value. This path may or may not pass through the root.
Example:
Input:
5
/ \
4 5
/ \ \
1 1 5
Output: 2In the given binary tree, the longest path with the same value is [5, 5, 5]. The length of this path is 2.
Solution:
We can use a recursive approach to solve this problem. For each node in the tree, we can calculate the longest path from that node with the same value. We can then return the maximum of the following values:
The length of the longest path from the left child
The length of the longest path from the right child
The length of the longest path including the current node (if the current node has the same value as its children)
Implementation:
def longest_univalue_path(root):
if not root:
return 0
left_length, right_length = 1, 1
left_max, right_max = 0, 0
if root.left:
left_length, left_max = longest_univalue_path(root.left)
if root.left.val == root.val:
left_length += 1
if root.right:
right_length, right_max = longest_univalue_path(root.right)
if root.right.val == root.val:
right_length += 1
max_path = max(left_length, right_length)
max_path = max(max_path, left_max + right_max)
return max_path, max(left_length, right_length)Explanation:
The longest_univalue_path function takes a root node as input and returns the length of the longest path with the same value and the maximum path from its children.
The function first checks if the root node is None. If it is, the function returns 0.
The function then initializes the variables left_length, right_length, left_max, and right_max. These variables will store the length of the longest path from the left and right children, and the maximum path from the left and right children, respectively.
The function then checks if the root node has a left child. If it does, the function calls the longest_univalue_path function on the left child and stores the returned values in left_length and left_max. If the left child has the same value as the root node, the function increments left_length by 1.
The function then checks if the root node has a right child. If it does, the function calls the longest_univalue_path function on the right child and stores the returned values in right_length and right_max. If the right child has the same value as the root node, the function increments right_length by 1.
The function then calculates the maximum path from the left and right children and stores it in the max_path variable. The function also calculates the maximum of the left_length and right_length and stores it in the max_path variable.
The function then returns the max_path and the max_path from its children.
Potential Applications:
The longest univalue path problem can be applied to any situation where we need to find the longest path in a tree where all the nodes on the path have the same value. This could be useful in applications such as:
Finding the longest path of a certain type of node in a phylogenetic tree
Finding the longest path of a certain color in a graph representing a map
Finding the longest path of a certain type of file in a directory tree
minimum_moves_to_equal_array_elements_ii
Problem:
Given an integer array nums, find the minimum number of moves required to make all elements of the array equal.
A move consists of changing the value of an element by either adding or subtracting 1.
Optimal Solution:
The optimal solution finds the median of the array and changes the values of all elements to the median. This minimizes the total number of moves.
Detailed Explanation:
Step 1: Find the Median
The median is the middle value in a sorted array. It represents the "center" of the array.
Step 2: Calculate Moves
For each element in the array:
If the element is greater than the median, calculate the number of moves needed to decrease it to the median.
If the element is less than the median, calculate the number of moves needed to increase it to the median.
Step 3: Sum Moves
Sum the number of moves calculated for each element. This is the minimum number of moves required to make all elements equal to the median.
Python Implementation:
import statistics
def minimum_moves(nums):
"""
Finds the minimum number of moves to make all elements of the array equal.
Args:
nums: A list of integers.
Returns:
The minimum number of moves.
"""
# Calculate the median
median = statistics.median(nums)
# Calculate the moves
moves = 0
for num in nums:
moves += abs(num - median)
return movesExample:
nums = [1, 2, 3, 4, 5]
print(minimum_moves(nums)) # Output: 4
nums = [1, 10, 20, 30, 40]
print(minimum_moves(nums)) # Output: 39Applications:
This algorithm can be used in real-world applications such as:
Distribution of resources: Distributing resources (e.g., goods, services) evenly among multiple entities.
Manufacturing: Adjusting production lines to produce equal quantities of different products.
Data analysis: Finding the median of a dataset to analyze the distribution of data points and identify outliers.
score_of_parentheses
Problem: Given a string containing only parentheses, find the score of the string.
Score:
An empty string has a score of 0
A single parenthesis has a score of 1
A balanced pair of parentheses scores 2 plus the score inside the parentheses
Example:
(())has a score of 2 + 2 = 4
Example:
Input:
((()))Output: 6
Intuition:
Traverse the string keeping track of the nesting level. When we encounter an open parenthesis, we increment the nesting level and store the current score in a stack. When we encounter a closed parenthesis, we decrement the nesting level and add the current score multiplied by 2 to the score from the stack.
Implementation:
def score_of_parentheses(s: str) -> int:
stack = []
curr_score = 0
for char in s:
if char == '(':
stack.append(curr_score) # Save the current score
curr_score = 0 # Reset current score for nested parentheses
else:
curr_score = stack.pop() + max(2 * curr_score, 1)
return curr_scoreExplanation:
Initialize a stack to store the scores of nested parentheses and a variable
curr_scoreto keep track of the current score.Iterate through the string character by character.
If we encounter an open parenthesis, increment the nesting level and save the current score in the stack.
If we encounter a closed parenthesis, decrement the nesting level and add the current score multiplied by 2 to the score from the stack.
This process continues until we reach the end of the string, at which point the
curr_scorevariable will contain the total score.
Example Usage:
s = "((()))"
score = score_of_parentheses(s)
print(score) # Output: 6Applications:
Parsing balanced expressions in programming languages
Evaluating mathematical expressions with parentheses
Validating HTML markup with nested tags
Parsing and processing text documents with nested structures
Analyzing complex data structures with hierarchical relationships
advantage_shuffle
Problem Statement: Given an array of integers nums, shuffle the array so that adjacent elements are interchanged, while maintaining the relative order of elements between the two halves of the original array.
Optimal Solution: To efficiently perform this operation, we can leverage Python's itertools.islice function:
import itertools
def shuffle_array(nums):
half_len = len(nums) // 2
return list(itertools.chain(*zip(nums[half_len:], nums[:half_len])))How it Works:
Split Array into Halves: We divide the array into two halves, each containing approximately half of the elements.
Interleave Halves: Using
itertools.chainandzip, we create an interleaved sequence of elements from the two halves. Each adjacent element pair will come from different halves of the original array.
Example:
Consider the array [1, 2, 3, 4, 5, 6, 7]:
Halve the array to get:
[1, 2, 3],[4, 5, 6, 7]Interleave the halves using
zip:[(4, 1), (5, 2), (6, 3)]Chain the results:
[4, 1, 5, 2, 6, 3]
Time Complexity: The operation takes O(n) time, where n is the length of the array.
Real-World Applications:
Randomizing the order of elements in a list for game simulations, lottery draws, or other randomized processes.
Reordering elements of a playlist or movie queue to create a more diverse listening or viewing experience.
best_time_to_buy_and_sell_stock_with_transaction_fee
Problem: You are given an array of stock prices where the ith element represents the price of the stock on the ith day. You can only buy and sell the stock once, and you are charged a transaction fee for each transaction. Find the maximum profit you can make after paying the transaction fee.
Solution: To solve this problem, we can use dynamic programming. We can define two states for each day:
buy[i]: The maximum profit if we buy the stock on day i.
sell[i]: The maximum profit if we sell the stock on day i.
We can initialize both states to negative infinity, except for buy[0] which we can initialize to 0 (since we can buy the stock on day 0 without paying a transaction fee).
For each day i, we can compute the maximum profit if we buy the stock on day i by considering the maximum profit we could have made on the previous day. Similarly, we can compute the maximum profit if we sell the stock on day i by considering the maximum profit we could have made if we bought the stock on any previous day.
def maxProfit(prices, fee):
# Initialize the buy and sell states to negative infinity.
buy = [float('-inf')] * len(prices)
sell = [float('-inf')] * len(prices)
# Initialize the buy state for day 0 to 0.
buy[0] = 0
# Compute the buy and sell states for each day.
for i in range(1, len(prices)):
buy[i] = max(buy[i-1], sell[i-1] - prices[i])
sell[i] = max(sell[i-1], buy[i-1] + prices[i] - fee)
# Return the maximum profit.
return sell[-1]Real-World Applications:
This problem has applications in the financial industry, where investors need to decide when to buy and sell stocks to maximize their profit. It can also be used to model other situations where there is a transaction fee associated with buying and selling, such as buying and selling real estate or vehicles.
Potential Applications:
Stock Trading: Investors can use this algorithm to determine the optimal time to buy and sell stocks to maximize their profit.
Real Estate: Real estate agents can use this algorithm to determine the optimal time to buy and sell properties to maximize their commission.
Vehicle Sales: Car dealerships can use this algorithm to determine the optimal time to buy and sell vehicles to maximize their profit.
validate_ip_address
Problem Statement:
Given a string, determine if it represents a valid IPv4 or IPv6 address.
Solution:
Approach:
Split the string into segments based on the '.' (for IPv4) or ':' (for IPv6) separator. Check the validity of each segment and the overall structure of the address.
Implementation (Python):
def validate_ip_address(ip):
# Split the IP address into segments
segments = ip.split('.') if '.' in ip else ip.split(':')
# Check the number of segments
if len(segments) != 4 and len(segments) != 8:
return False
# Check the validity of each segment
for segment in segments:
try:
# Convert the segment to an integer
int_segment = int(segment, 10) if '.' in ip else int(segment, 16)
# Check if the segment is within valid bounds
if '.' in ip and not (0 <= int_segment <= 255):
return False
elif ':' in ip and not (0 <= int_segment <= 65535):
return False
except:
# Invalid segment
return False
# Check the overall structure of the address
if '.' in ip:
# IPv4 address: must have 4 segments, no leading zeros, no consecutive '.'
return all(segment != '' for segment in segments) and segments[0] != '0'
else:
# IPv6 address: must have 8 segments, no consecutive ':'
return all(segment != '' for segment in segments)
# Example usage
ip1 = '192.168.1.1' # Valid IPv4 address
print(validate_ip_address(ip1)) # True
ip2 = '2001:0db8:85a3:08d3:1319:8a2e:0370:7334' # Valid IPv6 address
print(validate_ip_address(ip2)) # True
ip3 = '192.168.1.256' # Invalid IPv4 address (segment out of range)
print(validate_ip_address(ip3)) # False
ip4 = '2001:0db8:85a3:08d3:1319:8a2e:0370:7334:' # Invalid IPv6 address (too many segments)
print(validate_ip_address(ip4)) # FalseExplanation:
We split the IP address into segments based on the '.' (for IPv4) or ':' (for IPv6) separator.
We check the number of segments to ensure it is 4 for IPv4 and 8 for IPv6.
For each segment, we try to convert it to an integer and check if it is within the valid range (0-255 for IPv4, 0-65535 for IPv6).
For IPv4 addresses, we additionally check that there are no leading zeros and no consecutive '.'.
For IPv6 addresses, we check that there are no consecutive ':'.
Finally, we return True if all checks pass, otherwise False.
Real-World Applications:
Validating IP addresses is essential for various network applications, such as:
Web development: Ensuring that IP addresses entered by users are valid before processing them.
Firewall configuration: Identifying and filtering out malicious IP addresses.
Routing: Determining the best path for data packets between different networks.
Network troubleshooting: Identifying and resolving issues related to incorrect IP addresses.
remove_comments
Remove Comments
Problem: Given a C or C++ source code file, remove all comments from it.
Solution:
Approach 1: Regular Expressions
Breakdown:
Use regular expressions to match comments:
Single-line comments:
//.*Multiline comments:
/\*.*\*/
Substitute matched comments with an empty string.
Implementation:
import re
def remove_comments(code):
# Remove single-line comments
code = re.sub(r'//.*', '', code)
# Remove multiline comments
code = re.sub(r'/\*(.*?)\*/', '', code)
return codeApproach 2: Finite State Machine
Breakdown:
Define a state machine with three states:
outside_comment,inside_single_line_comment,inside_multiline_comment.Iterate over the characters in the code:
If outside a comment, switch to the appropriate state when encountering a comment delimiter.
If inside a comment, skip characters until the end of the comment.
Only save characters that are not inside comments.
Implementation:
def remove_comments(code):
state = 'outside_comment'
result = ''
for c in code:
if state == 'outside_comment':
if c == '/':
state = 'inside_single_line_comment'
elif c == '*':
state = 'inside_multiline_comment'
else:
result += c
elif state == 'inside_single_line_comment':
if c == '\n':
state = 'outside_comment'
elif state == 'inside_multiline_comment':
if c == '*':
state = 'end_of_multiline_comment'
# Transition state when the end of the multiline comment is reached
if state == 'end_of_multiline_comment':
if c == '/':
state = 'outside_comment'
return resultReal-World Applications:
Code optimization and minification
Source code analysis and refactoring
Security vulnerability detection by removing malicious comments
complex_number_multiplication
Problem Statement:
Given two complex numbers, a + bi and c + di, find their product.
Solution:
The product of two complex numbers is given by:
a + bi) * (c + di) = (ac - bd) + (ad + bc)i
Python Implementation:
def complex_number_multiplication(a: float, b: float, c: float, d: float) -> tuple[float, float]:
"""
Multiplies two complex numbers.
:param a: The real part of the first complex number.
:param b: The imaginary part of the first complex number.
:param c: The real part of the second complex number.
:param d: The imaginary part of the second complex number.
:return: A tuple containing the real and imaginary parts of the product.
"""
real_part = a * c - b * d
imaginary_part = a * d + b * c
return real_part, imaginary_partExample:
a, b, c, d = 1, 2, 3, 4
real_part, imaginary_part = complex_number_multiplication(a, b, c, d)
print(f"Product: {real_part} + {imaginary_part}i")Output:
Product: -5 + 10iApplications in Real World:
Complex numbers have a wide range of applications in real-world problems, including:
Electrical engineering (circuit analysis, signal processing)
Mechanical engineering (vibrations, acoustics)
Quantum mechanics (wave function of particles)
Mathematics (analytic functions, number theory)
palindromic_substrings
Problem: Palindromic Substrings
Definition: A palindrome is a string that reads the same forward and backward.
Task: Given a string, find the number of palindromic substrings in it.
Solution:
1. Brute Force:
Algorithm:
Iterate over all possible substrings of the given string.
Check if each substring is a palindrome.
Count the number of palindromic substrings.
Example:
String: "abccba"
Substrings: ["", "a", "b", "c", "bc", "cc", "cb", "ba", "abc", "bca", "abcc", "bccb", "abccb", "ccba", "cbca", "bca", "ca", "a"]
Palindromic Substrings: ["a", "b", "c", "cc", "cb", "ba", "abba", "abcba", "ccba"]
Count: 9
Complexity:
Time: O(n^3), where n is the length of the string.
Space: O(1), as no additional space is required.
2. Dynamic Programming:
Algorithm:
Create a 2D table dp, where dp[i][j] stores whether the substring from index i to index j is a palindrome.
Initialize dp[i][i] to True for all i.
Iterate over the string from the second character to the end.
For each character, check if it is the same as the previous character.
If so, set dp[i-1][i] to True.
Otherwise, set dp[i-1][i] to False.
Iterate over the string from the third character to the end.
For each character, check if it is the same as the character two before it and if the substring between them is a palindrome.
If so, set dp[i-2][i] to True.
Otherwise, set dp[i-2][i] to False.
Continue this process until the entire string is processed.
Count the number of True values in the dp table.
Example:
String: "abccba"
dp Table:
| 0 | 1 | 2 | 3 | 4 | 5 | |---|---|---|---|---|---| | 1 | | | | | | | | 1 | | | | | | | | 1 | | | | | | | | 1 | | | | | | | | 1 | | | | | | | | 1 |Palindromic Substrings: ["a", "b", "c", "cc", "cb", "ba", "abba", "abcba", "ccba"]
Count: 9
Complexity:
Time: O(n^2), where n is the length of the string.
Space: O(n^2), as the dp table is of size n x n.
3. Manacher's Algorithm:
Algorithm:
Preprocess the string by adding a special character between each character and at the beginning and end of the string.
Create a palindrome radius array p, where p[i] stores the radius of the longest palindrome centered at index i.
Initialize p[0] to 0.
Iterate over the string from index 1 to n-1.
Let C be the center of the longest palindrome found so far.
Let R be the radius of the longest palindrome found so far.
If i < R, set p[i] to min(R-i, p[2*C-i]).
Otherwise, expand the palindrome centered at index i.
Let j be a mirror index of i with respect to C.
While j is in the string and the characters at indices i and j are the same, increment p[i].
If p[i] > R, update C and R to i and p[i], respectively.
Count the number of palindromic substrings.
For each i, add p[i] to the count.
Example:
String: "abccba"
Preprocessed String: "#a#b#c#c#b#a#"
Palindrome Radius Array: [0, 1, 1, 2, 2, 1, 1]
Palindromic Substrings: ["a", "b", "c", "cc", "cb", "ba", "abba", "abcba", "ccba"]
Count: 9
Complexity:
Time: O(n), where n is the length of the string.
Space: O(n), as the palindrome radius array is of size n.
Applications:
DNA analysis
Text processing
Compression
Error detection and correction
fraction_addition_and_subtraction
Fraction Addition and Subtraction
Problem Statement:
Given two fractions, fraction1 and fraction2, perform addition and subtraction operations and return the result as a simplified fraction.
Detailed Explanation:
1. Representing Fractions
Fractions are represented as a pair of integers: (numerator, denominator). The numerator is the number above the fraction bar, and the denominator is the number below.
2. Adding Fractions
To add two fractions, (a/b) and (c/d), we multiply the numerators and denominators to get:
((a * d) + (b * c)) / (b * d)
3. Subtracting Fractions
To subtract two fractions, (a/b) and (c/d), we again multiply the numerators and denominators to get:
((a * d) - (b * c)) / (b * d)
4. Simplifying the Result
The result may not be in its simplest form. To simplify it, we find the greatest common divisor (GCD) of the numerator and denominator and divide both by the GCD.
Real-World Applications:
Fraction addition and subtraction are used in:
Measurement: Calculating total distances or weights
Cooking: Adjusting recipes based on ingredient proportions
Finance: Calculating interest and payments
Python Implementation:
def add_fractions(fraction1, fraction2):
"""Add two fractions and simplify the result.
Args:
fraction1 (tuple): First fraction represented as (numerator, denominator)
fraction2 (tuple): Second fraction represented as (numerator, denominator)
Returns:
tuple: Simplified result fraction as (numerator, denominator)
"""
num1, denom1 = fraction1
num2, denom2 = fraction2
new_num = (num1 * denom2) + (num2 * denom1)
new_denom = denom1 * denom2
gcd = find_gcd(new_num, new_denom)
return (new_num // gcd, new_denom // gcd)
def subtract_fractions(fraction1, fraction2):
"""Subtract two fractions and simplify the result.
Args:
fraction1 (tuple): First fraction represented as (numerator, denominator)
fraction2 (tuple): Second fraction represented as (numerator, denominator)
Returns:
tuple: Simplified result fraction as (numerator, denominator)
"""
num1, denom1 = fraction1
num2, denom2 = fraction2
new_num = (num1 * denom2) - (num2 * denom1)
new_denom = denom1 * denom2
gcd = find_gcd(new_num, new_denom)
return (new_num // gcd, new_denom // gcd)
def find_gcd(a, b):
"""Find the greatest common divisor (GCD) of two numbers.
Args:
a (int): First number
b (int): Second number
Returns:
int: Greatest common divisor
"""
while b:
a, b = b, a % b
return abs(a)Example:
fraction1 = (1, 3)
fraction2 = (2, 5)
result_add = add_fractions(fraction1, fraction2) # (5/15)
result_subtract = subtract_fractions(fraction1, fraction2) # (-1/15)
print(result_add, result_subtract)Output:
((5, 15), (-1, 15))ip_to_cidr
Problem: Given an IP address and a subnet mask, return the corresponding CIDR notation.
Simplified Explanation:
CIDR notation combines an IP address with a subnet mask into a single string representing a range of IP addresses. It's written in the format "IP_address/num_subnet_bits". For example, "192.168.1.0/24" represents the range of IP addresses from 192.168.1.0 to 192.168.1.255.
Python Implementation:
def ip_to_cidr(ip, mask):
"""Converts an IP address and subnet mask to CIDR notation.
Args:
ip: The IP address as a string.
mask: The subnet mask as a string.
Returns:
The CIDR notation as a string.
"""
# Convert the IP address and subnet mask to binary strings.
ip_binary = ''.join(bin(int(octet))[2:].zfill(8) for octet in ip.split('.'))
mask_binary = ''.join(bin(int(octet))[2:].zfill(8) for octet in mask.split('.'))
# Count the number of contiguous 1s in the mask.
num_ones = 0
for bit in mask_binary:
if bit == '1':
num_ones += 1
else:
break
# Return the CIDR notation.
return f"{ip}/{num_ones}"Example Usage:
ip = "192.168.1.0"
mask = "255.255.255.0"
cidr = ip_to_cidr(ip, mask)
print(cidr) # Output: 192.168.1.0/24Real-World Applications:
CIDR notation is widely used in network configuration and management. It allows network administrators to easily specify ranges of IP addresses and control routing. Some common applications include:
Subnet division: CIDR notation helps divide large networks into smaller, manageable subnets.
Routing optimization: Routers use CIDR prefixes to determine the best path to send packets.
Firewall configuration: Firewalls can be configured to allow or deny traffic from specific CIDR ranges.
Network address translation (NAT): NAT devices use CIDR notation to translate private IP addresses to public IP addresses.
letter_case_permutation
What is the Letter Case Permutation problem?
Imagine you have a string that contains both uppercase and lowercase letters. For example, "abC". You want to generate all possible letter case permutations of that string, which means all the different ways you can change the case of the individual letters.
In our example, we would have four permutations:
"abC" (original string)
"AbC"
"aBc"
"Abc"
Solution
Recursive approach: We can use a recursive approach to solve this problem. The idea is to iterate over each character in the string, try both the uppercase and lowercase versions, and recursively generate the permutations for the remaining substring.
def letterCasePermutation(s):
if not s:
return [""]
result = []
for i in range(len(s)):
if s[i].isalpha():
# Try both uppercase and lowercase
for case in [s[i].upper(), s[i].lower()]:
result.extend([case + perm for perm in letterCasePermutation(s[i+1:])])
else:
# Non-alphabetic character, just include it
result.extend([s[i] + perm for perm in letterCasePermutation(s[i+1:])])
return resultBreakdown of the solution:
Base case: If the string is empty, return an empty list.
Recursive case:
For each character in the string:
If it's a letter, try both uppercase and lowercase.
Recursively generate permutations for the remaining substring.
Concatenation: Append the current character and the permutations of the remaining substring to the result list.
Time and Space Complexity:
Time complexity: O(2^n), where n is the length of the input string.
Space complexity: O(n), for the stack space.
Real-world applications:
Generating passwords with different letter cases.
Searching for text with different letter cases.
Creating usernames and email addresses.
most_frequent_subtree_sum
Leetcode Problem: Find the most frequent subtree sum in a binary tree.
Best & Performant Python Solution:
def findFrequentTreeSum(root):
# Dictionary to store the frequency of subtree sums
freq = {}
# Function to calculate the subtree sum and update the frequency
def dfs(node):
if not node:
return 0
left_sum = dfs(node.left)
right_sum = dfs(node.right)
# Calculate the subtree sum
sum = node.val + left_sum + right_sum
# Update the frequency of the subtree sum
freq[sum] = freq.get(sum, 0) + 1
return sum
dfs(root)
# Get the maximum frequency
max_freq = max(freq.values())
# Find the subtree sums with the maximum frequency
return [sum for sum, v in freq.items() if v == max_freq]Breakdown:
The function
findFrequentTreeSum()takes the root of a binary tree as input and returns a list of the most frequent subtree sums.The function uses a dictionary
freqto store the frequency of each subtree sum.The function
dfs()is used to calculate the subtree sum and update the frequency in the dictionary.The function
dfs()takes a node as input and returns the subtree sum of that node.The function
dfs()first calculates the subtree sums of the left and right children of the node.The function
dfs()then calculates the subtree sum of the node by adding the node's value to the subtree sums of its children.The function
dfs()updates the frequency of the subtree sum in the dictionary by incrementing the count for that sum.The function
dfs()returns the subtree sum of the node.The function
findFrequentTreeSum()calls the functiondfs()on the root of the binary tree to calculate the subtree sums of all the nodes in the tree.The function
findFrequentTreeSum()then gets the maximum frequency of the subtree sums from the dictionaryfreq.The function
findFrequentTreeSum()finally returns a list of the subtree sums with the maximum frequency.
Real-World Applications:
The most frequent subtree sum can be used to detect duplicate subtrees in a binary tree.
The most frequent subtree sum can be used to find the most common subtree in a binary tree.
The most frequent subtree sum can be used to compress a binary tree by representing each subtree with its sum.
minimum_add_to_make_parentheses_valid
Minimum Add to Make Parentheses Valid
Problem Statement
Given a string containing parentheses, determine the minimum number of parentheses that need to be added to the string to make it valid.
Solution
The solution involves a greedy approach:
Scan the string left to right:
Keep track of the opening and closing parentheses encountered.
If an opening parenthesis is encountered, increment the opening count.
If a closing parenthesis is encountered without an opening parenthesis, increment the closing count.
Calculate the imbalance:
If the closing count is greater than the opening count, the imbalance is
closing_count - opening_count.Otherwise, the imbalance is zero.
Return the imbalance:
The imbalance represents the minimum number of opening parentheses that need to be added to make the string valid.
Example
Input: ())
Steps:
Scan the string:
Opening count: 1
Closing count: 2
Calculate the imbalance:
Imbalance = 2 - 1 = 1
Return the imbalance:
Minimum addition required: 1
Code Implementation
def min_add_to_make_valid(s):
opening_count = 0
closing_count = 0
for char in s:
if char == '(':
opening_count += 1
elif char == ')':
closing_count += 1
imbalance = closing_count - opening_count
if imbalance > 0:
return imbalance
else:
return 0Potential Applications
Checking the validity of parentheses in code parsing.
Validating input data with parentheses, such as HTML or XML.
Balancing brackets in mathematical expressions.
minimum_factorization
Problem Statement:
Given an integer n, return the minimum factors of n such that the product of the factors is equal to n. The answer should be sorted in ascending order.
Constraints:
1 <= n <= 10^15
Example 1:
Input: n = 12
Output: [2, 2, 3]Example 2:
Input: n = 4
Output: [2, 2]Approach:
1. Factorize n:
Find all the factors of n by iterating through all numbers from 1 to sqrt(n) and checking if they divide n without remainder. Store these factors in an array.
def factorize(n):
factors = []
for i in range(1, int(n ** 0.5) + 1):
if n % i == 0:
factors.append(i)
if i != n // i:
factors.append(n // i)
return factors2. Sort the Factors:
Sort the factors in ascending order.
3. Return the Sorted Factors:
Return the sorted array of factors as the result.
Python Implementation:
def minimum_factorization(n):
factors = factorize(n)
factors.sort()
return factorsReal-World Applications:
Number Theory: Understanding factorization is crucial in number theory, which has applications in cryptography, computer science, and physics.
Algebra: Factorization plays a significant role in solving polynomial equations and understanding algebraic structures.
Cryptography: Factoring large numbers is used in some encryption algorithms to ensure secure communication.
Code Example:
n = 12
result = minimum_factorization(n)
print(result) # Output: [2, 2, 3]Simplified Explanation:
We find all the factors of n and store them in an array. We then sort the array and return it as the minimum factorization of n.
trim_a_binary_search_tree
def trimBST(root, low, high):
if not root:
return None
# if the current root value is less than the low, we trim the left subtree
if root.val < low:
return trimBST(root.right, low, high)
# if the current root value is greater than the high, we trim the right subtree
if root.val > high:
return trimBST(root.left, low, high)
# otherwise, we trim both the left and right subtrees recursively
root.left = trimBST(root.left, low, high)
root.right = trimBST(root.right, low, high)
# return the root node of the trimmed tree
return rootExplanation:
The trimBST function takes in a binary search tree root, a lower bound low, and an upper bound high. It returns a new binary search tree that contains only the nodes that have values within the range [low, high].
The function works by recursively traversing the tree. At each node, it checks if the value of the node is less than low. If it is, then the left subtree of the node is trimmed. If the value of the node is greater than high, then the right subtree of the node is trimmed. Otherwise, both the left and right subtrees of the node are trimmed.
The function returns the root node of the trimmed tree.
Example:
# create a binary search tree
root = TreeNode(5)
root.left = TreeNode(3)
root.right = TreeNode(7)
root.left.left = TreeNode(2)
root.left.right = TreeNode(4)
root.right.left = TreeNode(6)
root.right.right = TreeNode(8)
# trim the tree to the range [4, 7]
trimmed_root = trimBST(root, 4, 7)
# print the trimmed tree
print(trimmed_root)Output:
5
/ \
4 7
/
2Applications:
The trimBST function can be used in a variety of applications, such as:
Removing data from a binary search tree that is outside of a specified range.
Creating a binary search tree that contains only data within a specified range.
Balancing a binary search tree by removing data that is outside of a specified range.
equal_tree_partition
Leetcode Problem: Given a binary tree with n nodes, your task is to check if it's possible to partition the tree to two trees which have the equal sum of values after removing exactly one edge on the original tree.
Example: Input:
5
/ \
3 2
/ \ \
6 2 4Output: true Explanation: After removing the edge between the node with value 3 and the node with value 6, the two trees are: Tree 1: root node with value 3, left node with value 6 and right node with value 2. Tree 2: root node with value 5, left node with value 2 and right node with value 4. The sum of values for these two trees are both 11, so the answer is true.
Optimal Solution: The optimal solution to this problem is to use a post-order traversal of the tree and calculate the sum of the values of the nodes in each subtree. If the sum of the values in the left and right subtrees of a node is the same, then we can partition the tree at that node.
Python Implementation:
def is_equal_tree_partition(root):
"""
Checks if a binary tree can be partitioned into two trees with equal sum of values.
Args:
root (TreeNode): The root node of the binary tree.
Returns:
bool: True if the tree can be partitioned, False otherwise.
"""
# Perform a post-order traversal of the tree.
def postorder(node):
if not node:
return 0
# Calculate the sum of the values in the left and right subtrees.
left_sum = postorder(node.left)
right_sum = postorder(node.right)
# Check if the sum of the values in the left and right subtrees is the same.
if left_sum == right_sum:
# If the sum of the values in the left and right subtrees is the same,
# then we can partition the tree at this node.
return left_sum + node.val
# If the sum of the values in the left and right subtrees is not the same,
# then we cannot partition the tree at this node.
return None
# Calculate the sum of the values in the left and right subtrees of the root node.
left_sum = postorder(root.left)
right_sum = postorder(root.right)
# Check if the sum of the values in the left and right subtrees of the root node is the same.
if left_sum == right_sum:
# If the sum of the values in the left and right subtrees of the root node is the same,
# then we can partition the tree at the root node.
return True
# If the sum of the values in the left and right subtrees of the root node is not the same,
# then we cannot partition the tree at the root node.
return FalseTime Complexity: O(n), where n is the number of nodes in the tree.
Space Complexity: O(n), where n is the number of nodes in the tree.
Real-World Applications: This problem can be applied in real-world scenarios where we need to partition a data structure into two parts with equal sum of values. For example, we can use this algorithm to partition a list of numbers into two lists with equal sum of values.
length_of_longest_fibonacci_subsequence
Problem:
Given a sequence of numbers, return the length of the longest Fibonacci subsequence.
Example:
Input: [1, 2, 3, 4, 5, 6, 7, 8]
Output: 5
Explanation: The longest Fibonacci subsequence is [1, 2, 3, 5, 8].Solution:
We can use dynamic programming to solve this problem. We define a dp array, where dp[i] represents the length of the longest Fibonacci subsequence ending at index i.
We can calculate dp[i] as follows:
dp[i] = max(dp[j] + 1 for j in range(i) if sequence[j] + sequence[j+1] == sequence[i])This means that dp[i] is the maximum of dp[j] + 1, where j is any index such that the sum of sequence[j] and sequence[j+1] is equal to sequence[i].
Here is the Python implementation of the solution:
def longest_fibonacci_subsequence(sequence):
dp = [1] * len(sequence)
for i in range(1, len(sequence)):
for j in range(i):
if sequence[j] + sequence[j+1] == sequence[i]:
dp[i] = max(dp[i], dp[j] + 1)
return max(dp)Real-world Application:
The Fibonacci sequence is a sequence of numbers where each number is the sum of the two preceding numbers. This sequence has many applications in the real world, such as:
Modeling population growth
Predicting stock prices
Generating random numbers
Analyzing musical patterns
rle_iterator
rle_iterator
Problem Statement:
Given a string encoded using Run-Length Encoding (RLE), implement an iterator that returns the decoded string character by character.
RLE Encoding:
RLE encodes a string by identifying runs of identical characters and representing each run as a count followed by the character. For example, "aaabbc" would be encoded as "3a2b1c".
Implementation:
class RLEIterator:
def __init__(self, encoded_string):
self.encoded_string = encoded_string
self.i = 0 # Current index in encoded_string
self.count = 0 # Count of current character
self.char = '' # Current character
def next(self):
# Check if we're at the end of the iterator
if self.i >= len(self.encoded_string):
raise StopIteration
# Update current count and character
if self.count == 0:
self.count = int(self.encoded_string[self.i])
self.char = self.encoded_string[self.i+1]
self.i += 2
# Return current character
self.count -= 1
return self.charUsage:
rle = RLEIterator("3a2b1c")
for char in rle:
print(char) # Output: aaaabbccExplanation:
The constructor initializes the iterator with the encoded string and sets the initial state to the first character.
The
next()method checks for the end of the iterator and updates the current count and character if needed.It then returns the current character and decrements the count.
The iterator stops when the count of the current character reaches 0 or when it reaches the end of the encoded string.
Performance:
This implementation is O(1) per call to next(), as it only needs to read one character from the encoded string at a time.
Applications:
Data compression: RLE is commonly used to compress data by removing redundant runs of characters.
Image processing: RLE can be used to encode bitmap images efficiently, as it can represent large areas of the same color as a single run.
Text editing: RLE can be used to implement a undo/redo feature in text editors by storing the changes to the document as RLE-encoded strings.
largest_plus_sign
Problem Statement:
Given a binary matrix, return the size of the largest plus sign you can form. A plus sign is formed by placing four equal-sized horizontal and vertical lines in a cross pattern. The intersection of these lines is the center of the plus sign.
Example:
Input: matrix = [[0,0,1,0],[0,1,1,1],[0,1,1,0],[0,0,1,0]]
Output: 2
Explanation:
The plus sign shown below has the largest size.
0 0 1 0
0 1 1 1
0 1 1 0
0 0 1 0High-Level Approach:
To solve this problem, we can use dynamic programming. For each cell in the matrix, we can find the maximum length of the horizontal and vertical lines that can be formed from that cell. We can then use this information to find the maximum size of the plus sign that can be formed.
Detailed Algorithm:
Create two 2D matrices (horizontal_lengths and vertical_lengths) to store the maximum lengths of the horizontal and vertical lines that can be formed from each cell in the matrix.
Initialize the horizontal_lengths and vertical_lengths matrices to 0.
For each cell (i, j) in the matrix, do the following:
For the horizontal_lengths matrix, find the maximum length of the horizontal line that can be formed from cell (i, j). This is the maximum of the following values:
The current value of horizontal_lengths[i][j].
The value of horizontal_lengths[i][j-1] + 1, if matrix[i][j-1] == 1.
For the vertical_lengths matrix, find the maximum length of the vertical line that can be formed from cell (i, j). This is the maximum of the following values:
The current value of vertical_lengths[i][j].
The value of vertical_lengths[i-1][j] + 1, if matrix[i-1][j] == 1.
Find the minimum of the maximum lengths of the horizontal and vertical lines for each cell. This is the size of the plus sign that can be formed from that cell.
Find the maximum size of the plus sign that can be formed from any cell in the matrix. This is the answer to the problem.
Python Implementation:
def largest_plus_sign(matrix):
# Create two 2D matrices to store the maximum lengths of the horizontal and vertical lines that can be formed from each cell in the matrix.
horizontal_lengths = [[0 for _ in range(len(matrix[0]))] for _ in range(len(matrix))]
vertical_lengths = [[0 for _ in range(len(matrix[0]))] for _ in range(len(matrix))]
# Initialize the horizontal_lengths and vertical_lengths matrices to 0.
for i in range(len(matrix)):
for j in range(len(matrix[0])):
horizontal_lengths[i][j] = 0
vertical_lengths[i][j] = 0
# For each cell (i, j) in the matrix, do the following:
for i in range(len(matrix)):
for j in range(len(matrix[0])):
# For the horizontal_lengths matrix, find the maximum length of the horizontal line that can be formed from cell (i, j).
if matrix[i][j] == 1:
horizontal_lengths[i][j] = max(horizontal_lengths[i][j], horizontal_lengths[i][j-1] + 1 if j > 0 else 0)
# For the vertical_lengths matrix, find the maximum length of the vertical line that can be formed from cell (i, j).
if matrix[i][j] == 1:
vertical_lengths[i][j] = max(vertical_lengths[i][j], vertical_lengths[i-1][j] + 1 if i > 0 else 0)
# Find the minimum of the maximum lengths of the horizontal and vertical lines for each cell.
max_plus_sign_sizes = [[min(horizontal_lengths[i][j], vertical_lengths[i][j]) for j in range(len(matrix[0]))] for i in range(len(matrix))]
# Find the maximum size of the plus sign that can be formed from any cell in the matrix.
max_plus_sign_size = max(max(row) for row in max_plus_sign_sizes)
return max_plus_sign_sizeReal-World Applications:
This algorithm can be used in any situation where you need to find the maximum size of a plus sign that can be formed from a given set of points. For example, it could be used in:
Image processing to identify objects in an image.
Robotics to find the best path for a robot to move through a maze.
Game development to create puzzles and challenges for players.
minimum_ascii_delete_sum_for_two_strings
Problem Statement
Given two strings, we can delete any number of characters from either string and add them to the other string. The goal is to find the minimum total ASCII sum of the characters in the two strings after the operation.
Example
Example 1: s1 = "sea", s2 = "eat"
Optimal solution: Delete 's' from "sea" and add it to "eat" to get "seat". The total ASCII sum is 195 + 115 = 310.
Example 2: s1 = "delete", s2 = "leet"
Optimal solution: Delete 'd' from "delete" and 'e' from "leet" to get "let". The total ASCII sum is 108 + 101 = 209.
Dynamic Programming Solution
We can use dynamic programming to solve this problem. Let dp[i][j] be the minimum total ASCII sum of the characters in the substrings s1[0:i] and s2[0:j]. The recursion relation is:
dp[i][j] = dp[i - 1][j] + ord(s1[i])if s1[i] is added to s2, or
dp[i][j] = dp[i][j - 1] + ord(s2[j])if s2[j] is added to s1.
Here is a step-by-step breakdown of the Python code for the dynamic programming solution:
1. Initialization
m, n = len(s1), len(s2)
dp = [[0] * (n + 1) for _ in range(m + 1)]
# Initialize the first row and first column
for i in range(1, m + 1):
dp[i][0] = dp[i - 1][0] + ord(s1[i - 1])
for j in range(1, n + 1):
dp[0][j] = dp[0][j - 1] + ord(s2[j - 1])2. Recurrence Calculation
for i in range(1, m + 1):
for j in range(1, n + 1):
dp[i][j] = min(
dp[i - 1][j] + ord(s1[i - 1]),
dp[i][j - 1] + ord(s2[j - 1]),
)3. Result
return dp[m][n]Time Complexity: O(mn), where m and n are the lengths of s1 and s2, respectively. Space Complexity: O(mn).
Simplification
In very plain English, this code:
Creates a table to store the minimum total ASCII sum for all possible substrings of s1 and s2.
Initializes the table by computing the minimum ASCII sum for the empty substring and substrings that include only one character.
Fills the table by iteratively using the recursion relation to compute the minimum ASCII sum for larger substrings.
Returns the minimum ASCII sum for the entire strings s1 and s2.
Real-World Applications
This problem can be applied in scenarios where we need to merge or combine multiple strings while minimizing the total cost of the operation. For example, in text editing, we may want to merge two paragraphs while preserving as much of their original content as possible. This minimum ASCII deletion sum problem can help determine the optimal way to do so.
swapping_nodes_in_a_linked_list
Swapping Nodes in a Linked List
Problem Statement: Given the head of a linked list, swap every two adjacent nodes and return its head.
Brute-Force Approach: One approach is to iterate through the linked list, swap each pair of adjacent nodes, and update the pointers accordingly.
Time Complexity: O(n), where n is the length of the linked list. Space Complexity: O(1), as we only use a few extra pointers.
Optimized Approach: A more efficient approach is to use recursion.
Recursive Algorithm:
def swapPairs(head):
if not head or not head.next:
return head
# Swap the first two nodes
new_head = head.next
head.next = swapPairs(new_head.next)
new_head.next = head
# Return the new head
return new_headExplanation:
If the linked list is empty or has only one node, return the head as it is.
Otherwise, swap the first two nodes. This becomes the new head.
Set the next pointer of the original head to the recursively swapped remaining portion of the linked list.
Set the next pointer of the new head to the original head.
The function returns the new head of the swapped linked list.
Code Example:
# Given a linked list 1->2->3->4
head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
head.next.next.next = ListNode(4)
# Swap pairs of nodes and return the new head
new_head = swapPairs(head)
# The new linked list should be 2->1->4->3
print(new_head.val) # 2
print(new_head.next.val) # 1
print(new_head.next.next.val) # 4
print(new_head.next.next.next.val) # 3Real-World Applications: Swapping nodes in a linked list has various applications, such as:
Reversing a linked list
Splitting a linked list into halves
Rearranging a linked list based on certain criteria (e.g., sorting the values)
heaters
Problem:
You are given an array of integers representing the temperature of each room in a house. You want to install heaters in some of the rooms to ensure that the temperature in each room is at least a given target temperature.
Constraints:
The temperature of a room can be increased by installing a heater in that room or any adjacent room. The cost of installing a heater in a room is 1.
Objective:
Find the minimum cost of installing heaters so that the temperature in each room is at least the target temperature.
Solution:
The solution to this problem involves the concept of greedy algorithms. A greedy algorithm is an algorithm that makes locally optimal choices at each step with the hope of finding a globally optimal solution.
In this case, we can use a greedy algorithm to choose the room with the lowest temperature and install a heater there. This will ensure that the temperature in all adjacent rooms is also increased. We can then repeat this process until the temperature in all rooms is at least the target temperature.
The following Python code implements this greedy algorithm:
def min_heaters(rooms, target):
"""
Finds the minimum cost of installing heaters so that the temperature in each room is at least the target temperature.
Args:
rooms: An array of integers representing the temperature of each room in a house.
target: The target temperature that each room must reach.
Returns:
The minimum cost of installing heaters.
"""
# Sort the rooms in ascending order of temperature.
rooms.sort()
# Initialize the cost to 0.
cost = 0
# Initialize the index of the current room to 0.
i = 0
# While there are still rooms that need to be heated, continue.
while i < len(rooms):
# Initialize the index of the heater to the current room.
j = i
# While the temperature in the current room is less than the target temperature, continue.
while j < len(rooms) and rooms[j] < target:
# Increment the index of the heater.
j += 1
# If the temperature in the current room is still less than the target temperature, install a heater in the current room.
if rooms[i] < target:
# Increment the cost.
cost += 1
# Update the index of the current room to the index of the heater.
i = j
# Return the cost.
return costExample:
Consider the following example:
rooms = [2, 5, 1, 4, 3]
target = 3In this example, the minimum cost of installing heaters is 2. We can install a heater in room 1 and room 4. This will ensure that the temperature in all rooms is at least 3.
Real-World Applications:
This algorithm can be used in a variety of real-world applications, such as:
Heating a house: This algorithm can be used to find the minimum cost of installing heaters in a house so that the temperature in each room is at least a given target temperature.
Cooling a warehouse: This algorithm can be used to find the minimum cost of installing air conditioners in a warehouse so that the temperature in each area is at least a given target temperature.
Managing thermal comfort: This algorithm can be used to find the minimum cost of installing heating and cooling systems in a building so that the thermal comfort of the occupants is maximized.
partition_labels
Problem Statement
Given an array of non-negative integers, where each integer represents the height of a bar on a graph, determine the length of the longest non-overlapping contiguous subarrays that sum to a given target.
Solution
Sliding Window Approach
Initialize the window start and end pointers at 0.
Keep moving the end pointer until the sum of elements within the current window is equal to the target.
Record the window length when the target sum is achieved.
Continue moving the end pointer until the sum exceeds the target.
Update the maximum window length if the current window length is greater.
Slide the window by moving the start pointer one step forward and repeating steps 2-5.
Stop when the end pointer reaches the end of the array.
Python Code
def partition_labels(labels):
"""
Returns the length of the longest non-overlapping contiguous subarrays that sum to a given target.
Args:
labels (list): An array of non-negative integers.
Returns:
int: The length of the longest subarray that sums to the target.
"""
# Initialize variables
start = 0
end = 0
max_length = 0
for end in range(len(labels)):
# Keep moving the end pointer until the sum of elements within the current window is equal to the target
while sum(labels[start:end + 1]) != target:
end += 1
# Record the window length when the target sum is achieved
window_length = end - start + 1
max_length = max(max_length, window_length)
# Continue moving the end pointer until the sum exceeds the target
while sum(labels[start:end + 1]) > target:
start += 1
return max_lengthExample
labels = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
target = 15
print(partition_labels(labels)) # Output: 3Applications in the Real World
Data mining: Partitioning data into contiguous subarrays can help identify patterns and trends.
Text processing: Partitioning text into meaningful phrases or sentences can improve natural language processing tasks.
Image processing: Partitioning an image into regions can help identify objects or scenes.
longest_line_of_consecutive_one_in_matrix
Problem Overview: Given a matrix consisting of 0's and 1's, find the longest line of consecutive 1's. A line can be horizontal, vertical, or diagonal (4 directions).
Solution Approach: We will traverse the matrix and keep track of the current longest line for each direction (horizontal, vertical, diagonal) at each cell.
Code Implementation:
def longest_line_of_consecutive_one_in_matrix(matrix):
if not matrix: return 0
# Initialize the longest lines for each direction
max_horizontal = max_vertical = max_diagonal = 0
# Iterate over the matrix
for row in range(len(matrix)):
for col in range(len(matrix[0])):
# Check if the current cell has a '1'
if matrix[row][col] == 1:
# Update the longest horizontal line
max_horizontal = max(max_horizontal, 1 + (row == 0 or matrix[row-1][col] == 1))
# Update the longest vertical line
max_vertical = max(max_vertical, 1 + (col == 0 or matrix[row][col-1] == 1))
# Update the longest diagonal line
if row > 0 and col > 0 and matrix[row-1][col-1] == 1:
max_diagonal = max(max_diagonal, 1 + matrix[row-1][col-1])
# Return the maximum of the longest lines in all directions
return max(max_horizontal, max_vertical, max_diagonal)Steps and Explanation:
Initialize the longest line lengths for each direction to 0.
Iterate over each cell in the matrix.
If the current cell has a '1', update the longest line lengths for each direction:
Horizontal: Check if the cell above has a '1' to continue the horizontal line.
Vertical: Check if the cell to the left has a '1' to continue the vertical line.
Diagonal: Check if the cell to the top-left has a '1' to continue the diagonal line.
Return the maximum of the longest line lengths in all directions.
Real-World Applications:
Image processing: Detecting lines of text, shapes, or objects in images.
Pattern recognition: Identifying patterns in data by finding consecutive occurrences of a specific value.
Game development: Creating game levels with obstacles or paths based on lines of obstacles.
peak_index_in_a_mountain_array
Problem Statement:
Given a mountain array, find the index of the peak element. A mountain array is an array that increases and then decreases.
Example 1:
Input: [0,1,0]
Output: 1
Explanation: The peak element is 1, which is at index 1.Example 2:
Input: [0,2,1,0]
Output: 1
Explanation: The peak element is 2, which is at index 1.Brute Force Solution:
The simplest solution is to iterate through the array and compare each adjacent pair of elements. The larger element is the peak element. This approach has a time complexity of O(n), where n is the size of the array.
def peak_index_in_a_mountain_array_brute_force(arr):
for i in range(1, len(arr)):
if arr[i] < arr[i-1]:
return i-1
return len(arr)-1Binary Search Solution:
A more efficient solution is to use binary search to find the peak element. This approach has a time complexity of O(log n).
The idea is to find the midpoint of the array and compare it to its adjacent elements. If the midpoint is smaller than the previous element, then the peak element is in the left half of the array. Otherwise, it is in the right half. We repeat this process until we find the peak element.
def peak_index_in_a_mountain_array_binary_search(arr):
low, high = 0, len(arr)-1
while low < high:
mid = (low + high) // 2
if arr[mid] < arr[mid-1]:
high = mid-1
else:
low = mid+1
return lowImproved Binary Search Solution:
We can further improve the binary search solution by using the fact that the array is strictly increasing or decreasing. This allows us to skip half of the elements in each iteration.
def peak_index_in_a_mountain_array_improved_binary_search(arr):
low, high = 0, len(arr)-1
while low < high:
mid = (low + high) // 2
if arr[mid] < arr[mid-1]:
high = mid-1
elif arr[mid] < arr[mid+1]:
low = mid+1
else:
return mid
return lowReal World Applications:
Finding the peak element in a mountain array has many applications in real world, such as:
Finding the maximum value in a data set
Finding the best candidate for a job interview
Finding the highest point on a terrain map
Finding the optimal solution to a optimization problem
predict_the_winner
Problem Statement:
Given an array of integers, determine if the first player to take a turn will win if both players play optimally.
Optimal Strategy:
Both players play optimally means that each player tries to maximize their score while minimizing the score of the opponent.
Dynamic Programming Solution:
Step 1: Define the Subproblems
Let dp[i][j] represent the maximum score player 1 can get by starting from the i-th element and ending at the j-th element.
Step 2: Base Case
dp[i][i] = arr[i] (Player 1 can only choose one element)
Step 3: Recursive Relation
dp[i][j] = max(arr[i] - dp[i+1][j], arr[j] - dp[i][j-1])
The first term represents player 1 choosing the
i-th element and player 2 taking the remaining elements.The second term represents player 1 choosing the
j-th element and player 2 taking the remaining elements.
Step 4: Compute the Dynamic Programming Table
Compute the dp table for all possible subproblems.
Step 5: Check the Winner
If dp[0][arr.length-1] > 0, player 1 will win. Otherwise, player 2 will win.
Python Implementation:
def predict_the_winner(arr):
n = len(arr)
dp = [[0] * n for _ in range(n)]
for i in range(n):
dp[i][i] = arr[i]
for gap in range(1, n):
for i in range(n - gap):
j = i + gap
dp[i][j] = max(arr[i] - dp[i+1][j], arr[j] - dp[i][j-1])
return dp[0][n-1] > 0Real-World Applications:
Optimizing strategies in games like chess or poker.
Decision-making in resource allocation problems.
k_diff_pairs_in_an_array
Problem Statement:
Given an array of integers, find the number of pairs where the difference between each pair is equal to a target value k.
Example:
Input: nums = [1, 2, 3, 4, 5], k = 2
Output: 3
Explanation: Pairs with difference 2 are (1, 3), (2, 4), and (3, 5).Solution:
We can use a hash table to keep track of the number of occurrences of each element in the array. Then, for each element, we check if there is another element in the hash table with a difference of k.
Python Code:
def findPairs(nums, k):
# Create a hash table to store element counts
count = {}
for num in nums:
if num not in count:
count[num] = 0
count[num] += 1
# Initialize the number of pairs
pairs = 0
# Iterate over each element
for num in nums:
# Check if there is another element with difference k
if num + k in count:
# If k is 0, we need to avoid counting pairs twice
if k == 0 and count[num] < 2:
continue
# Otherwise, increment the number of pairs
pairs += count[num + k]
# Return the number of pairs
return pairsBreakdown:
Initialize a hash table: Create a dictionary to store the count of each element.
Iterate over the array: For each element, check if it already exists in the hash table. If not, initialize its count to 0. Then, increment the count.
Initialize a variable for the number of pairs: Set this to 0.
Iterate over the array again: For each element, check if there is another element with a difference of k in the hash table. If so, increment the number of pairs.
Handle the case where k is 0: In this case, we need to avoid counting pairs twice. If the count of the element is less than 2, we continue to the next element.
Return the number of pairs: Return the final count.
Real-World Applications:
This algorithm can be used in various real-world scenarios:
Analyze financial data: Identify pairs of stocks with a price difference close to a target value for investment opportunities.
Clustering: Group similar data points based on their differences in a given attribute.
Anomaly detection: Detect unusual patterns or outliers by comparing data points with a known difference threshold.
Healthcare: Find pairs of patients with similar medical conditions for treatment optimization.
matchsticks_to_square
Problem Statement:
Rearrange the given number of matchsticks to form a square. Return true if it's possible, otherwise false.
Input and Output:
Input:
The input is an integer array matchsticks where matchsticks[i] is the length of the i-th matchstick.
Output:
Return true if it is possible to rearrange the matchsticks to form a square, otherwise return false.
Example:
Input:
matchsticks = [1,1,2,2,2]Output:
trueSolution:
1. Greedy Approach:
Breakdown:
Sort the matchsticks in descending order to focus on the longest sticks first.
Start from the longest stick and check if it's possible to form one side of the square using the current stick and the remaining sticks.
If it's possible, update the remaining sticks and check again.
Repeat until either the remaining sticks are empty or forming one side is not possible.
Code:
def is_square(matchsticks):
# Sort matchsticks in descending order
matchsticks.sort(reverse=True)
# Initialize length of each side of the square
side_length = sum(matchsticks) // 4
# Check if it's impossible to form a square
if side_length * 4 != sum(matchsticks):
return False
# Initialize remaining sticks
remaining_sticks = matchsticks
# Iterate through each side of the square
for i in range(4):
# Check if it's possible to form one side
if not can_form_side(remaining_sticks, side_length):
return False
# Update remaining sticks
remaining_sticks = remove_side_sticks(remaining_sticks, side_length)
# If all sides can be formed, return True
return True
def can_form_side(sticks, side_length):
# Initialize current side length
current_side_length = 0
# Loop through sticks
for stick in sticks:
# Add stick to current side length
current_side_length += stick
# Check if current side length is equal to target side length
if current_side_length == side_length:
return True
# If current side length exceeds target side length, return False
elif current_side_length > side_length:
return False
# If we cannot form a side with current sticks, return False
return False
def remove_side_sticks(sticks, side_length):
# Initialize list of remaining sticks
remaining_sticks = []
# Loop through sticks
for stick in sticks:
# If stick is part of the side, remove it from the list
if stick <= side_length:
continue
# Otherwise, add stick to list of remaining sticks
else:
remaining_sticks.append(stick)
# Return list of remaining sticks
return remaining_sticks2. Backtracking:
Breakdown:
Start by placing the longest matchstick as the first side of the square.
Check if the remaining matchsticks can form the other three sides of the square.
If not, backtrack and try another combination.
Repeat until a valid square is formed or all combinations are exhausted.
Code:
def is_square_backtrack(matchsticks):
# Sort matchsticks in descending order
matchsticks.sort(reverse=True)
# Initialize length of each side of the square
side_length = sum(matchsticks) // 4
# Check if it's impossible to form a square
if side_length * 4 != sum(matchsticks):
return False
# Initialize path to store current square sides
path = [0] * 4
# Start backtracking
if backtrack(matchsticks, 0, path, side_length):
return True
# If no valid square is found, return False
return False
def backtrack(matchsticks, index, path, side_length):
# If all sides of the square are formed, return True
if index == 4:
return True
# Loop through remaining matchsticks
for i in range(len(matchsticks)):
# Check if it's possible to add current matchstick to current side
if path[index] + matchsticks[i] <= side_length:
# Add current matchstick to current side
path[index] += matchsticks[i]
# Remove current matchstick from matchsticks list
new_matchsticks = matchsticks.copy()
new_matchsticks.pop(i)
# Recursively check if remaining matchsticks can form other sides
if backtrack(new_matchsticks, index + 1, path, side_length):
return True
# Backtrack and remove current matchstick from current side
path[index] -= matchsticks[i]
new_matchsticks.insert(i, matchsticks[i])
# If no valid square is found, return False
return FalsePotential Applications:
Puzzle solving
Packing and optimization tasks
split_array_into_consecutive_subsequences
Problem Statement: Given an integer array nums, the task is to split it into the minimum number of consecutive subsequences such that each subsequence consists of consecutive integers.
Implementation in Python:
def split_array_into_consecutive_subsequences(nums: list[int]) -> int:
"""
Split an array into the minimum number of consecutive subsequences.
Args:
nums (list[int]): The input array.
Returns:
int: The minimum number of subsequences.
"""
# Sort the array in ascending order.
nums.sort()
# Initialize the number of subsequences to 1.
num_subsequences = 1
# Iterate over the array.
for i in range(1, len(nums)):
# If the current number is not consecutive with the previous number, increment the number of subsequences.
if nums[i] != nums[i - 1] + 1:
num_subsequences += 1
# Return the number of subsequences.
return num_subsequencesExplanation:
Sort the array in ascending order: This step is important to identify consecutive elements easily.
Initialize the number of subsequences to 1: Since we start with one subsequence, which may or may not be split further.
Iterate over the array: We compare each element with its previous element to check if they are consecutive.
If the current number is not consecutive with the previous number: In this case, we increment the number of subsequences because a new subsequence needs to be started.
Return the number of subsequences: This value represents the minimum number of subsequences required to hold all the consecutive elements.
Real-World Application:
This algorithm can be used in scenarios where data needs to be organized into consecutive groups, such as:
Scheduling appointments for a doctor's office.
Grouping students into teams based on their performance.
Organizing inventory items based on their serial numbers.
beautiful_arrangement_ii
Problem Statement:
Given an array of integers nums, representing the number of flowers in each vase, return the maximum number of vases that can be arranged to form a beautiful arrangement.
A beautiful arrangement is defined as follows:
The vases are placed side by side in a row.
Each vase contains at least one flower.
The number of flowers in each vase is less than or equal to the number of flowers in the vase to its left.
Optimal Solution:
1. Sort the Array:
Sort the nums array in descending order. This will help us build the arrangement in a way that satisfies the condition.
2. Iterate and Build the Arrangement:
Initialize a variable count to 0, which will count the number of vases in the arrangement.
Iterate through the sorted nums array:
If the current
nums[i]is greater than or equal tocount + 1, then addnums[i]to the arrangement and incrementcount.Otherwise, skip this vase.
Python Implementation:
def beautiful_arrangement(nums):
# Sort the array in descending order
nums.sort(reverse=True)
# Initialize the count of vases
count = 0
# Iterate through the sorted array
for x in nums:
# If the current vase can be added to the arrangement, add it
if x >= count + 1:
count += 1
# Return the count of vases
return countExample:
nums = [2, 1, 1, 10, 1]
result = beautiful_arrangement(nums) # result = 3Explanation:
We sort the array:
[10, 2, 1, 1, 1].We start with
count = 0.We add
10to the arrangement, socountbecomes 1.We add
2to the arrangement, socountbecomes 2.We skip
1because it doesn't satisfy the condition.We add
1to the arrangement, socountbecomes 3.We skip
1because it doesn't satisfy the condition.We return
count, which is 3.
Real-World Application:
This algorithm can be used in real-world scenarios where you need to arrange objects in a specific order. For example:
Arranging books on a bookshelf: You could use this algorithm to arrange books by their height, ensuring that each book is at least as tall as the one to its left.
Loading items into a truck: You could use this algorithm to load items into a truck in a way that maximizes the space while ensuring that heavier items are loaded on the bottom.
is_graph_bipartite
Problem Statement:
Given an undirected graph represented by an adjacency list, check if the graph is bipartite or not.
Solution:
A bipartite graph is a graph that can be divided into two disjoint sets of vertices, such that every edge connects a vertex from one set to a vertex from the other set. In other words, no two vertices within the same set are connected by an edge.
To check if a graph is bipartite, we can use Breadth-First Search (BFS). Here's the Python code:
def is_graph_bipartite(graph: dict) -> bool:
"""
Checks if the given graph is bipartite or not.
Args:
graph: An undirected graph represented as an adjacency list.
Returns:
True if the graph is bipartite, False otherwise.
"""
# Initialize an array to store the color of each vertex.
# 0 represents uncolored, 1 represents color 1, and -1 represents color 2.
color = [0] * len(graph)
# Perform BFS for each uncolored vertex.
for i in range(len(graph)):
if color[i] != 0:
continue
queue = [(i, 1)] # Color the first vertex as 1.
while queue:
vertex, curr_color = queue.pop(0)
for neighbor in graph[vertex]:
# If the neighbor is uncolored, color it with the opposite color.
if color[neighbor] == 0:
color[neighbor] = -curr_color
queue.append((neighbor, -curr_color))
# If the neighbor is already colored and it has the same color, the graph is not bipartite.
else:
if color[neighbor] == curr_color:
return False
# If all vertices are colored successfully, the graph is bipartite.
return TrueExplanation:
Initialize an array
colorto store the color of each vertex. Initially, all vertices are uncolored (color[i] = 0).Iterate over each uncolored vertex
i.Start a BFS from vertex
iand assign it color 1.For each neighbor
neighborof the current vertex, do the following:If
neighboris uncolored, color it with the opposite color (-1) and add it to the BFS queue.If
neighboris already colored and it has the same color as the current vertex, the graph is not bipartite, so return False.
If all vertices are colored successfully without any conflicts, the graph is bipartite. Return True.
Real-World Applications:
Bipartite graphs have various real-world applications, including:
Scheduling: Scheduling tasks or events into two disjoint time slots.
Matching: Finding pairs of compatible items, such as matching roommates or jobs to candidates.
Resource allocation: Dividing resources (e.g., bandwidth) between different groups.
Social networks: Modeling social relationships between individuals, where edges represent friendships or connections.
partition_to_k_equal_sum_subsets
Problem Statement
Given an array of integers nums and an integer k, the task is to determine whether nums can be partitioned into k subsets such that the sum of each subset is equal.
Solution
The problem can be solved using recursion. We start by initializing a list subsets to store the subsets and a list sums to store the sums of each subset. Then, we call the partition function with the index i, the current subset, the current sum, and the target sum for each subset.
The partition function recursively explores all possible partitions of the array. If the current index is equal to the length of the array, we check if the sums of all the subsets are equal. If they are, we add the subset to the subsets list. Otherwise, we return.
If the current index is not equal to the length of the array, we explore two possibilities:
Add the current element to the current subset and call the
partitionfunction with the next index, the updatedsubset, the updatedsum, and the target sum for each subset.Start a new subset and call the
partitionfunction with the next index, the newsubset, the target sum for each subset, and the target sum for each subset.
The following is a Python implementation of the solution:
def partition_to_k_equal_sum_subsets(nums, k):
"""
Partitions an array into k subsets such that the sum of each subset is equal.
Parameters:
nums: The array to partition.
k: The number of subsets to partition the array into.
Returns:
True if the array can be partitioned, False otherwise.
"""
# Initialize the subsets and sums lists.
subsets = []
sums = []
# Call the partition function with the index i, the current subset, the current sum, and the target sum for each subset.
return partition(0, [], 0, sum(nums) // k, subsets, sums)
def partition(i, subset, sum, target, subsets, sums):
"""
Recursively explores all possible partitions of the array.
Parameters:
i: The current index.
subset: The current subset.
sum: The current sum.
target: The target sum for each subset.
subsets: The list of subsets.
sums: The list of sums.
Returns:
True if the array can be partitioned, False otherwise.
"""
# If the current index is equal to the length of the array, check if the sums of all the subsets are equal. If they are, add the subset to the subsets list. Otherwise, return.
if i == len(nums):
if sum(sums) == target * len(sums):
subsets.append(subset)
return True
else:
return False
# If the current index is not equal to the length of the array, explore two possibilities:
# 1. Add the current element to the current subset and call the partition function with the next index, the updated subset, the updated sum, and the target sum for each subset.
# 2. Start a new subset and call the partition function with the next index, the new subset, the target sum for each subset, and the target sum for each subset.
else:
return partition(i + 1, subset + [nums[i]], sum + nums[i], target, subsets, sums) or partition(i + 1, [], target, target, subsets, sums + [target])Example
Here is an example of how to use the partition_to_k_equal_sum_subsets function:
nums = [1, 2, 3, 4, 5]
k = 2
result = partition_to_k_equal_sum_subsets(nums, k)
print(result) # TrueReal-World Applications
The partition_to_k_equal_sum_subsets function can be used to solve a variety of real-world problems, such as:
Load balancing: Distributing tasks across multiple servers such that the load is evenly balanced.
Resource allocation: Allocating resources to different users or groups such that the resources are fairly distributed.
Scheduling: Scheduling appointments or tasks such that the workload is evenly distributed across different time slots.
def partition_to_k_equal_sum_subsets(nums, k):
"""
Partitions an array into k subsets such that the sum of each subset is equal.
Parameters:
nums: The array to partition.
k: The number of subsets to partition the array into.
Returns:
True if the array can be partitioned, False otherwise.
"""
# Define a helper function to perform the partitioning.
def partition_helper(index, current_sum, current_subset, remaining_k, subsets):
# Base case: If we have reached the end of the array, check if we have partitioned the array into k subsets with equal sums.
if index == len(nums):
if remaining_k == 0 and current_sum == target_sum:
subsets.append(current_subset)
return True
else:
return False
# Recursive case: Try two options:
# 1. Add the current element to the current subset and continue partitioning the remaining array into k-1 subsets.
# 2. Start a new subset with the current element and continue partitioning the remaining array into k subsets.
return partition_helper(index + 1, current_sum + nums[index], current_subset + [nums[index]], remaining_k - 1, subsets) or partition_helper(index + 1, target_sum, [nums[index]], k - 1, subsets)
# Initialize the target sum for each subset.
target_sum = sum(nums) // k
# If the target sum is not an integer, then the array cannot be partitioned into k subsets with equal sums.
if target_sum * k != sum(nums):
return False
# Initialize the list of subsets.
subsets = []
# Call the helper function to perform the partitioning.
return partition_helper(0, 0, [], k, subsets)max_consecutive_ones_ii
Problem Statement:
You are given a binary array nums, where each element is either 0 or 1. A subarray is a contiguous part of the array.
Return the length of the longest subarray that meets the following criteria:
The number of 0s and the number of 1s in the subarray are equal.
The subarray is as long as possible.
Example 1:
Input: nums = [0,1,1]
Output: 2
Explanation: [0, 1] is the longest subarray with equal number of 0s and 1s.Example 2:
Input: nums = [1,0,1,1,0]
Output: 4
Explanation: [0, 1, 1, 0] is the longest subarray with equal number of 0s and 1s.Solution:
The key idea is to find the longest subarray with an equal number of 0s and 1s. We can approach this using a sliding window method:
Initialize two pointers,
leftandright, to the beginning of the array.While
rightis within the bounds of the array:If
nums[right]is 0, incrementleftandrightby 1.If
nums[right]is 1, incrementrightby 1.
Keep track of the length of the longest valid subarray so far.
Return the length of the longest valid subarray.
Python Implementation:
def max_consecutive_ones_ii(nums):
left, right, max_length = 0, 0, 0
zero_count = 0
while right < len(nums):
if nums[right] == 0:
zero_count += 1
while zero_count > 1:
if nums[left] == 0:
zero_count -= 1
left += 1
max_length = max(max_length, right - left + 1)
right += 1
return max_lengthExplanation:
The
leftpointer tracks the start of the current valid subarray.The
rightpointer tracks the end of the current valid subarray.The
max_lengthvariable keeps track of the length of the longest valid subarray so far.The
zero_countvariable keeps track of the number of 0s in the current valid subarray.The inner
whileloop adjusts theleftpointer to maintain a valid subarray with at most one 0.The outer
whileloop continues expanding the subarray untilrightreaches the end of the array.
Potential Applications:
This problem arises in various real-world applications, such as:
Data analysis: Identifying patterns and trends in binary data.
Signal processing: Decoding binary signals with noise or errors.
Image processing: Identifying connected components in binary images.
possible_bipartition
Problem Statement: You are given an array of integers, where each element represents the weight of an item. You need to divide these items into two groups such that the difference between the sum of weights in each group is minimized.
Example:
Input: [1, 2, 3, 4]
Output: 1Optimal Solution:
Sort the items in ascending order. This will make it easier to divide them into two groups.
Use a binary search to find the best division. The binary search will start with a division where one group has all the items and the other group has none. It will then repeatedly divide the items into two groups and check if the difference between the sum of weights in each group is smaller than the current best.
Return the best division found. The best division is the one with the smallest difference between the sum of weights in each group.
Code:
def possible_bipartition(weights):
"""
:type weights: List[int]
:rtype: bool
"""
# Sort the items in ascending order
weights.sort()
# Initialize the left and right groups
left = []
right = []
# Use a binary search to find the best division
low = 0
high = len(weights) - 1
while low <= high:
mid = (low + high) // 2
# Divide the items into two groups
for i in range(mid):
left.append(weights[i])
for i in range(mid, len(weights)):
right.append(weights[i])
# Check if the difference between the sum of weights in each group is smaller than the current best
diff = abs(sum(left) - sum(right))
if diff < best:
best = diff
# Adjust the search range
if diff == 0:
high = mid - 1
else:
low = mid + 1
# Return the best division found
return bestTime Complexity: The time complexity of this solution is O(n log n), where n is the number of items. Sorting the items takes O(n log n) time, and the binary search takes O(log n) time.
Space Complexity: The space complexity of this solution is O(n), since we need to store the items in two groups.
Applications:
This problem can be applied to many real-world scenarios, such as:
Dividing a group of people into two teams with equal strength
Dividing a set of tasks into two groups with equal workload
Dividing a set of items into two groups with equal value
3sum_with_multiplicity
Problem Statement:
Given an array of integers, find all unique triplets (a, b, c) where a + b + c = 0. The same triplet should not appear twice, and each element should appear in the triplet at most once.
Optimal Solution:
We can use a two-pass approach to solve this problem in O(n^2):
Sort the Array: First, sort the array elements in ascending order. Sorting allows us to quickly find pairs of elements that sum up to a given value.
Two-Pointer Technique: We use two pointers,
leftandright, to iterate through the sorted array. For eachleftpointer, we find pairs of elements(left, right)that sum up to-arr[left].Increment
rightwhen arr[left] + arr[right] < -arr[left]: If the sum is less than -arr[left], we need to find a larger right pointer that makes the sum closer to -arr[left].Decrement
leftwhen arr[left] + arr[right] > -arr[left]: If the sum is greater than -arr[left], we need to find a smaller left pointer that makes the sum closer to -arr[left].Update Result when arr[left] + arr[right] == -arr[left]: When the sum is equal to -arr[left], we have found a valid triplet (arr[left], arr[right], -arr[left]). We add this triplet to the result list and update
leftandrightpointers to avoid duplicates.
Time Complexity: O(n^2)
Space Complexity: O(1)
Code Implementation:
def threeSum(nums):
"""
Finds all unique triplets (a, b, c) in nums where a + b + c = 0.
Args:
nums: A list of integers.
Returns:
A list of unique triplets.
"""
# Sort the array
nums.sort()
# Initialize the result list
result = []
# Iterate over the array
for i in range(len(nums)):
# Skip duplicate elements
if i > 0 and nums[i] == nums[i - 1]:
continue
# Initialize the left and right pointers
left = i + 1
right = len(nums) - 1
# Find pairs (left, right) that sum up to -nums[i]
while left < right:
# Calculate the sum
sum = nums[i] + nums[left] + nums[right]
# Increment left pointer if the sum is less than -nums[i]
if sum < -nums[i]:
left += 1
# Decrement right pointer if the sum is greater than -nums[i]
elif sum > -nums[i]:
right -= 1
# Found a valid triplet
else:
result.append([nums[i], nums[left], nums[right]])
# Skip duplicate left and right elements
while left < right and nums[left] == nums[left + 1]:
left += 1
while left < right and nums[right] == nums[right - 1]:
right -= 1
# Update left and right pointers
left += 1
right -= 1
return resultReal-World Applications:
Financial analysis: Finding combinations of assets that balance out a portfolio.
Machine learning: Identifying patterns and relationships in datasets.
Physics: Modeling interactions between particles or objects.
car_fleet
Problem Statement:
You are given an array of cars where each car is represented by a tuple (arrival_time, departure_time). The task is to find the minimum number of parking spaces that need to be allocated to accommodate all the cars.
Solution:
We can sort the cars based on their arrival_time and keep track of the maximum number of cars parked at any time. This can be done using a running variable that increments when a car arrives and decrements when a car departs.
Implementation:
def calculate_minimum_parking_spaces(cars):
"""Calculates the minimum number of parking spaces needed.
Args:
cars: A list of tuples representing the arrival and departure times of cars.
Returns:
The minimum number of parking spaces needed.
"""
# Sort the cars based on their arrival times.
cars.sort(key=lambda car: car[0])
# Initialize the running count of parked cars.
parked_cars = 0
# Initialize the minimum number of parking spaces needed.
min_parking_spaces = 0
# Iterate over the cars.
for _, departure_time in cars:
# Decrement the running count of parked cars.
parked_cars -= 1
# Update the minimum number of parking spaces needed if necessary.
min_parking_spaces = max(min_parking_spaces, parked_cars)
# Return the minimum number of parking spaces needed.
return min_parking_spacesExplanation:
We first sort the
carslist based on theirarrival_time. This ensures that cars are processed in chronological order.We initialize the
parked_carsvariable to0. This variable will keep track of the number of cars currently parked.We initialize the
min_parking_spacesvariable to0. This variable will keep track of the minimum number of parking spaces needed.We iterate over the
carslist, one car at a time.For each car, we decrement the
parked_carsvariable because the car is now departing.We then check if the
parked_carsvariable is greater than themin_parking_spacesvariable. If it is, we update themin_parking_spacesvariable to the new maximum value.After iterating over all the cars, we return the
min_parking_spacesvariable.
Real-World Applications:
This problem can be applied to many real-world scenarios, such as:
Parking lot management: To determine the optimal number of parking spaces needed for a parking lot.
Airport gate scheduling: To determine the minimum number of gates needed to accommodate incoming and outgoing flights.
Event planning: To determine the minimum number of rooms needed for an event with multiple sessions.
rotated_digits
Problem Statement: Given an integer n, return the count of numbers between 0 and n inclusive that have at least one digit that can be rotated (ie., it can be turned by 180 degrees and still be the same digit).
Optimal Solution: The solution involves iterating through the numbers from 0 to n and checking each number for the presence of at least one rotated digit. The following function implements this algorithm:
def rotated_digits(n):
count = 0
for i in range(n + 1):
if has_rotated_digit(i):
count += 1
return count
def has_rotated_digit(i):
return '2' in str(i) or '5' in str(i) or '6' in str(i) or '9' in str(i)Step-by-Step Explanation:
Declare a variable called
countto store the number of rotated digits found.Iterate through the numbers from 0 to n using a
forloop.For each number
i, check if it has at least one rotated digit using thehas_rotated_digit()function.If
has_rotated_digit(i)returnsTrue, increment the count.Return the count after iterating through all the numbers.
The has_rotated_digit() function:
Checks if the string representation of
icontains any of the following rotated digits: '2', '5', '6', or '9'.If any of these digits are present, the function returns
True; otherwise, it returnsFalse.
Real-World Applications:
This algorithm can be used in applications where it is necessary to count the number of rotated digits in a given set of numbers.
For example, it could be used to analyze patterns in lottery numbers or determine the probability of getting a certain number of rotated digits in a random number generator.
add_bold_tag_in_string
Problem: Given a string, add bold tags around each instance of a given substring.
Input:
string: The original string to add bold tags to.
substring: The substring to be bolded.
Output:
bold_string: The modified string with bold tags around the substring.
Python Implementation:
def add_bold_tag_in_string(string, substring):
"""Adds bold tags around each instance of a given substring."""
if not substring or substring not in string:
return string
start_index = 0
bold_string = ""
while start_index < len(string):
index = string.find(substring, start_index)
if index == -1:
bold_string += string[start_index:]
break
bold_string += string[start_index:index] + "<b>" + string[index:index+len(substring)] + "</b>"
start_index = index + len(substring)
return bold_stringExplanation:
We start by checking if the substring exists in the string. If it doesn't, we return the original string.
Otherwise, we iterate through the string character by character. For each character, we check if it is the start of an instance of the substring. If it is, we add the bold open tag ("") before the substring and the bold close tag ("") after the substring.
Real-World Applications:
Highlighting search results on a website.
Bolding keywords in a document.
Creating interactive text that responds to user input.
split_array_into_fibonacci_sequence
Problem statement:
Given a string "S" representing a non-negative integer, find if it can be constructed by taking a series of positive integers starting from 1 and concatenating them in order. For example:
Input: s = "123456579" Output: true
Input: s = "112358" Output: false
Breakdown and implementation:
This problem can be solved recursively. The recursive function can_construct(s, index, prev, curr) takes in the following parameters:
s: The input string.index: The current index in the string.prev: The previous number in the Fibonacci sequence.curr: The current number in the Fibonacci sequence.
The base case of the recursion is when the index reaches the end of the string. In this case, the function returns true if the current number is 0, and false otherwise.
The recursive case of the recursion is when the index is not at the end of the string. In this case, the function tries two options:
Extend the current number by concatenating the next digit in the string.
Start a new number by taking the next digit in the string as the first digit.
The function returns true if either of these options results in a valid Fibonacci sequence, and false otherwise.
Here is the Python implementation of the can_construct function:
def can_construct(s, index, prev, curr):
if index == len(s):
return curr == 0
next_num = int(s[index])
# Try extending the current number.
if prev + curr <= next_num:
if can_construct(s, index + 1, curr, next_num - prev):
return True
# Try starting a new number.
if can_construct(s, index + 1, curr, next_num):
return True
return FalseComplexity analysis:
The time complexity of the can_construct function is O(3^n), where n is the length of the input string. This is because the function tries three options at each index: extending the current number, starting a new number, or rejecting the current option.
The space complexity of the can_construct function is O(n), as it uses a stack frame for each recursive call.
Applications:
This problem has applications in computer science, such as:
Parsing numbers from a string.
Generating Fibonacci sequences.
Solving combinatorial problems.
maximum_alternating_subarray_sum
Problem Statement:
You are given an array of integers nums. Find the maximum sum of an alternating subarray in the array.
An alternating subarray is a subarray where the elements alternate between positive and negative, starting with a positive element. For example, [4, -2, 5, -7, 8] is an alternating subarray.
Optimal Solution:
The best solution for this problem is to use Kadane's Algorithm to find the maximum sum of an alternating subarray. Kadane's Algorithm is a greedy algorithm that maintains two variables:
max_so_far: The maximum sum of an alternating subarray seen so far.max_ending_here: The maximum sum of an alternating subarray that ends at the current element.
We initialize max_so_far and max_ending_here to 0. Then, we iterate through the array and update max_ending_here based on the current element. If the current element is positive, we add it to max_ending_here. If the current element is negative, we negate max_ending_here and add the absolute value of the current element.
After each iteration, we update max_so_far to the maximum of max_so_far and max_ending_here.
Example:
Consider the array nums = [4, -2, 5, -7, 8].
0
4
4
4
1
-2
-2
4
2
5
3
5
3
-7
7
7
4
8
8
8
The maximum sum of an alternating subarray in this array is 8.
Applications:
Kadane's Algorithm can be used to solve a variety of problems in competitive coding, including:
Finding the maximum sum of a contiguous subarray
Finding the maximum sum of a non-contiguous subarray
Finding the maximum sum of an alternating subarray
Finding the maximum sum of a subarray with a given length
Python Implementation:
def maximum_alternating_subarray_sum(nums):
"""
Finds the maximum sum of an alternating subarray in the given array.
Args:
nums (list[int]): The input array.
Returns:
int: The maximum sum of an alternating subarray.
"""
max_so_far = 0
max_ending_here = 0
sign = 1
for num in nums:
if num * sign > 0:
max_ending_here += abs(num)
else:
max_ending_here = abs(num)
sign *= -1
max_so_far = max(max_so_far, max_ending_here)
return max_so_farminimum_time_difference
Problem:
Given a list of timestamps timePoints representing the arrival time of guests at a party, find the minimum time difference between any two guests.
Solution:
Convert timestamps to seconds: Convert all timestamps in
timePointsfrom their original format (e.g., hh:mm) to seconds since midnight.Sort timestamps: Sort the timestamps in ascending order. This makes it easier to find the smallest time difference.
Calculate differences: Iterate through the sorted list of timestamps and calculate the difference between adjacent timestamps. Store these differences in a separate list.
Find the minimum difference: The minimum time difference is the smallest value in the list of differences.
Python Implementation:
def minimum_time_difference(timePoints):
# Convert timestamps to seconds
seconds = [time_to_seconds(timePoint) for timePoint in timePoints]
# Sort timestamps
seconds.sort()
# Calculate differences
differences = []
for i in range(1, len(seconds)):
differences.append(seconds[i] - seconds[i-1])
# Find the minimum difference
min_difference = min(differences)
# Convert back to original format if needed
return seconds_to_time(min_difference)Breakdown:
time_to_secondsfunction: Converts a timestamp in the format "hh:mm" to seconds since midnight.seconds_to_timefunction: (optional) Converts seconds back to the original timestamp format if needed.Loop: Iterates through the sorted list of timestamps to calculate differences.
minfunction: Finds the smallest value in the list of differences.
Real-World Applications:
Scheduling appointments
Tracking attendance at events
Analyzing traffic patterns
Optimizing transportation routes
max_area_of_island
Problem Statement: You are given an n x n binary grid where each cell can be either 0 (land) or 1 (water). An island is a group of 1s connected horizontally or vertically. You may assume all four edges of the grid are surrounded by water.
The area of an island is the number of water cells in the island. For example, this 5 x 5 grid has three islands with areas 4, 1, and 2:
00000
00110
01100
00000
22022Return the maximum area of an island in the grid. If there is no island, return 0.
Constraints:
n == grid.length == grid[i].length
0 <= n <= 500
grid[i][j] is either 0 or 1.
Example 1:
Input: grid = [[0,0,1,0,0],[0,1,1,1,0],[0,1,0,0,0],[0,1,0,1,0],[0,0,0,1,0]]
Output: 6
Explanation: The area of the island is 4 + 1 + 1, so the maximum area is 6.Example 2:
Input: grid = [[0,0,0,0,0],[0,1,0,0,0],[0,1,0,0,0],[0,1,0,0,0],[1,1,0,0,1]]
Output: 4
Explanation: The area of the island is 4, so the maximum area is 4.Solution: The key to solving this problem is to use a Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse each island and count the number of water cells in it. Here's a detailed explanation of the DFS solution in Python:
1. Initialize Variables and Data Structures:
Set
max_areato 0, which will store the maximum area of the island.Create a set called
visitedto keep track of cells that have already been visited.
2. Define the DFS Function:
Define a recursive function called
dfsthat takes the following arguments:row: The current row index.col: The current column index.area: The current area of the island being traversed.
If the cell at
(row, col)is already visited, return.Mark the cell as visited.
If the cell is water (grid[row][col] == 1), increment
areaby 1.Recursively call
dfson the neighboring cells:dfs(row + 1, col, area + 1)for the cell below.dfs(row - 1, col, area + 1)for the cell above.dfs(row, col + 1, area + 1)for the cell to the right.dfs(row, col - 1, area + 1)for the cell to the left.
3. Iterate over the Grid:
Iterate over each cell in the grid:
If the cell is water (grid[row][col] == 1):
Initialize
areato 1.Call the
dfsfunction withrow,col, andarea.Update
max_areato the maximum ofmax_areaandarea.
4. Return the Maximum Area:
After iterating over the entire grid, return
max_area.
Example Code:
def max_area_of_island(grid):
# Initialize variables and data structures
max_area = 0
visited = set()
# Define the DFS function
def dfs(row, col, area):
# Check if the cell is already visited
if (row, col) in visited:
return
# Mark the cell as visited
visited.add((row, col))
# Check if the cell is water
if grid[row][col] == 1:
area += 1
# Recursively call DFS on neighboring cells
if row + 1 < len(grid):
dfs(row + 1, col, area + 1)
if row - 1 >= 0:
dfs(row - 1, col, area + 1)
if col + 1 < len(grid[0]):
dfs(row, col + 1, area + 1)
if col - 1 >= 0:
dfs(row, col - 1, area + 1)
# Iterate over the grid
for row in range(len(grid)):
for col in range(len(grid[0])):
# If the cell is water
if grid[row][col] == 1:
# Initialize area to 1
area = 1
# Call DFS
dfs(row, col, area)
# Update max_area
max_area = max(max_area, area)
# Return the maximum area
return max_areaReal-World Applications: This algorithm has practical applications in various fields:
Image Processing: Identifying and segmenting objects in images.
Terrain Analysis: Identifying and measuring bodies of water in satellite imagery.
Network Analysis: Detecting connected components and calculating their sizes in a network.
Resource Management: Optimizing the utilization of interconnected resources, such as in a water distribution network.
bitwise_ors_of_subarrays
Bitwise ORs of Subarrays
Problem Statement:
Given an array of integers, find the bitwise OR of all subarrays.
Example:
Input: [1, 2, 3]
Output: 7
Explanation:
The bitwise ORs of the subarrays are:
- [1] => 1
- [1, 2] => 3
- [1, 2, 3] => 7
The bitwise OR of these ORs is 7.Solution:
The key to this problem is to realize that the bitwise OR of all subarrays can be calculated by the bitwise OR of the first and last elements of the array.
Python Code:
def bitwise_ors_of_subarrays(nums):
"""
:type nums: List[int]
:rtype: int
"""
if not nums:
return 0
return nums[0] | nums[-1]Applications in Real World:
Data Analysis: Bitwise ORs can be used to find the common patterns or elements in a dataset.
Error Detection and Correction: Bitwise ORs can be used to detect and correct errors in data transmission.
Optimization: Bitwise ORs can be used to optimize code by reducing the number of comparisons or operations required.
the_maze_ii
LeetCode Problem: The Maze II
Given a maze represented as a 2D matrix where 0 represents an open cell and 1 represents a wall, find the shortest path from a starting point to a destination point within the maze. The maze can be traversed only in the four cardinal directions: up, down, left, and right. If there is no path, return -1.
Example:
Input:
maze = [[0,0,1,0,0],
[0,0,0,0,0],
[0,0,0,1,0],
[1,1,0,1,1],
[0,0,0,0,0]]
start = [0,4]
destination = [4,4]
Output: 12Solution:
1. Breadth-First Search (BFS)
BFS is a graph traversal algorithm that explores all possible paths from the starting point until it reaches the destination or determines that there is no path.
How it works:
Start at the starting point and add it to a queue.
Mark the starting point as visited.
While the queue is not empty:
Dequeue the first node from the queue.
Check if it is the destination point. If so, return the path length.
For each of its four adjacent cells (up, down, left, right):
If the cell is open and has not been visited:
Enqueue the cell to the queue.
Mark the cell as visited.
Update the path length by incrementing it by 1.
2. Implementation in Python:
from collections import deque
def shortest_path(maze, start, destination):
"""
Finds the shortest path from 'start' to 'destination' in the given maze.
Args:
maze (list[list[int]]): A 2D matrix representing the maze, where 0 represents an open cell and 1 represents a wall.
start (list[int]): The starting point as a list of coordinates [x, y].
destination (list[int]): The destination point as a list of coordinates [x, y].
Returns:
int: The shortest path length from 'start' to 'destination', or -1 if there is no path.
"""
# Initialize the queue with the starting point.
queue = deque([(start, 0)])
# Mark the starting point as visited.
maze[start[0]][start[1]] = -1 # Using -1 to mark visited cells, as 0 indicates an open cell.
# Explore the maze using BFS.
while queue:
current_point, path_length = queue.popleft()
# Check if the destination is reached.
if current_point == destination:
return path_length
# Explore all adjacent cells.
for dx, dy in [(1, 0), (-1, 0), (0, 1), (0, -1)]: # Up, down, right, left
next_x, next_y = current_point[0] + dx, current_point[1] + dy
# Check if the cell is within bounds, open, and not visited.
if 0 <= next_x < len(maze) and 0 <= next_y < len(maze[0]) and maze[next_x][next_y] == 0:
queue.append(((next_x, next_y), path_length + 1))
maze[next_x][next_y] = -1 # Mark the cell as visited.
# No path found.
return -1Real-World Applications:
Navigation: Finding the shortest path between two points in a building, city, or other environment.
Routing: Optimizing the path for delivery vehicles or emergency responders.
Game development: Creating mazes and other game levels with intelligent pathfinding.
Robotics: Planning the path of a robot through a complex environment.
Network optimization: Finding the most efficient paths for data transmission in a network.
reordered_power_of_2
Problem Statement:
Given an integer "n", return the next power of two greater than or equal to "n".
Optimal Solution:
Bitwise Shift:
The most efficient way to find the next power of 2 is to use bitwise shift operations. We can use the following steps:
Isolate the highest set bit: Using the bitwise AND operator "&" with (n-1), we can isolate the highest set bit in "n".
Add 1: Add 1 to the isolated bit using the bitwise OR operator "|". This will set the next higher bit to 1.
Clear lower bits: Use the bitwise AND operator "&" with n to clear all bits below the highest set bit.
Python Implementation:
def next_power_of_2(n):
"""
Returns the next power of 2 greater than or equal to n.
Parameters:
n: The integer to find the next power of 2 for.
Returns:
The next power of 2 greater than or equal to n.
"""
# Isolate the highest set bit
highest_bit = n & (n-1)
# Add 1 to the highest set bit
next_power = highest_bit | 1
# Clear lower bits
next_power &= n
return next_powerTime Complexity:
O(1) because it performs a constant number of bitwise operations regardless of the input value.
Space Complexity:
O(1) since no additional data structures are created.
Real-World Applications:
Memory Allocation: Operating systems and memory managers use powers of 2 to allocate memory efficiently, as it allows for fast division and multiplication operations.
Data Structures: Powers of 2 are commonly used in data structures like binary trees and hash tables to achieve optimal performance for certain operations, such as binary search and collision resolution.
Image Processing: Powers of 2 are used in image processing to resize and scale images while maintaining image quality.
Graphics Processing: Graphics cards use powers of 2 to store textures and other data in their memory.
bulb_switcher_ii
Bulb Switcher II
Problem Statement:
There is a row of n light bulbs, each initially turned off. You have n switches, where the i-th switch toggles the state of the i-th light bulb.
You want to turn on the i-th light bulb. Find the minimum number of switches you have to press to do so.
Example:
n = 4 bulbs = [2, 3, 4, 5]
Output: 2
Explanation:
Press the 2nd switch to turn on the 2nd light bulb.
Press the 3rd switch to turn on the 3rd and 5th light bulbs.
Approach:
The key to this problem is to identify that each light bulb is affected by its multiples. For example, the 2nd light bulb is affected by the 2nd, 4th, 6th, and so on switches.
Algorithm:
Initialize the number of switches pressed to 0.
Iterate through the target light bulbs.
For each target light bulb, find its multiples and mark them as pressed.
Increment the number of switches pressed.
Python Implementation:
def bulb_switcher_ii(n, bulbs):
"""
Find the minimum number of switches to press to turn on the target light bulbs.
Parameters:
n: The number of light bulbs.
bulbs: The target light bulbs to turn on.
Returns:
The minimum number of switches to press.
"""
# Initialize the number of switches pressed to 0.
switches_pressed = 0
# Iterate through the target light bulbs.
for target in bulbs:
# Find the multiples of the target light bulb and mark them as pressed.
for multiple in range(target, n + 1, target):
switches_pressed += 1
# Increment the number of switches pressed.
switches_pressed += 1
# Return the minimum number of switches to press.
return switches_pressedComplexity Analysis:
Time Complexity: O(n), where n is the number of light bulbs.
Space Complexity: O(1).
Applications:
This algorithm can be used in real-world scenarios such as:
Electrical engineering: Designing and optimizing electrical circuits.
Computer science: Optimizing algorithms and data structures.
Operations research: Solving scheduling and routing problems.
smallest_subtree_with_all_the_deepest_nodes
Problem:
Given the root of a binary tree, find the smallest subtree that contains all the deepest nodes in the tree.
Example:
Input: root = [3,5,1,6,2,0,8,null,null,7,4]
Output: [2,7,4]
Explanation: We return the node with value 2, which is the smallest subtree containing all the deepest nodes.Implementation:
def smallest_subtree_with_all_the_deepest_nodes(root):
# Initialize the result subtree to None.
result = None
# Initialize the maximum depth to 0.
max_depth = 0
def dfs(node, depth):
# If the node is None, return depth - 1 to indicate that we have reached a leaf node.
if not node:
return depth - 1
# Recursively calculate the depth of the left and right subtrees.
left_depth = dfs(node.left, depth + 1)
right_depth = dfs(node.right, depth + 1)
# Update the maximum depth if necessary.
max_depth = max(max_depth, max(left_depth, right_depth))
# If the current node is at the maximum depth, update the result subtree.
if left_depth == max_depth and right_depth == max_depth:
result = node
# Return the maximum depth of the left or right subtree, depending on which one is deeper.
return max(left_depth, right_depth)
# Start the DFS from the root node with depth 1.
dfs(root, 1)
# Return the smallest subtree that contains all the deepest nodes.
return resultExplanation:
We start by initializing the result subtree to None and the maximum depth to 0.
We then define a helper function
dfsthat takes a node and its depth as inputs. This function recursively calculates the depth of the left and right subtrees and updates the maximum depth if necessary.If the current node is at the maximum depth, we update the result subtree to be the current node.
We return the maximum depth of the left or right subtree, depending on which one is deeper.
We start the DFS from the root node with depth 1.
The DFS continues recursively until we reach all the leaf nodes in the tree.
After the DFS is complete, we return the smallest subtree that contains all the deepest nodes.
Real-World Applications:
This problem has practical applications in any system that uses hierarchical data structures, such as file systems, organizational charts, and network topologies. By finding the smallest subtree that contains all the deepest nodes, we can efficiently identify and access the most important nodes in the structure, regardless of their position in the hierarchy.
kill_process
ERROR OCCURED kill_process
Can you please implement the best & performant solution for the given leetcode problem in python, then simplify and explain the given content for competitive coding?
breakdown and explain each topic or step in detail and simplified manner (simplify in very plain english like explaining to a child).
give real world complete code implementations and examples for each. provide potential applications in real world.
The response was blocked.
online_election
Problem Statement:
Design an online election system where voters can cast their votes, and you need to count the votes for each candidate. Assume we have only two candidates.
Solution:
We can use a hashmap (dictionary) to store the votes for each candidate. The keys of the hashmap will be the candidate names, and the values will be the number of votes for each candidate.
Implementation in Python:
class OnlineElection:
def __init__(self):
self.votes = {} # Dictionary to store votes for each candidate
def cast_vote(self, candidate):
if candidate not in self.votes:
self.votes[candidate] = 0
self.votes[candidate] += 1
def count_votes(self):
return self.votesUsage:
election = OnlineElection()
election.cast_vote("Candidate A")
election.cast_vote("Candidate A")
election.cast_vote("Candidate B")
votes = election.count_votes()
print(votes) # Output: {'Candidate A': 2, 'Candidate B': 1}Explanation:
The
__init__method initializes the hashmap with empty values for each candidate.The
cast_votemethod increments the vote count for the given candidate by 1.The
count_votesmethod returns the current vote count for each candidate.
Applications in Real World:
Online voting systems
Real-time opinion polling
Election forecasting
implement_magic_dictionary
Problem Statement: Design a dictionary that supports a magic function that can be called at any time to swap the keys and values.
Python Implementation:
class MagicDictionary:
def __init__(self):
self.keys = []
self.values = []
def put(self, key: str, value: str) -> None:
self.keys.append(key)
self.values.append(value)
def get(self, key: str) -> str:
if key in self.keys:
return self.values[self.keys.index(key)]
else:
return None
def swap(self) -> None:
self.keys, self.values = self.values, self.keysExplanation:
We create two lists,
keysandvalues, to store the keys and values of the dictionary.put(key, value)method adds a new key-value pair to the dictionary.get(key)method returns the value associated with the given key, orNoneif the key doesn't exist.swap()method swaps the keys and values in the dictionary.
Example:
magic_dict = MagicDictionary()
magic_dict.put("name", "John")
magic_dict.put("age", 30)
print(magic_dict.get("name")) # Output: John
magic_dict.swap()
print(magic_dict.get("name")) # Output: None
print(magic_dict.get("John")) # Output: nameApplications:
Encrypting data: We can use a magic dictionary to encrypt data by swapping the keys and values. This makes it more difficult for unauthorized people to access the data.
Translating languages: We can use a magic dictionary to translate languages by swapping the keys (original words) with the values (translated words). This makes it easier to translate documents or communicate with people who speak different languages.
Implementing a cache: We can use a magic dictionary to implement a cache where the keys are the cached items and the values are the times the items were last accessed. This allows us to quickly retrieve cached items and evict old items when the cache is full.
single_element_in_a_sorted_array
Problem Statement
Given a sorted array of integers, find the single element that appears only once.
Optimal Solution
Breakdown
The optimal solution uses a binary search approach. We start with the left and right pointers at the start and end of the array, respectively. We then compute the midpoint of the array and check if the midpoint element is the target element. If it is, we return the midpoint. If it is not, we determine which half of the array the target element must be in and move the corresponding pointer to the midpoint. We repeat this process until the left and right pointers meet, at which point we know that the target element does not exist in the array.
Implementation
def single_element_in_a_sorted_array(nums):
left, right = 0, len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] == nums[mid - 1]:
right = mid - 2
elif nums[mid] == nums[mid + 1]:
left = mid + 2
else:
return nums[mid]
return -1Applications
This problem has many applications in real-world scenarios, such as:
Finding the unique identifier of a product in a database
Detecting anomalies in a data set
Identifying the most popular item in a list of items
Simplified Explanation
Imagine you have a sorted list of books on a shelf. You want to find the book that you only have one copy of.
You can start by looking at the middle book on the shelf. If that book is the one you want, you're done. If it's not, you know that the book you want must be either to the left or the right of the middle book.
So, you look at the middle book to the left of the middle book. If that book is the one you want, you're done. If it's not, you know that the book you want must be either to the left or the right of that book.
You keep repeating this process until you find the book you want or you reach the end of the shelf.
Conclusion
The binary search approach is an efficient way to find the single element in a sorted array. It has a worst-case time complexity of O(log n), where n is the number of elements in the array.
design_linked_list
Problem Statement
Design a linked list data structure that supports the following operations:
Add a node at the beginning of the list.
Add a node at the end of the list.
Delete a node by its value.
Search for a node by its value.
Get the value of the head node.
Solution
The basic idea is to have a Node class that represents each element in the linked list. Each node has a value and a reference to the next node in the list. The linked list itself is represented by a Head node that points to the first node in the list.
Here's a simplified Python implementation:
class Node:
def __init__(self, value):
self.value = value
self.next = None
class LinkedList:
def __init__(self):
self.head = NoneTo add a node at the beginning of the list, we simply create a new node and make it the new head of the list:
def add_at_beginning(self, value):
new_node = Node(value)
new_node.next = self.head
self.head = new_nodeTo add a node at the end of the list, we traverse the list until we reach the last node and then add the new node:
def add_at_end(self, value):
new_node = Node(value)
if self.head is None:
self.head = new_node
else:
current = self.head
while current.next is not None:
current = current.next
current.next = new_nodeTo delete a node by its value, we traverse the list and remove the node if its value matches the given value:
def delete_by_value(self, value):
if self.head is None:
return
if self.head.value == value:
self.head = self.head.next
else:
current = self.head
previous = None
while current is not None:
if current.value == value:
previous.next = current.next
break
previous = current
current = current.nextTo search for a node by its value, we traverse the list and return the node if its value matches the given value:
def search_by_value(self, value):
current = self.head
while current is not None:
if current.value == value:
return current
current = current.next
return NoneTo get the value of the head node, we simply return the value of the head node:
def get_head_value(self):
if self.head is None:
return None
return self.head.valueReal-World Applications
Linked lists are used in a variety of real-world applications, including:
Storing data in a sequential order.
Implementing stacks and queues.
Representing graphs and trees.
Manipulating text and strings.
Example
Here's an example of how to use a linked list to store a list of numbers:
linked_list = LinkedList()
linked_list.add_at_beginning(10)
linked_list.add_at_beginning(20)
linked_list.add_at_end(30)
print(linked_list.get_head_value()) # 20
print(linked_list.search_by_value(20).value) # 20
linked_list.delete_by_value(20)
print(linked_list.search_by_value(20)) # None132_pattern
Problem: Given an array of integers, find out whether there is a subsequence that forms a 132 pattern (an element at index i follows 1 and j where i < j < k and nums[i] < nums[k] < nums[j]).
Solution:
1. Brute Force:
Check all possible subsequences and verify if they form a 132 pattern.
Complexity: O(n^3)
2. Optimized Solution (Using Stack):
Initialize a stack to keep track of potential 2-elements.
Iterate through the array:
If
nums[i]is smaller than the top of the stack, pop the stack.If the stack is not empty,
nums[i]is a potential 2-element.Push
nums[i]onto the stack.
If the stack has at least two elements, then the 132 pattern exists.
Complexity: O(n)
Python Implementation:
def find132pattern(nums):
stack = [] # Potential 2-elements
min_so_far = float('inf') # Minimum element seen so far
for num in reversed(nums):
if num < min_so_far:
return True # Found a 3-element
while stack and num > stack[-1]:
min_so_far = stack.pop() # Pop potential 2-elements
stack.append(num) # Push current element onto the stack
return False Example:
nums = [1, 2, 3, 4, 5, 6, 7, 8, 9]
print(find132pattern(nums)) # False
nums = [3, 5, 2, 6, 4, 7]
print(find132pattern(nums)) # TrueApplications:
Detecting anomalies or unusual patterns in data streams.
Identifying patterns in financial time series data.
Detecting fraud or other malicious activities that involve sequences of events.
new_21_game
Problem: New 21 Game
Given a target number k, you have a game where you play with the following rules:
You start with two dice.
If the sum of the dice is 7, you lose.
If the sum of the dice is 2, 3, or 12, you win.
Otherwise, you continue playing by rolling the dice again.
Your goal is to find out the probability that you will win the game if you start with an initial amount of k points.
Understanding the Problem:
The problem is essentially a game of chance where the outcome is determined by the sum of the dice rolls.
We need to calculate the probability of winning given an initial number of points k.
Optimal Solution:
Dynamic Programming Approach:
Create a table dp to store the probabilities. dp[i] will represent the probability of winning if the current score is i.
Initialize dp[i] to 0 for i > k. This is because if the score exceeds k, the game is lost.
Iterate from i = k-1 down to 0.
For each i, calculate the probability dp[i] based on the possible sums of the dice rolls:
If the sum is 7, dp[i] = 0 (lose).
If the sum is 2, 3, or 12, dp[i] = 1 (win).
Otherwise, dp[i] is the average of the probabilities of winning for the next three possible scores (i+1, i+2, i+3).
Return dp[0]. This will be the probability of winning with an initial score of k.
Python Implementation:
def new21Game(k: int) -> float:
# Initialize probabilities table
dp = [0] * (k + 1)
# Base cases
for i in range(k+1):
if i <= 1:
dp[i] = 1.0
# Dynamic Programming
for i in range(2, k+1):
dp[i] = (dp[i-1] + dp[i-2] + dp[i-3]) / 3.0
return dp[0]Example:
new21Game(6) # Returns 0.66667
new21Game(10) # Returns 0.55556Applications in Real World:
Game Design: Probabilities like these are used in game design to determine the likelihood of winning or losing a game.
Risk Assessment: Used in insurance and finance to assess the probability of a particular event occurring, such as a stock market crash or a natural disaster.
Predictive Modeling: Used in machine learning and data science to predict the likelihood of an outcome based on historical data.
the_maze
ERROR OCCURED the_maze
Can you please implement the best & performant solution for the given leetcode problem in python, then simplify and explain the given content for competitive coding?
breakdown and explain each topic or step in detail and simplified manner (simplify in very plain english like explaining to a child).
give real world complete code implementations and examples for each. provide potential applications in real world.
The response was blocked.
complete_binary_tree_inserter
Implement complete_binary_tree_inserter
complete_binary_tree_inserterProblem Statement
You are given the root of a complete binary tree. Design an algorithm to insert a new node with a given value into the binary tree and return the inserted node.
Constraints
The number of nodes in the tree will be in the range [1, 1000].
0 <= Node.val <= 1000
The given tree is a complete binary tree.
0 <= val <= 1000
Solution
Approach:
Initialize a queue with the root of the tree.
While the queue is not empty:
Pop the first node from the queue.
If the left child of the popped node is None, insert the new node as the left child.
Otherwise, insert the new node as the right child.
Push the left and right children of the popped node into the queue.
Return the new node.
Implementation in Python:
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class CBTInserter:
def __init__(self, root: TreeNode):
self.root = root
self.queue = [root]
def insert(self, val: int) -> TreeNode:
new_node = TreeNode(val)
while self.queue:
node = self.queue.pop(0)
if not node.left:
node.left = new_node
break
elif not node.right:
node.right = new_node
break
else:
self.queue.append(node.left)
self.queue.append(node.right)
self.queue.append(new_node)
return new_node
def get_root(self) -> TreeNode:
return self.rootExample:
# Create a complete binary tree
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
root.left.right = TreeNode(5)
root.right.left = TreeNode(6)
root.right.right = TreeNode(7)
# Insert a new node with value 8
inserter = CBTInserter(root)
new_node = inserter.insert(8)
# Print the new tree
print(inserter.get_root())Output:
1
/ \
2 3
/ \ / \
4 5 6 7
/
8Applications in Real World:
Balancing trees: Inserting nodes into a binary tree in a way that maintains balance can help improve the performance of search and insert operations.
Storing hierarchical data: A complete binary tree can be used to represent data with a hierarchical structure, such as a file system or an organization chart.
Managing queues: A complete binary tree can be used to implement a priority queue, where nodes with higher priority are inserted at lower levels of the tree.
minimum_moves_to_equal_array_elements
Problem Statement
Given an integer array nums of length n, return the minimum number of moves to make all elements of the array equal.
In one move, you can increment or decrement an element of the array by 1.
Example 1:
Input: nums = [1,2,3]
Output: 2
Explanation:
One possible operation sequence is:
- Increment nums[0] by 1 to get nums = [2,2,3].
- Increment nums[0] by 1 again to get nums = [3,3,3].
We need a total of 2 operations to make all elements equal.Example 2:
Input: nums = [1,10,2,9]
Output: 16
Explanation:
One possible operation sequence is:
- Increment nums[0] by 9 to get nums = [10,10,2,9].
- Increment nums[2] by 8 to get nums = [10,10,10,9].
- Increment nums[0] by 1 to get nums = [11,11,11,9].
- Increment nums[3] by 2 to get nums = [11,11,11,11].
We need a total of 16 operations to make all elements equal.Solution
Breakdown
The problem can be broken down into the following steps:
Find the median of the array. The median is the middle value when the array is sorted.
For each element in the array, calculate the difference between the element and the median.
The minimum number of moves is the sum of the absolute values of the differences.
Implementation
Here is a Python implementation of the solution:
def min_moves(nums):
# Sort the array
nums.sort()
# Find the median
median = nums[len(nums) // 2]
# Calculate the difference between each element and the median
differences = [abs(num - median) for num in nums]
# The minimum number of moves is the sum of the differences
return sum(differences)Real-World Applications
The minimum number of moves to equalize an array problem has many real-world applications, including:
Balancing workloads: In a distributed computing system, the goal is to distribute the workload evenly across multiple machines. To do this, the system can use the minimum moves algorithm to determine how many tasks to move from one machine to another to equalize the load.
Scheduling appointments: In a scheduling system, the goal is to schedule appointments so that there are no conflicts. To do this, the system can use the minimum moves algorithm to determine how many appointments to move from one time slot to another to minimize the number of conflicts.
Inventory management: In an inventory management system, the goal is to maintain an optimal level of inventory. To do this, the system can use the minimum moves algorithm to determine how many items to order from a supplier to minimize the total cost of inventory.
pour_water
Problem Statement: Given an array representing the heights of buildings, calculate the amount of water that can be trapped between them.
Best & Performant Solution: The best solution is to use two pointers, one starting from the left and one from the right. We keep track of the maximum height seen so far on both sides.
def pour_water(heights):
"""
Calculates the amount of water trapped between buildings.
Args:
heights: A list representing the heights of buildings.
Returns:
The amount of water trapped.
"""
# Initialize left and right pointers.
left = 0 # pointer from the left
right = len(heights) - 1 # pointer from the right
# Initialize maximum heights seen so far on both sides.
max_left = 0
max_right = 0
# Iterate until the pointers cross each other.
while left <= right:
# Check if the current height on the left is less than or equal to the maximum height seen so far.
if heights[left] <= max_left:
# Update the amount of water trapped.
amount_left += heights[left] - max_left
# Move the left pointer to the next building.
left += 1
# Otherwise, update the maximum height seen so far on the left.
else:
max_left = heights[left]
# Check if the current height on the right is less than or equal to the maximum height seen so far.
if heights[right] <= max_right:
# Update the amount of water trapped.
amount_right += heights[right] - max_right
# Move the right pointer to the previous building.
right -= 1
# Otherwise, update the maximum height seen so far on the right.
else:
max_right = heights[right]
# Return the total amount of water trapped.
return amount_left + amount_rightReal-World Applications:
Civil engineering: Designing reservoirs and dams.
Water management: Predicting flood risks.
Agriculture: Optimizing irrigation systems.
word_subsets
Problem: Given a list of strings, find all subsets of strings where each string is a valid subset of another string in the list.
Example:
Input: ["a", "b", "c", "ab", "ac", "bc", "abc"]
Output: [["a", "ab", "abc"], ["b", "bc", "abc"], ["c", "ac", "abc"]]Approach:
Sort the list of strings by their length in ascending order.
Initialize an empty hash map to store the subsets.
Iterate over the list of strings:
For each string, find all its substrings.
For each substring, check if it exists in the hash map. If it does, then add the current string to the subset of that substring.
If the substring does not exist in the hash map, then create a new entry in the hash map with the substring as the key and an empty list as the value. Add the current string to the list.
Return the values of the hash map.
Implementation:
def word_subsets(words):
words.sort(key=lambda x: len(x))
subsets = {}
for word in words:
substrings = set()
for i in range(1, len(word) + 1):
for j in range(len(word) - i + 1):
substrings.add(word[j:j + i])
for substring in substrings:
if substring in subsets:
subsets[substring].append(word)
else:
subsets[substring] = [word]
return list(subsets.values())
# Example usage
words = ["a", "b", "c", "ab", "ac", "bc", "abc"]
result = word_subsets(words)
print(result) # Output: [["a", "ab", "abc"], ["b", "bc", "abc"], ["c", "ac", "abc"]]Explanation:
Sorting the list: Sorting the list of strings by their length allows us to handle the substrings more efficiently.
Hash map for subsets: The hash map is used to store the subsets of strings. Each key in the hash map is a substring, and the corresponding value is a list of strings that contain that substring.
Finding substrings: For each string, we find all its substrings using nested loops.
Checking for substrings in hash map: For each substring, we check if it exists in the hash map. If it does, then we add the current string to the subset of that substring. If it doesn't, we create a new entry in the hash map for that substring.
Returning subsets: Finally, we return the values of the hash map, which contain the subsets of strings.
Real-world Applications:
Finding commonalities among a set of text documents or articles.
Identifying synonyms or related terms in a thesaurus or dictionary.
Analyzing user queries or search terms to identify common patterns.
score_after_flipping_matrix
LeetCode Problem: Score After Flipping Matrix
Problem Statement: Given a binary matrix matrix of size m x n, you can flip any row or column by changing all 0s to 1s and all 1s to 0s. Determine the maximum possible score you can achieve after flipping rows and columns.
The score of a matrix is calculated as the number of 1s in the matrix.
Example 1:
Input: matrix = [[0,0,1],[1,1,0],[1,0,1]]
Output: 6
Explanation: Flip the 0th row, and 1st column:
[[1,0,1],[1,1,0],[1,0,1]]
Score is the number of 1's: 6.Example 2:
Input: matrix = [[1,1,1],[0,0,0],[0,1,1]]
Output: 9
Explanation: Flip the 0th row and 2nd column:
[[1,1,1],[1,1,0],[1,1,1]]
Score is the number of 1's: 9.Breakdown and Explanation:
Step 1: Understand the Problem The goal is to maximize the total number of 1s in the matrix. You can flip entire rows or columns to change the value of all elements in that row or column.
Step 2: Analyze the Matrix For each row and column, calculate the number of 1s it contains. Store these counts in two lists, row_ones and col_ones.
Step 3: Determine Best Flips To maximize the score, flip rows and columns that have more 0s than 1s. This will convert more 0s to 1s.
Step 4: Calculate Maximum Score Sum up the counts of 1s in all the rows and columns that were not flipped. This is the maximum possible score.
Simplified Python Implementation:
def matrix_score(matrix):
m, n = len(matrix), len(matrix[0])
# Calculate row and column 1s counts
row_ones = [0] * m
col_ones = [0] * n
for i in range(m):
for j in range(n):
row_ones[i] += matrix[i][j]
col_ones[j] += matrix[i][j]
# Flip rows and columns with more 0s
score = 0
for i in range(m):
if row_ones[i] < m - row_ones[i]:
for j in range(n):
matrix[i][j] ^= 1
row_ones[i] = m - row_ones[i]
for j in range(n):
if col_ones[j] < n - col_ones[j]:
for i in range(m):
matrix[i][j] ^= 1
col_ones[j] = n - col_ones[j]
# Calculate maximum score
for i in range(m):
for j in range(n):
score += matrix[i][j]
return scoreReal-World Applications:
Game Development: Optimizing the layout of obstacles and rewards in a game by maximizing the number of "hit" points earned by players.
Image Processing: Enhancing images by flipping regions to improve contrast or highlight specific features.
Data Analysis: Maximizing the number of positive or negative responses in a survey by flipping rows or columns that contain more of one type of response.
longest_mountain_in_array
Problem Statement:
Given an array of integers, find the length of the longest "mountain" in the array. A mountain is defined as a sequence of numbers that first increases and then decreases, without any flat segments.
Example:
Input: [2,1,4,7,3,2,5]
Output: 5
Explanation: The longest mountain is [1,4,7,3,2].Algorithm and Implementation:
The solution involves two passes through the array. In the first pass, we find the index of the peak of the mountain. In the second pass, we expand the mountain from the peak to the left and right until we reach a point where the array stops increasing or decreasing.
Here's the Python code:
def longest_mountain_in_array(nums):
"""
Finds the length of the longest mountain in an array.
Args:
nums: list of integers
Returns:
int: length of the longest mountain
"""
if len(nums) < 3:
return 0
# Find the peak index
peak_index = -1
increasing = True
for i in range(1, len(nums)):
if increasing:
if nums[i] < nums[i-1]:
increasing = False
peak_index = i-1
else:
if nums[i] >= nums[i-1]:
return max(longest_mountain_in_array(nums[:peak_index+1]),
longest_mountain_in_array(nums[peak_index:]))
# Expand left and right from the peak
left_index = peak_index-1
right_index = peak_index+1
while left_index >= 0 and nums[left_index] < nums[left_index+1]:
left_index -= 1
while right_index < len(nums) and nums[right_index] > nums[right_index-1]:
right_index += 1
# Calculate the length of the mountain
return right_index - left_index - 1Real World Applications:
This algorithm can be used in various real-world applications, such as:
Financial analysis: Identifying trends and patterns in stock prices
Weather forecasting: Predicting the path of storms and hurricanes
Medical diagnostics: Analyzing patient data to identify diseases and health conditions
serialize_and_deserialize_bst
Problem: Serialize and Deserialize a Binary Search Tree
Solution:
Serialization:
Recursively traverse the BST in pre-order traversal (visit node, then left subtree, then right subtree).
For each node, append its value to a string.
Use a null value (e.g., "N") as a placeholder for nodes that are None.
Example: Consider the BST below:
1
/ \
0 2Serialization: "1,0,N,N,2,N,N"
Deserialization:
Split the serialized string into individual values.
Use a stack to hold the values in pre-order traversal order.
Pop the top value from the stack and create the root node.
If the next value is not "N", it represents the left child of the current node. Push it onto the stack.
If the next value is "N", the current node has no left child.
Repeat steps 3-4 for the right child.
Example:
Deserialize("1,0,N,N,2,N,N"):
Pop 1 and create the root node.
Push 0 onto the stack.
Pop 0 and create the left child of the root.
Since the next value is "N", the root has no right child.
Pop 2 and create the right child of the root.
Code:
class Node:
def __init__(self, val):
self.val = val
self.left = None
self.right = None
def serialize(root):
if not root:
return "N"
return f"{root.val},{serialize(root.left)},{serialize(root.right)}"
def deserialize(data):
vals = iter(data.split(","))
stack = []
root = create_node(next(vals))
stack.append(root)
while stack:
node = stack.pop()
left = create_node(next(vals))
node.left = left
if left:
stack.append(left)
right = create_node(next(vals))
node.right = right
if right:
stack.append(right)
return root
def create_node(val):
if val == "N":
return None
return Node(int(val))Applications:
Saving a BST to disk or database.
Sending a BST over a network.
Copying a BST without having to traverse the entire tree.
number_of_distinct_islands
Problem Statement:
Given a binary grid where 0 represents water and 1 represents land, find the number of distinct islands. Two islands are considered distinct if they have different shapes.
Solution Breakdown:
1. Depth-First Search (DFS)
We perform DFS starting from each unvisited land cell ("1") on the grid. As we traverse each cell, we mark the unique shape of the island using a unique identifier.
Recursive Approach:
def dfs(grid, i, j, id):
if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] != 1:
return
grid[i][j] = id
dfs(grid, i+1, j, id) # Right
dfs(grid, i-1, j, id) # Left
dfs(grid, i, j+1, id) # Down
dfs(grid, i, j-1, id) # UpIterative Approach:
def dfs_iter(grid, i, j, id):
stack = [(i, j)]
while stack:
i, j = stack.pop()
if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] != 1:
continue
grid[i][j] = id
stack.extend([(i+1, j), (i-1, j), (i, j+1), (i, j-1)])2. Identifying Distinct Shapes
To identify distinct shapes, we generate a "signature" for each island. We define the signature as the sequence of direction changes while traversing the boundary of the island.
Direction Changes:
'R' (right)
'L' (left)
'D' (down)
'U' (up)
Signature Generation:
Start at the top-left corner of the island.
Follow the boundary clockwise.
Record the direction changes.
3. Set of Distinct Signatures
We maintain a set to store the distinct signatures. For each new island, we calculate its signature and add it to the set if it's not already there.
Example:
Input Grid:
[[1,1,0]
[0,1,1]
[0,1,1]]
Output: 2Island 1: Signature = "RDRD"
Island 2: Signature = "DRD"
Applications:
Image processing: Detecting and counting objects in an image.
Game development: Identifying distinct islands in a game map for level design.
GIS (Geographic Information Systems): Analyzing landforms and water bodies for environmental planning.