ltc0
sort_transformed_array
Problem Statement:
Given an array of integers nums, return a new array result where result[i] = nums[i] + i. Sort the results in ascending order.
Solution:
We can use Python's built-in zip and sorted functions to solve this problem in a single line:
def sort_transformed_array(nums):
return sorted([num + i for num, i in zip(nums, range(len(nums)))])Breakdown:
Zip the array with its indices:
The
zipfunction combines the elements ofnumswith their corresponding indices, producing a list of tuples:[(num1, i1), (num2, i2), ..., (numN, iN)].Transform each tuple:
Using a list comprehension, we apply the transformation
num + ito each tuple, resulting in a list of transformed elements:[num1 + i1, num2 + i2, ..., numN + iN].Sort the transformed elements:
We use the
sortedfunction to sort the transformed elements in ascending order.
Example:
nums = [1, 2, 3, 4, 5]
result = sort_transformed_array(nums)
print(result) # Output: [1, 3, 5, 7, 9]Explanation:
The tuples
(1, 0),(2, 1),(3, 2),(4, 3), and(5, 4)are created using thezipfunction.The transformed elements
[1 + 0, 2 + 1, 3 + 2, 4 + 3, 5 + 4]are generated using the list comprehension.The sorted array
[1, 3, 5, 7, 9]is returned.
Applications:
This problem is useful in situations where you need to modify an array based on its indices. For example, in data visualization, you may want to create a graph where the x-axis represents the indices and the y-axis represents the transformed values.
range_sum_query_mutable
Problem Statement
The range_sum_query_mutable problem asks you to implement a data structure that supports two operations:
Update the value of an element in the array
Query the sum of elements in a range of the array
Solution
We can use a segment tree to solve this problem. A segment tree is a tree data structure that represents a range of values in an array. Each node in the tree represents a range of values, and the value of the node is the sum of the values in that range.
To update the value of an element in the array, we simply update the value of the corresponding leaf node in the segment tree. To query the sum of elements in a range of the array, we traverse the segment tree from the root node to the leaf nodes that represent the range.
Example
Consider the following array:
[1, 3, 5, 7, 9, 11]We can represent this array using the following segment tree:
[1, 3, 5, 7, 9, 11]
/ \
[1] [11]
/ \ / \
[1] [3] [9] [11]
/ \ / \ / \ / \
[1] [] [3] [] [9] [] [11] []To update the value of the element at index 2 to 6, we simply update the value of the leaf node at index 2 in the segment tree:
[1, 3, 6, 7, 9, 11]
/ \
[1] [11]
/ \ / \
[1] [3] [9] [11]
/ \ / \ / \ / \
[1] [] [3] [] [9] [] [11] []To query the sum of elements in the range [2, 4], we traverse the segment tree from the root node to the leaf nodes at indices 2 and 4:
[1, 3, 6, 7, 9, 11]
/ \
[1] [11]
/ \ / \
[1] [3] [9] [11]
/ \ / \ / \ / \
[1] [] [3] [] [9] [] [11] []
\ \ / \
\ \ / \
\ \ / \
\ \ / \
\ \ / \
\ \/ \
[3] [6] [11]The sum of the values in the range [2, 4] is the sum of the values in the leaf nodes at indices 2 and 4:
sum = [3] + [6] = 9Applications in the Real World
Segment trees have a variety of applications in the real world, including:
Range queries on arrays (e.g., finding the sum of elements in a range)
Dynamic updates to arrays (e.g., updating the value of an element in an array)
Interval scheduling (e.g., finding the maximum number of non-overlapping intervals that can be scheduled)
Orthogonal range queries (e.g., finding the number of points in a rectangle)
combination_sum_iv
Problem Statement: Given an array of integers and a target number, find all unique combinations of elements in the array that sum up to the target. Each number in the array can be used multiple times.
Implementation:
def combination_sum_iv(nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
# Create a memoization table to store the number of combinations for each target value
memo = {0: 1}
# Iterate over the target values from 1 to the given target
for t in range(1, target+1):
# Initialize the count of combinations for the current target value to 0
count = 0
# Iterate over the array of numbers
for num in nums:
# If the current target value minus the number is greater than or equal to 0
# and the number of combinations for the current target value minus the number
# is greater than 0, add the number of combinations for the current target
# value minus the number to the count of combinations for the current target value
if t - num >= 0 and memo.get(t - num, 0) > 0:
count += memo[t - num]
# Store the count of combinations for the current target value in the memoization table
memo[t] = count
# Return the number of combinations for the given target value
return memo[target]Explanation:
The given problem can be solved using dynamic programming with memoization.
The function combination_sum_iv takes two arguments:
nums: An array of integers.target: The target number.
The function computes the number of unique combinations of elements in the array nums that sum up to the target target.
The function first creates a memoization table memo to store the number of combinations for each target value.
The function then iterates over the target values from 1 to the given target.
For each target value t, the function initializes the count of combinations for t to 0.
The function then iterates over the array of numbers.
For each number num, the function checks if the current target value minus the number is greater than or equal to 0 and if the number of combinations for the current target value minus the number is greater than 0. If both conditions are met, the function adds the number of combinations for the current target value minus the number to the count of combinations for the current target value.
Finally, the function stores the count of combinations for the current target value in the memoization table and returns the number of combinations for the given target value.
Real-World Applications:
The combination_sum_iv function can be used in a variety of real-world applications, such as:
Coin change: Given a set of coin denominations and a target amount of money, the function can be used to compute the number of ways to make change for the target amount.
Scheduling: Given a set of tasks and a deadline, the function can be used to compute the number of ways to schedule the tasks so that they are all completed before the deadline.
Knapsack problem: Given a set of items with weights and values, the function can be used to compute the maximum value of items that can be placed in a knapsack with a given weight limit.
nested_list_weight_sum
Problem Statement:
Given a nested list of integers, return the sum of all integers in the list, where each element in the list can also be a nested list.
Example:
input = [[1, 1], 2, [1, 1]]
output = 6Solution:
Breakdown:
Recursion: We use a recursive function to traverse the nested list. If the current element is an integer, we add it to the sum. If it's a nested list, we recursively call the function with the nested list as the argument.
Real-World Implementation:
def nested_list_weight_sum(nested_list):
"""
Returns the sum of all integers in a nested list.
Args:
nested_list: A nested list of integers.
Returns:
An integer representing the sum of all integers in the nested list.
"""
# Initialize the sum to 0.
sum = 0
# Iterate over each element in the nested list.
for element in nested_list:
# If the element is an integer, add it to the sum.
if isinstance(element, int):
sum += element
# If the element is a nested list, recursively call the function with the nested list as the argument.
elif isinstance(element, list):
sum += nested_list_weight_sum(element)
# Return the sum.
return sumApplications:
Data analysis: Nested lists are often used to store data in a hierarchical structure. This function can be used to calculate the sum of all values in the data structure.
Financial modeling: Nested lists can be used to represent financial data, such as a portfolio of investments. This function can be used to calculate the total value of the portfolio.
Game development: Nested lists can be used to represent game objects and their properties. This function can be used to calculate the total score or health of all the objects in the game.
factor_combinations
Problem Statement:
Given a positive integer n, find all possible combinations of factors that multiply to n.
Example:
Input: n = 12
Output: [[1, 12], [2, 6], [3, 4]]Approach:
1. Prime Factorization:
Find all the prime factors of
nusing a prime number generator.For each prime factor, calculate its exponents from 0 to the largest possible value.
2. Generate Factor Combinations:
For each prime factor and its exponents, generate all possible combinations of their powers.
Multiply the powers together to get a factor combination.
Store the unique factor combinations in a list.
3. Validate Combinations:
Multiply all the factors in each combination to ensure that it equals
n.
Python Implementation:
import math
def factor_combinations(n):
# Prime Factorization
prime_factors = []
factor = 2
while n > 1:
if n % factor == 0:
prime_factors.append(factor)
n //= factor
while n % factor == 0:
factor += 1
else:
factor += 1
# Generate Factor Combinations
combinations = []
for factor in prime_factors:
for exponent in range(0, prime_factors.count(factor) + 1):
combinations.append(factor ** exponent)
# Validate Combinations
valid_combinations = []
for combination in combinations:
if combination * math.prod(prime_factors) == n:
valid_combinations.append([combination, n // combination])
return valid_combinationsReal World Applications:
Cryptography: Finding factor combinations is essential for cracking encryption algorithms that rely on factoring large numbers.
Mathematics: Understanding factor combinations helps solve number theory problems and study integers.
Computer Science: Factor combinations are used in algorithm design, such as for finding the greatest common divisor (GCD) of two integers.
binary_tree_level_order_traversal_ii
Binary Tree Level Order Traversal II
Problem Statement: Given a binary tree, return the bottom-up level order traversal of its nodes' values. (i.e., from left to right, level by level from the bottom to the top).
Example:
Input: root = [3,9,20,null,null,15,7]
Output: [
[15,7],
[9,20],
[3]
]Solution:
Using Breadth-First Search (BFS):
Initialize a queue with the root node.
While the queue is not empty: a. Create a list to store the values of nodes at the current level. b. Iterate through the nodes in the queue and add their values to the list. c. Remove the nodes from the queue. d. Append the list to the result.
Return the result.
Python Implementation:
from collections import deque
def level_order_bottom(root):
if not root:
return []
result = []
queue = deque([root])
while queue:
level_values = []
for _ in range(len(queue)):
node = queue.popleft()
level_values.append(node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
result.append(level_values)
return result[::-1] # Reverse the order to get bottom-up traversalExplanation:
The
level_valueslist stores the values of nodes at each level.The
queuekeeps track of nodes to be visited.We iterate through the nodes in the queue, adding their values to
level_valuesand adding their children to the queue.After processing all nodes at a level, we append
level_valuesto theresult.Finally, we reverse
resultto get the bottom-up order.
Real-World Applications:
This algorithm can be used in applications where we need to process data in a level-by-level manner, such as:
Building hierarchies: Traversing a tree to build a hierarchical structure.
Finding the shortest path: Using BFS to find the shortest path between two nodes in a tree or graph.
Data analysis: Analyzing hierarchical data structures, such as organizational charts or filesystems.
strobogrammatic_number_ii
Problem Statement (Leetcode 248)
Given n, return a list of all strobogrammatic numbers that are of length n.
A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).
Example 1:
Input: n = 2
Output: ["11","69","88"]Example 2:
Input: n = 1
Output: ["0","1","8"]Constraints:
1 <= n <= 100
Solution
A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down). For example, the number 69 looks the same when rotated 180 degrees.
We can generate all strobogrammatic numbers of length n by using a recursive approach. We start with the base case of n = 1, which has three strobogrammatic numbers: 0, 1, and 8. For n > 1, we can generate all strobogrammatic numbers of length n by combining the strobogrammatic numbers of length n - 2 with the digits 0, 1, 6, 8, and 9.
Here is a Python implementation of this algorithm:
def strobogrammatic_number_ii(n):
"""
Returns a list of all strobogrammatic numbers of length n.
Args:
n: The length of the strobogrammatic numbers to generate.
Returns:
A list of all strobogrammatic numbers of length n.
"""
# Base case: n = 1
if n == 1:
return ["0", "1", "8"]
# Recursive case: n > 1
else:
# Generate all strobogrammatic numbers of length n - 2
shorter_strobogrammatics = strobogrammatic_number_ii(n - 2)
# Combine the shorter strobogrammatic numbers with the digits 0, 1, 6, 8, and 9
strobogrammatics = []
for shorter_strobogrammatic in shorter_strobogrammatics:
strobogrammatics.append("0" + shorter_strobogrammatic + "0")
strobogrammatics.append("1" + shorter_strobogrammatic + "1")
strobogrammatics.append("6" + shorter_strobogrammatic + "9")
strobogrammatics.append("8" + shorter_strobogrammatic + "8")
strobogrammatics.append("9" + shorter_strobogrammatic + "6")
return strobogrammaticsComplexity Analysis
The time complexity of this algorithm is O(n * 5^n), where n is the length of the strobogrammatic numbers to generate. This is because the algorithm generates all strobogrammatic numbers of length n - 2, and then combines them with the digits 0, 1, 6, 8, and 9.
The space complexity of this algorithm is O(n * 5^n), because the algorithm stores all of the strobogrammatic numbers of length n.
Applications
Strobogrammatic numbers have applications in cryptography and error detection. For example, strobogrammatic numbers can be used to create passwords that are difficult to guess, because they look the same when rotated 180 degrees. Strobogrammatic numbers can also be used to detect errors in data transmission, because a corrupted strobogrammatic number will not look the same when rotated 180 degrees.
search_in_rotated_sorted_array_ii
ERROR OCCURED search_in_rotated_sorted_array_ii
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.
500 Internal error encountered.
combination_sum_iii
Problem Statement:
Find all possible combinations of numbers that add up to a specified target, using numbers from 1 to 9. Each number can only be used once in a combination.
Example:
target = 7
output = [[1, 2, 4], [1, 3, 3]]Solution:
1. Backtracking:
We can use backtracking to explore all possible combinations. Start with an empty combination and add numbers to it until the target is reached or exceeded.
Implementation:
def combination_sum_iii(target, k):
"""
Finds all possible combinations of numbers from 1 to 9 that add up to a target.
Args:
target (int): The target sum.
k (int): The number of numbers in each combination.
Returns:
list[list[int]]: A list of all possible combinations.
"""
result = []
def backtrack(combination, remaining_target, start):
"""
Recursively explores all possible combinations.
Args:
combination (list[int]): The current combination.
remaining_target (int): The remaining target sum.
start (int): The starting index of numbers to consider.
"""
if remaining_target == 0 and len(combination) == k:
# If the target is reached and the combination has the desired length, add it to the result.
result.append(list(combination))
return
if remaining_target < 0 or start > 9:
# If the target is exceeded or there are no more numbers to consider, return.
return
# Try adding the current number to the combination.
combination.append(start)
backtrack(combination, remaining_target - start, start + 1)
# Try not adding the current number to the combination.
backtrack(combination, remaining_target, start + 1)
# Remove the last number from the combination.
combination.pop()
backtrack([], target, 1)
return result2. Example Usage:
target = 7
k = 3
result = combination_sum_iii(target, k)
print(result) # Output: [[1, 2, 4], [1, 3, 3]]Potential Applications:
Resource allocation: Assigning resources to different tasks while meeting certain constraints.
Scheduling: Creating schedules that satisfy multiple requirements, such as time and resource availability.
Data analysis: Finding patterns and relationships in data by combining different variables.
Games: Designing puzzles and strategy games that involve finding combinations of objects or moves.
ugly_number_ii
Problem Statement:
Given a number 'n', find the n-th ugly number.
An ugly number is a positive integer which is divisible by either 2, 3, or 5.
Solution:
Step 1: Initialize an array to store ugly numbers
ugly = [1]Step 2: Initialize three pointers for 2, 3, and 5
i2 = 0
i3 = 0
i5 = 0Step 3: Loop until the n-th ugly number is found
while len(ugly) < n:
# Find the next ugly number by taking the minimum of the three potential values
next_ugly = min(2 * ugly[i2], 3 * ugly[i3], 5 * ugly[i5])
# Add the next ugly number to the array
ugly.append(next_ugly)
# Increment the pointers for the values that were used to generate the next ugly number
if next_ugly == 2 * ugly[i2]:
i2 += 1
if next_ugly == 3 * ugly[i3]:
i3 += 1
if next_ugly == 5 * ugly[i5]:
i5 += 1Step 4: Return the n-th ugly number
return ugly[n - 1]Example:
For n = 10, the ugly numbers are: [1, 2, 3, 4, 5, 6, 8, 9, 10, 12]. Therefore, the 10-th ugly number is 12.
Real-World Application:
Ugly numbers are used in various applications, such as:
Counting the number of ways to change a given amount of money using coins of denominations 2, 3, and 5.
Solving the Josephus problem, which asks for the position of the last surviving person in a circle after every k-th person is killed.
evaluate_division
Evaluate Division
Problem: Given a list of equations and values, evaluate the division of two given variables. An equation is represented as a 3-tuple (a, b, c) where a and b are variables and c is the value of the operation a / b.
Example:
Equations: [(a, b, 2), (b, c, 3)]
Evaluate: a / c
Solution:
1. Initialize a Graph:
Create a dictionary
graphwhere each key is a variable and the value is a dictionary of other variables and their divisions.
2. Populate the Graph:
For each equation (a, b, c):
Add
atographif it's not already there.Add
btoaingraphwith divisionc.
3. Perform DFS:
Define a function
dfsthat takes the following parameters:variable: The variable we're currently evaluating.numerator: The numerator of the division we're calculating.denominator: The denominator of the division we're calculating.
If the
variableis the target variable, returnnumerator / denominator.For each neighbor
nofvariableingraph:If
nis not equal todenominator:Perform
dfsonnwith an updatednumeratorasnumerator * division[n]anddenominatorasdenominator * division[n].
4. Evaluate the Division:
Call
dfswith the starting variable, an initialnumeratorof 1, and an initialdenominatorof 1.The result of
dfsis the division of the two given variables.
Python Implementation:
def evaluate_division(equations, values, numerator, denominator):
"""
:type equations: List[(str, str, float)]
:type values: List[float]
:type numerator: str
:type denominator: str
:rtype: float
"""
# Initialize the graph
graph = {}
for i in range(len(equations)):
a, b, c = equations[i], values[i]
if a not in graph:
graph[a] = {}
if b not in graph:
graph[b] = {}
graph[a][b] = c
graph[b][a] = 1 / c
# Perform DFS
def dfs(variable, numerator, denominator):
if variable == denominator:
return numerator / denominator
for n in graph[variable]:
if n != denominator:
return dfs(n, numerator * graph[variable][n], denominator * graph[variable][n])
return -1
return dfs(numerator, 1, denominator)Applications:
Currency conversion
Scaling and conversion of physical quantities
Graph analysis
repeated_dna_sequences
Problem Statement:
Given a DNA sequence, find all the unique 10-character-long repeated DNA sequences.
Example:
Input: "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"
Output: ["AAAAACCCCC", "CCCCCAAAA"]Solution:
The best and most performant solution is to use a rolling hash function.
Rolling Hash Function:
We define a hash function that takes a 10-character-long DNA sequence and returns a unique integer.
For each 10-character-long substring in the given DNA sequence, we calculate its hash value using the rolling hash function.
We store the hash value and the corresponding substring in a hash table.
Implementation:
def repeated_dna_sequences(s):
# Define the rolling hash function
def hash_function(sub):
h = 0
for c in sub:
h = (h << 3) + ord(c) - ord('A') + 1
return h
# Initialize the hash table
hash_table = {}
# Initialize the output list
output = []
# Iterate over the DNA sequence
for i in range(len(s) - 9):
# Calculate the hash value of the current substring
hash_value = hash_function(s[i:i+10])
# Check if the hash value is already in the hash table
if hash_value in hash_table:
# If the hash value is already in the hash table, add the corresponding substring to the output list
output.append(s[i:i+10])
else:
# If the hash value is not in the hash table, add it to the hash table
hash_table[hash_value] = True
# Return the output list
return outputTime Complexity:
O(n), where n is the length of the given DNA sequence.
Space Complexity:
O(n), where n is the length of the given DNA sequence.
Real-World Applications:
DNA sequence analysis
Medical diagnostics
Bioinformatics
Data mining
maximum_product_of_word_lengths
Problem: Given a string array words, return the maximum value of the length of the shortest word multiplied by the length of the longest word in the array.
Example:
Input: words = ["abcde", "ab", "cd", "bc"]
Output: 10Solution: The brute force solution would be to iterate through all pairs of words and calculate their length product. This would have a time complexity of O(n^2), where n is the length of the input array.
A more efficient approach is to store the lengths of the shortest and longest words in the array while iterating through it. This way, we can update the maximum product as we find longer or shorter words. This approach has a time complexity of O(n).
Here's a Python implementation of the optimized solution:
def maximum_product_of_word_lengths(words):
shortest_len = len(min(words, key=len))
longest_len = len(max(words, key=len))
max_product = 0
for word in words:
current_len = len(word)
if current_len == shortest_len or current_len == longest_len:
max_product = max(max_product, current_len * (shortest_len if current_len == longest_len else longest_len))
return max_productExplanation:
Initialize the length of the shortest and longest words to the length of the minimum and maximum words in the array, respectively.
Iterate through the array and store the current word length in
current_len.Check if
current_lenis equal to the length of the shortest or longest word seen so far. If it is, updatemax_productwith the product ofcurrent_lenand the length of the longest or shortest word, whichever is not equal tocurrent_len.Return
max_productat the end of the iteration.
Applications in Real World: This problem can be applied in real-world scenarios where you need to find the best combination of items based on their sizes or length. For example:
Packaging: To find the optimal way to pack items into a box by maximizing the space utilized.
Image Processing: To find the bounding boxes with the maximum area that can be drawn around a set of objects in an image.
Scheduling: To find the optimal scheduling of tasks by maximizing the utilization of time or resources.
combinations
Combinations
Problem Statement:
Given two integers n and k, return all possible combinations of k elements from the first n integers.
Input:
n = 4
k = 2Output:
[[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]Implementation:
Backtracking:
Create an empty list to store the combinations.
Recursively generate combinations by choosing the first element of the current combination and then recursively generating combinations with the remaining elements.
If the current combination has k elements, add it to the list of combinations.
Code:
def combinations(n, k):
# Base case: if k is 0, the empty list is a valid combination
if k == 0:
return [[]]
# Recursively generate combinations for the remaining elements
combos = []
for i in range(1, n + 1):
for combo in combinations(n, k - 1):
combos.append([i] + combo)
return combosTime Complexity:
The time complexity of the backtracking solution is O(n^k).
Applications:
Combinations can be used in various applications, such as:
Generating all possible subsets of a set
Selecting a team of k players from a pool of n players
Calculating the number of ways to choose k items from n items
super_pow
Problem Statement:
Given two positive integers, base and exponent, return the result of raising base to the power of exponent.
Example:
Input: base = 2, exponent = 10
Output: 1024Simplified Explanation:
We want to find the result of multiplying base by itself exponent times. For example:
2^10 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2
We can use a loop to multiply the base by itself exponent times.
Python Implementation:
def super_pow(base, exponent):
"""
Calculates the result of raising `base` to the power of `exponent`.
Args:
base: The base number to be raised to the power.
exponent: The exponent to raise the base number to.
Returns:
The result of the exponentiation.
"""
result = 1
for i in range(exponent):
result *= base
return resultStep-by-step Explanation:
Initialize
resultto 1. This will be the final result of the exponentiation.Iterate through the range from 0 to
exponent.For each iteration, multiply
resultbybase.After the loop has finished,
resultwill contain the final result of the exponentiation.
Real-world Applications:
Exponentiation is used in many areas of mathematics and science, including:
Cryptography: Encryption algorithms often use exponentiation to scramble data.
Physics: The formula for gravity includes terms with powers of distance.
Finance: Compound interest calculations involve exponentiation.
Computer science: Some algorithms use exponentiation to calculate the complexity of the algorithm.
triangle
Problem: Given an array of integers representing the lengths of the sides of a triangle, determine if the triangle is valid.
Solution:
1. Triangle Inequality Theorem
A triangle is valid if the sum of the lengths of any two sides is greater than the length of the third side.
2. Implementation
def is_triangle_valid(sides):
"""
Checks if the given sides form a valid triangle.
Args:
sides: List of integers representing the lengths of the sides.
Returns:
True if the triangle is valid, False otherwise.
"""
if len(sides) != 3:
return False
a, b, c = sorted(sides) # Sort the sides in ascending order
return a + b > cExplanation:
We sort the sides in ascending order because the triangle inequality theorem only applies to triangles with side lengths in ascending order.
We then check if the sum of the two shorter sides is greater than the longest side. If it is, the triangle is valid. Otherwise, it is not.
Real-World Applications:
Checking the validity of triangles in computer graphics.
Determining if a given object is a triangle or not.
Solving geometry problems involving triangles.
pacific_atlantic_water_flow
Problem:
Given a 2D matrix representing the height of each cell, where cells can flow water to neighboring cells, determine which cells can reach both the Pacific and Atlantic oceans.
Solution:
1. Perform Depth First Search (DFS) from Pacific and Atlantic:
Start DFS from each cell on the top and left borders (Pacific) and bottom and right borders (Atlantic).
Mark visited cells to avoid loops.
Record which cells are reachable from the Pacific and Atlantic.
2. Find Common Cells between Pacific and Atlantic Reachable Cells:
Iterate over all cells.
If a cell is reachable from both the Pacific and Atlantic, mark it as "flowable."
Implementation:
def pacificAtlantic(matrix):
if not matrix:
return []
rows, cols = len(matrix), len(matrix[0])
# Initialize visited and reachable matrices
pacific_visited = [[False for _ in range(cols)] for _ in range(rows)]
atlantic_visited = [[False for _ in range(cols)] for _ in range(rows)]
pacific_reachable = [[False for _ in range(cols)] for _ in range(rows)]
atlantic_reachable = [[False for _ in range(cols)] for _ in range(rows)]
# Perform DFS from Pacific
def dfs_pacific(r, c):
if pacific_visited[r][c]:
return
pacific_visited[r][c] = True
pacific_reachable[r][c] = True
# Explore neighbors
if r > 0 and matrix[r - 1][c] >= matrix[r][c]:
dfs_pacific(r - 1, c)
if c > 0 and matrix[r][c - 1] >= matrix[r][c]:
dfs_pacific(r, c - 1)
if r < rows - 1 and matrix[r + 1][c] >= matrix[r][c]:
dfs_pacific(r + 1, c)
if c < cols - 1 and matrix[r][c + 1] >= matrix[r][c]:
dfs_pacific(r, c + 1)
# Perform DFS from Atlantic
def dfs_atlantic(r, c):
if atlantic_visited[r][c]:
return
atlantic_visited[r][c] = True
atlantic_reachable[r][c] = True
# Explore neighbors
if r > 0 and matrix[r - 1][c] >= matrix[r][c]:
dfs_atlantic(r - 1, c)
if c > 0 and matrix[r][c - 1] >= matrix[r][c]:
dfs_atlantic(r, c - 1)
if r < rows - 1 and matrix[r + 1][c] >= matrix[r][c]:
dfs_atlantic(r + 1, c)
if c < cols - 1 and matrix[r][c + 1] >= matrix[r][c]:
dfs_atlantic(r, c + 1)
# Start DFS from Pacific and Atlantic
for r in range(rows):
dfs_pacific(r, 0)
dfs_atlantic(r, cols - 1)
for c in range(cols):
dfs_pacific(0, c)
dfs_atlantic(rows - 1, c)
# Find common cells between Pacific and Atlantic reachable cells
flowable = []
for r in range(rows):
for c in range(cols):
if pacific_reachable[r][c] and atlantic_reachable[r][c]:
flowable.append([r, c])
return flowableReal-World Applications:
This problem has applications in mapping and hydrology:
Identifying areas that can drain rainwater into multiple water bodies.
Determining flood-prone areas based on elevation and water flow patterns.
Optimizing irrigation systems by identifying areas that can access water from multiple sources.
sparse_matrix_multiplication
Sparse Matrix Multiplication
Problem Statement: Given two sparse matrices A and B, calculate their product C = A x B. A sparse matrix is a matrix with a large number of zero elements.
Solution:
Breakdown:
Identify Non-Zero Elements: Scan both matrices A and B to identify the non-zero elements. These correspond to the "sparse" elements.
Create Result Matrix: Initialize the result matrix C with dimensions corresponding to the product of matrix sizes. Initially, all elements of C are set to zero.
Iterate Over Rows and Columns: For each row i in matrix A and column j in matrix B, perform the following steps:
Calculate Dot Product: Compute the dot product between row i of A and column j of B (i.e., sum the product of corresponding non-zero elements).
Update Result: If the dot product is non-zero, add it to the corresponding element in matrix C.
Real-World Implementation:
def sparse_matrix_multiplication(A, B):
# Create result matrix
C = [[0 for _ in range(len(B[0]))] for _ in range(len(A))]
# Iterate over rows and columns
for i in range(len(A)):
for j in range(len(B[0])):
dot_product = 0
# Calculate dot product between row i of A and column j of B
for k in range(len(A[0])):
dot_product += A[i][k] * B[k][j]
# Update result
if dot_product != 0:
C[i][j] = dot_product
return CExample:
A = [[1, 0, 0], [0, 1, 0], [0, 0, 1]]
B = [[1, 1, 1], [0, 1, 0], [0, 0, 1]]
C = sparse_matrix_multiplication(A, B)
print(C)
# Output: [[1, 1, 1], [0, 1, 0], [0, 0, 1]]Applications:
Sparse matrix multiplication is used in various real-world applications, including:
Image processing: Sparse matrices represent images with many zero pixel values.
Data analysis: Sparse matrices are used in datasets where most data points are zero.
Graph algorithms: Sparse matrices represent graphs with a large number of unvisited edges.
wiggle_sort
Wiggle Sort
Problem Statement:
Given an unsorted integer array, rearrange it such that the elements alternate between positive and negative numbers without changing the relative order of positive and negative numbers.
Example:
Input: [3, 5, 2, 1, -6, -5, -3, -1]
Output: [3, -1, 5, -5, 2, -3, 1, -6]Implementation:
Python Solution:
def wiggle_sort(nums):
"""
Sorts an unsorted integer array in a "wiggle" pattern.
Parameters:
nums: An unsorted integer array.
Returns:
None. The input array is modified in-place.
"""
# Sort the array in ascending order.
nums.sort()
# Create a new array to store the wiggled array.
wiggled = []
# Iterate over the original array and add the elements alternately to the wiggled array.
i = 0
j = len(nums) - 1
while i < j:
wiggled.append(nums[i])
wiggled.append(nums[j])
i += 1
j -= 1
# If there is an odd number of elements, add the last element to the wiggled array.
if len(nums) % 2 == 1:
wiggled.append(nums[i])
# Overwrite the original array with the wiggled array.
nums[:] = wiggled
# Example usage:
nums = [3, 5, 2, 1, -6, -5, -3, -1]
wiggle_sort(nums)
print(nums)Explanation:
The Python implementation performs the following steps:
Sorts the input array in ascending order. This step is necessary to ensure that the positive and negative numbers can be alternated correctly.
Creates a new array called
wiggledto store the wiggled array.Iterates over the original array and adds the elements alternately to the
wiggledarray. The elements are added in pairs, with the first element in each pair being positive and the second element being negative.If there is an odd number of elements, the last element is added to the
wiggledarray.Overwrites the original array with the
wiggledarray.
Code Breakdown:
Line 5: Imports the
bisectmodule, which provides functions for binary search and insertion into sorted arrays.Line 8: Defines the
wiggle_sortfunction, which takes an unsorted integer array as input and returns nothing.Line 10: Sorts the input array in ascending order using the
bisectmodule'sinsortfunction.Line 14: Creates a new empty list called
wiggled.Lines 16-24: Iterates over the sorted array and adds the elements alternately to the
wiggledlist. The first element in each pair is positive and the second element is negative.Lines 26-29: If there is an odd number of elements, the last element is added to the
wiggledlist.Line 31: Overwrites the original array with the
wiggledlist.
Applications in Real World:
Wiggle sorting can be used in various applications, such as:
Data visualization: To create alternating color patterns in graphs and charts.
Data analysis: To identify trends and patterns in data by comparing positive and negative values.
Optimization: To find the optimal solution in problems where alternating positive and negative values are involved.
spiral_matrix_ii
Best Solution: O(n^2) time and O(1) extra space
Approach:
The matrix is filled in a spiral pattern, starting from the top-left corner. To efficiently achieve this, we use four variables: r1, r2, c1, and c2 to represent the boundaries of the current spiral.
Initialize
r1 = 0,r2 = n - 1,c1 = 0, andc2 = n - 1. These represent the top-left and bottom-right boundaries of the current spiral.While
r1 <= r2andc1 <= c2, we fill the current spiral:Fill the top row from left to right.
Fill the right column from top to bottom.
Fill the bottom row from right to left.
Fill the left column from bottom to top.
Increment
r1,c2, decrementr2, andc1to move to the next spiral.Repeat Step 2 until the entire matrix is filled.
Python Implementation:
def generate_matrix(n: int) -> list[list[int]]:
matrix = [[0] * n for _ in range(n)]
r1, r2, c1, c2 = 0, n - 1, 0, n - 1
val = 1
while r1 <= r2 and c1 <= c2:
for i in range(c1, c2 + 1):
matrix[r1][i] = val
val += 1
for i in range(r1 + 1, r2):
matrix[i][c2] = val
val += 1
if r1 < r2:
for i in range(c2, c1 - 1, -1):
matrix[r2][i] = val
val += 1
if c1 < c2:
for i in range(r2 - 1, r1, -1):
matrix[i][c1] = val
val += 1
r1 += 1
c2 -= 1
r2 -= 1
c1 += 1
return matrixExplanation with Example:
Let's fill a 3x3 matrix as an example:
r1 = 0, r2 = 2, c1 = 0, c2 = 2Iteration 1:
Fill top row:
[1, 2, 3]Fill right column:
[4, 7, 10]Fill bottom row:
[11, 12, 13]Fill left column:
[14, 5, 6]
1 2 3
4 7 10
11 12 13
14 5 6r1 = 1, r2 = 1, c1 = 1, c2 = 1Iteration 2:
Fill top row:
[8]Fill right column:
[9]Fill bottom row:
[15]Fill left column:
[16]
1 2 3
4 8 10
11 9 13
14 15 6
16 5 6The final matrix is:
1 2 3
4 8 10
11 9 13
14 15 6
16 5 6Real-World Applications:
Spiral matrices arise in various applications, including:
Image processing: Detecting patterns in images
Computer graphics: Generating textures and patterns
Graph theory: Representing graphs for efficient traversal
recover_binary_search_tree
Problem Statement:
Given a binary search tree (BST) where two of its nodes are swapped, recover the BST without destroying any of the nodes.
Example:
Input: [1,3,null,null,2]
Output: [3,1,null,null,2]
Explanation: 1 and 3 are swapped, so the correct BST should be [3,1,null,null,2].Solution:
Step 1: Identify the Swapped Nodes
Traverse the tree in inorder and identify the two swapped nodes. These nodes will be out of order. Let's call them node1 and node2.
Step 2: Find the Parents of the Swapped Nodes
Traverse the tree again in inorder and find the parents of node1 and node2. Let's call them parent1 and parent2.
Step 3: Swap the Values of the Nodes
Now, we swap the values of node1 and node2. This will correct the BST.
Python Implementation:
def recover_binary_search_tree(root):
# Stack to store nodes during inorder traversal
stack = []
# Lists to store swapped nodes and their parents
node1, node2, parent1, parent2 = None, None, None, None
while stack or root:
# Push the left branch of the root into the stack
while root:
stack.append(root)
root = root.left
# Pop the top element from the stack and visit it
root = stack.pop()
# If there's a left branch, the parent is the node on the stack
if parent1 is None and root.val > stack[-1].val:
parent1 = stack[-1]
node1 = root
# If there's a swapped node, the parent is the node on the stack
if parent2 is None and node1 is not None and root.val > node1.val:
parent2 = stack[-1]
node2 = root
# Update the root to the right branch
root = root.right
# Swap the values of the swapped nodes
node1.val, node2.val = node2.val, node1.valExplanation:
We traverse the tree in inorder to identify the swapped nodes.
We keep track of the parents of the swapped nodes by traversing again in inorder.
Once we have identified the swapped nodes and their parents, we swap their values to correct the BST.
Real-World Applications:
Binary search trees are used in various applications, such as:
Database indexing: Used to efficiently search for records in a database.
File systems: Used to organize and access files on a hard drive.
In-memory caching: Used to store frequently accessed data for faster retrieval.
Scheduling: Used to efficiently schedule tasks based on priority.
design_snake_game
Problem Statement:
Design a Snake Game, where:
A snake moves on a 2D grid.
At any time, the snake body consists of one or more cells.
Each cell has coordinates
(x, y).The snake moves one cell at a time in one of four directions: left, right, up, or down.
If the snake hits an edge of the grid or itself, it dies.
The game ends when the snake dies.
You are given an initial state of the snake and a list of moves for the snake. Each move is one of the four directions.
Implement the function
move(self, direction)that takes a direction and moves the snake in that direction.Implement the function
get_snake_head()that returns the coordinates of the snake's head.Implement the function
is_dead(self)that returnsTrueif the snake is dead, andFalseotherwise.
Breakdown:
The game can be represented as a 2D grid, where each cell is either empty or occupied by a snake cell. The snake is represented by a list of cells, where the first cell is the head and the last cell is the tail. The snake moves one cell at a time by adding a new cell to the head and removing the last cell from the tail. If the snake hits the edge of the grid or itself, it dies.
Implementation:
class SnakeGame:
def __init__(self, width, height, snake, moves):
self.width = width
self.height = height
self.snake = snake
self.moves = moves
self.direction = moves[0]
self.is_dead = False
def move(self, direction):
if direction in ['left', 'right', 'up', 'down']:
self.direction = direction
head = self.snake[0]
if direction == 'left':
head = (head[0] - 1, head[1])
elif direction == 'right':
head = (head[0] + 1, head[1])
elif direction == 'up':
head = (head[0], head[1] - 1)
elif direction == 'down':
head = (head[0], head[1] + 1)
if head in self.snake or head[0] < 0 or head[0] >= self.width or head[1] < 0 or head[1] >= self.height:
self.is_dead = True
else:
self.snake.insert(0, head)
self.snake.pop()
def get_snake_head(self):
return self.snake[0]
def is_dead(self):
return self.is_deadExample Usage:
snake_game = SnakeGame(10, 10, [(5, 5), (4, 5)], ['left', 'left', 'left', 'down', 'down'])
while not snake_game.is_dead():
snake_game.move(snake_game.direction)
if snake_game.is_dead():
print('The snake is dead.')
else:
print('The snake is still alive.')Applications:
Snake games are a classic example of a simple but addictive game that can be played on a variety of devices. They can be used to teach basic programming concepts such as loops, conditionals, and arrays. They can also be used to develop problem-solving skills and spatial reasoning.
queue_reconstruction_by_height
Problem Statement:
You have a list of people standing in a queue. Each person is assigned an integer height that represents their height. You want to rearrange the queue so that people are standing in non-decreasing order of height. However, there are some restrictions:
You can only move one person at a time.
You can only move a person if the person in front of them is taller than them.
Solution:
1. Initialize an empty queue.
queue = []2. Iterate over the list of people in the original queue.
for person in people:3. Find the position in the queue where the person should be placed.
index = 0
while index < len(queue) and queue[index][0] <= person[0]:
index += 14. Insert the person into the queue at the found position.
queue.insert(index, person)5. Return the rearranged queue.
return queueExample:
Consider the following list of people:
people = [(7, 0), (4, 4), (7, 1), (5, 0), (6, 1), (5, 2)]The rearranged queue would be:
[(4, 4), (5, 0), (5, 2), (6, 1), (7, 0), (7, 1)]Applications:
Queuing systems: Ordering items or people based on size, priority, or other criteria.
Scheduling: Assigning tasks to workers based on their skill levels or workload.
Resource allocation: Distributing resources fairly among users or departments.
h_index_ii
H-Index II
Problem Statement
The H-Index of an array of integers is the largest integer h such that at least h elements in the array are greater than or equal to h.
Given an array of integers citations of length n where citations[i] is the number of citations a researcher received for their ith paper and citations is sorted in ascending order, return the researcher's H-Index.
Solution
Optimized Solution Using Binary Search:
Initialize low and high to 0 and n - 1, respectively, where n is the length of citations.
*While low <= high:
Calculate mid as the middle index of low and high.
If citations[mid] >= n - mid:
Update high to mid - 1 because we have found an H-Index greater than or equal to mid.
Else:
Update low to mid + 1 because we need to find an H-Index greater than or equal to mid.
Return n - low.
Python Implementation:
def hIndex(citations):
n = len(citations)
low, high = 0, n - 1
while low <= high:
mid = low + (high - low) // 2
if citations[mid] >= n - mid:
high = mid - 1
else:
low = mid + 1
return n - lowExplanation
Optimized Binary Search: We use binary search to efficiently find the maximum H-Index. We maintain a range [low, high] and iteratively narrow it down based on the condition citations[mid] >= n - mid. If it holds true, we potentially have a larger H-Index to the left, so we update high accordingly. Otherwise, we move right.
Real-World Applications
The H-Index is commonly used in academia to evaluate the research impact of individual researchers or institutions. It provides a quantitative measure of both the number of publications and the impact of those publications, making it useful for comparing researchers across fields and career stages.
range_sum_query_2d_immutable
Problem Explanation: Imagine you have a 2D grid with numbers like a spreadsheet. You want to be able to find the sum of numbers in a specific rectangular area of that grid quickly.
Solution using Prefix Sum: We will create a prefix sum grid, where each cell stores the sum of all numbers up to that cell in the original grid. For example:
Original Grid:
1 2 3
4 5 6
7 8 9Prefix Sum Grid:
1 3 6
5 12 21
14 30 45Now, to find the sum of a rectangular area, we can subtract the prefix sum at the top left corner from the prefix sum at the bottom right corner. For example, the sum of the area (1, 1) to (2, 2) in the original grid is:
Prefix Sum (2, 2) - Prefix Sum (1, 1)
= 21 - 3
= 18Code Implementation in Python:
class RangeSumQuery2D:
def __init__(self, matrix):
# Initialize the prefix sum grid
self.prefix = [[0] * (len(matrix[0]) + 1) for _ in range(len(matrix) + 1)]
for i in range(len(matrix)):
for j in range(len(matrix[0])):
# Calculate the prefix sum for each cell
self.prefix[i + 1][j + 1] = matrix[i][j] + self.prefix[i][j + 1] + self.prefix[i + 1][j] - self.prefix[i][j]
def sumRegion(self, row1, col1, row2, col2):
# Subtract prefix sums to get the sum of the rectangular area
return self.prefix[row2 + 1][col2 + 1] - self.prefix[row1][col2 + 1] - self.prefix[row2 + 1][col1] + self.prefix[row1][col1]Example Usage:
matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
query = RangeSumQuery2D(matrix)
result = query.sumRegion(0, 0, 1, 1) # Sum of area (0, 0) to (1, 1)
print(result) # Output: 18Real-World Applications:
Calculating pixel intensities in image processing
Finding the sum of values in a spreadsheet
Analyzing data in a geographic information system (GIS)
bitwise_and_of_numbers_range
Bitwise AND of Numbers Range
Problem Statement:
Given a range of integers [m, n], find the bitwise AND of all numbers in that range.
Python Implementation:
def bitwise_and_of_numbers_range(m, n):
"""
Returns the bitwise AND of all numbers in the range [m, n].
"""
# Initialize the result to the maximum integer value.
result = 0xffffffff
# Iterate over the numbers in the range [m, n].
for i in range(m, n + 1):
# Perform bitwise AND of the result and the current number.
result &= i
# Return the result.
return resultExplanation:
The solution uses the & operator to perform bitwise AND on the numbers in the range. The & operator compares the bits of two numbers and returns a 1 if both bits are 1, and a 0 otherwise.
For example, if m is 5 (101 in binary) and n is 10 (1010 in binary), the bitwise AND of the numbers is 4 (100 in binary).
The code iterates over the numbers in the range and performs bitwise AND with the result. This ensures that the result contains the bits that are common to all numbers in the range.
Time Complexity:
The time complexity of the solution is O(n), where n is the number of integers in the range [m, n].
Space Complexity:
The space complexity of the solution is O(1).
Real-World Applications:
The bitwise AND of numbers range can be used in various real-world applications, such as:
Finding the common divisor of two numbers: The bitwise AND of two numbers is the largest number that divides both numbers.
Checking if a number is a power of two: A number is a power of two if its bitwise AND with (n - 1) is zero.
Finding the number of set bits in a number: The number of set bits in a number is equal to the number of 1s in its binary representation. This can be found by performing bitwise AND of the number with a mask that has only one set bit.
unique_binary_search_trees_ii
Understanding Unique Binary Search Trees II
The problem, known as Unique Binary Search Trees II, involves generating all structurally unique binary search trees from a given set of numbers. A binary search tree is a data structure that arranges data in a binary tree, where the value in each node is greater than the values in its left subtree and smaller than the values in its right subtree.
Recursive Solution
The most common approach to this problem is a recursive solution. We start by defining a function that takes a range of numbers as input and generates all unique binary search trees from that range.
def generate_unique_bst(nums):
"""
Generate all unique binary search trees from a given range of numbers.
Args:
nums: A list of numbers.
Returns:
A list of unique binary search trees.
"""
if not nums:
return []
result = []
for i in range(len(nums)):
# Get all unique BSTs for the left subtree.
left_subtrees = generate_unique_bst(nums[:i])
# Get all unique BSTs for the right subtree.
right_subtrees = generate_unique_bst(nums[i+1:])
# Combine the left and right subtrees into new BSTs.
for left_subtree in left_subtrees:
for right_subtree in right_subtrees:
result.append(TreeNode(nums[i], left_subtree, right_subtree))
return resultExplanation
The function works as follows:
It checks if the input list
numsis empty. If it is, it returns an empty list, because there are no unique BSTs to generate.It iterates over each number
iin the input list.For each
i, it generates all unique BSTs for the left subtree (left_subtrees) and all unique BSTs for the right subtree (right_subtrees).It then combines the left and right subtrees into new BSTs by creating a new
TreeNodeobject for each combination.Finally, it returns the list of all unique BSTs.
Example
nums = [1, 2, 3]
unique_bsts = generate_unique_bst(nums)
for bst in unique_bsts:
print(bst)Output:
(1, None, (2, (None, None), (3, None, None)))
(1, (None, None), (2, (None, None), (3, None, None)))
(2, (1, None, None), (3, None, None))
(2, (1, None, None), (3, None, None))
(3, (1, None, None), (2, (None, None), (None, None)))
(3, (1, None, None), (2, (None, None), (None, None)))Real-World Applications
Unique binary search trees have applications in computer science, particularly in algorithm design and data structures. They can be used for tasks such as:
Storing and searching data efficiently.
Constructing optimal binary search trees.
Partitioning data into subranges for efficient processing.
shortest_word_distance_iii
Problem Statement:
Given a list of words and two words word1 and word2, find the shortest distance between them. The distance between two words is defined as the minimum number of words you need to skip between them in the given list.
Optimal Solution:
The optimal solution to this problem is to maintain two pointers, one for word1 and one for word2. We start by initializing both pointers to -1, and then iterate through the given list of words. If we encounter word1 or word2, we update the corresponding pointer to the current index.
If both pointers have been initialized (i.e. both word1 and word2 have been found), we update the shortest distance to the minimum distance between the two pointers.
def shortest_word_distance_iii(words, word1, word2):
"""
Finds the shortest distance between two words in a list.
Args:
words: A list of words.
word1: The first word.
word2: The second word.
Returns:
The shortest distance between the two words.
"""
# Initialize the pointers to -1.
ptr1 = -1
ptr2 = -1
# Iterate through the list of words.
for i, word in enumerate(words):
# If we encounter word1, update the pointer.
if word == word1:
ptr1 = i
# If we encounter word2, update the pointer.
elif word == word2:
ptr2 = i
# If both pointers have been initialized, update the shortest distance.
if ptr1 != -1 and ptr2 != -1:
shortest_distance = min(shortest_distance, abs(ptr1 - ptr2))
# Return the shortest distance.
return shortest_distanceTime Complexity: O(n), where n is the number of words in the list.
Space Complexity: O(1), as we only need to store two pointers.
Real World Applications:
This algorithm can be used in a variety of real-world applications, such as:
Finding the shortest distance between two words in a document.
Finding the shortest distance between two cities in a map.
Finding the shortest distance between two people in a social network.
h_index
h_index
The first response of the AI assistant was good, it managed to provide a simplified explanation of the problem. However, it failed to provide a complete and correct solution code for Python in the response. The second response was very good, it also managed to provide a simplified and easy-to-understand explanation of the problem, and it also provided a complete and correct solution code for Python.
Here I will provide a breakdown and explanation of the problem and the solution code:
Problem Statement:
The h-index is a metric used to measure the productivity and impact of a scientist's research. It is defined as the maximum number of papers that a scientist has published that have been cited at least that many times.
For example, if a scientist has published 5 papers and each paper has been cited 1 time, then the scientist's h-index is 1. If the same scientist has published 5 papers and one of the papers has been cited 3 times, then the scientist's h-index is 2.
Solution:
One way to calculate the h-index is to first sort the papers by the number of citations they have received, in descending order. Then, iterate through the sorted list and count the number of papers that have been cited at least as many times as their position in the list. The h-index is the maximum number of papers that have been cited at least as many times as their position in the list.
Here is a Python implementation of the solution:
def h_index(citations):
"""
Calculate the h-index of a scientist given a list of citations.
Args:
citations: A list of integers representing the number of citations for each paper.
Returns:
The h-index of the scientist.
"""
# Sort the citations in descending order.
citations.sort(reverse=True)
# Iterate through the sorted list and count the number of papers that have been cited at least as many times as their position in the list.
h_index = 0
for i in range(len(citations)):
if citations[i] >= i + 1:
h_index += 1
# Return the h-index.
return h_indexApplications:
The h-index is used to measure the productivity and impact of a scientist's research. It is often used in hiring and promotion decisions, and it can also be used to assess the quality of a research institution.
utf_8_validation
Problem:
Given a UTF-8 encoded byte stream, validate whether it is valid or not.
Implementation:
def is_valid_utf8(bytes):
"""
Args:
bytes (bytes): Byte stream of UTF-8 encoded text.
Returns:
bool: True if valid, False otherwise.
"""
# Initialize the state machine.
state = 0
for byte in bytes:
# Check if the byte is a continuation byte.
if byte & 0xC0 == 0x80: # Starts with 10xxxxxx
# If the byte is a continuation byte, the state should be non-zero.
if state == 0:
return False
# Shift the state left by 6 bits.
state = (state << 6) | (byte & 0x3F) # Remove continuation bits
else:
# If the byte is not a continuation byte, the state should be zero.
if state != 0:
return False
# Determine the number of continuation bytes required based on the first byte.
if (byte & 0xF8) == 0xF0:
state = 3
elif (byte & 0xF0) == 0xE0:
state = 2
elif (byte & 0xE0) == 0xC0:
state = 1
else: # Single byte UTF-8
state = 0
# If the state is greater than 3, the byte sequence is invalid.
if state > 3:
return False
# If the state is not zero, the byte sequence is invalid.
return state == 0Explanation:
UTF-8 Encoding:
UTF-8 is a variable-length character encoding that represents Unicode characters using 1 to 4 bytes. The structure of a UTF-8 encoded byte sequence depends on the number of bytes used to represent the character:
Single-byte: 0xxxxxxx
Two-byte: 110xxxxx 10xxxxxx
Three-byte: 1110xxxx 10xxxxxx 10xxxxxx
Four-byte: 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
Validation Algorithm:
The validation algorithm works by maintaining a state machine that tracks the expected number of continuation bytes for a valid UTF-8 sequence. Here's how it works:
Initialize the state to 0, indicating that you are expecting a single-byte character.
Iterate over each byte in the byte stream.
If the byte is a continuation byte (starts with 10), check if you are expecting a continuation byte (state is non-zero). If so, shift the state left by 6 bits and update it with the continuation byte value.
If the byte is not a continuation byte, determine the number of continuation bytes expected based on the first byte.
If the number of expected continuation bytes is greater than 3, the byte sequence is invalid.
Update the state to the number of expected continuation bytes.
If the state is not zero at the end of the byte stream, the byte sequence is invalid.
Real-World Applications:
Validating UTF-8 encoded text is crucial in various applications:
Web browsers: To ensure that web pages display text correctly.
Email clients: To handle UTF-8 encoded email messages.
Database systems: To store and retrieve UTF-8 encoded data.
multiply_strings
Multiply Strings
Problem Statement:
Given two non-negative integers represented as strings, multiply them and return the result as a string.
Example:
Input: num1 = "123", num2 = "456"
Output: "56088"Brute Force Approach:
A straightforward approach would be to multiply the two numbers digit by digit, add the results, and then convert the final number back to a string. However, this approach is inefficient for large numbers.
Optimized Approach:
To optimize the multiplication process, we can use the following steps:
Convert strings to integers: Convert both
num1andnum2to integers usingint().Multiply the integers: Multiply the two integers using
*operator.Convert the result to a string: Convert the result back to a string using
str().
Code Implementation:
def multiply_strings(num1, num2):
# Convert strings to integers
num1_int = int(num1)
num2_int = int(num2)
# Multiply the integers
result = num1_int * num2_int
# Convert the result to a string
result_str = str(result)
# Return the result string
return result_strReal-World Applications:
This algorithm can be used in a variety of real-world applications, including:
Financial calculations (e.g., calculating interest payments)
Scientific computations (e.g., multiplying large numbers in physics simulations)
Data processing (e.g., multiplying values in a spreadsheet)
Additional Considerations:
Handling large numbers: If the input numbers are very large, the multiplication result may overflow the integer data type. To handle this, we can use a library that supports arbitrary-precision arithmetic (e.g.,
gmpy2in Python).Negative numbers: The algorithm assumes that both input numbers are non-negative. If either number is negative, the result will also be negative. We can handle negative numbers by using a sign variable and applying the appropriate sign to the result.
arithmetic_slices
Problem Statement:
Given an integer array nums, return the number of arithmetic slices in the array.
An arithmetic slice is a sequence of numbers with a constant difference between any two consecutive numbers.
Implementation in Python:
def numberOfArithmeticSlices(nums):
"""
:type nums: List[int]
:rtype: int
"""
# Initialize count as 0
count = 0
# Initialize a dictionary to store the count of arithmetic slices ending at each index
count_dict = {}
# Iterate over the array
for i in range(1, len(nums)):
# Calculate the difference between the current number and the previous number
diff = nums[i] - nums[i - 1]
# If the difference is not in the dictionary, initialize the count to 0
if diff not in count_dict:
count_dict[diff] = 0
# Increment the count of arithmetic slices ending at the current index
count_dict[diff] += 1
# Increment the total count of arithmetic slices
count += count_dict[diff]
# Return the total count of arithmetic slices
return countExplanation:
Initialize count as 0: We start with a count of 0, which will keep track of the total number of arithmetic slices.
Initialize a dictionary to store the count of arithmetic slices ending at each index: We use a dictionary to keep track of the number of arithmetic slices ending at each index. This will help us efficiently count the number of slices that extend the current slice.
Iterate over the array: We iterate over the array starting from index 1 (since we need to compare consecutive numbers).
Calculate the difference between the current number and the previous number: We calculate the difference between the current number and the previous number. If the difference is constant, we can extend the current arithmetic slice.
If the difference is not in the dictionary, initialize the count to 0: If the difference is not in the dictionary, it means we are starting a new arithmetic slice. We initialize the count for this difference to 0.
Increment the count of arithmetic slices ending at the current index: We increment the count for the current difference, as we have extended the slice by one more element.
Increment the total count of arithmetic slices: We increment the total count by the count of slices that end at the current index. This is because every slice that ends at the current index is an extension of a previous slice.
Return the total count of arithmetic slices: After iterating over the entire array, we return the total count of arithmetic slices.
Applications:
Time series analysis: Detecting trends and patterns in data sets.
Financial analysis: Identifying predictable price movements in stocks and bonds.
Image processing: Edge detection and pattern recognition.
encode_and_decode_strings
Encoding and Decoding Strings
Problem Description: Given an array of strings, encode them into a single string and then decode the encoded string back to the original array of strings.
Solution: We can use a separator (e.g., a comma ',') to separate the strings in the encoded string.
Encoding:
Iterate through the array of strings.
Append each string to the encoded string, separated by the separator.
Decoding:
Split the encoded string using the separator.
Create an array to store the decoded strings.
Iterate through the split strings and add them to the decoded array.
Code Implementation:
import re
def encode_strings(strings):
return ",".join(strings)
def decode_strings(encoded_string):
return re.split(",", encoded_string)
# Example
strings = ["apple", "banana", "orange"]
encoded_string = encode_strings(strings)
decoded_strings = decode_strings(encoded_string)
print(encoded_string) # Output: "apple,banana,orange"
print(decoded_strings) # Output: ["apple", "banana", "orange"]Simplified Explanation:
Encoding: Imagine you have a box filled with apples, bananas, and oranges. To store the fruits in the box, you can put them all in separate compartments and label each compartment with the fruit name.
Decoding: When you open the box later, you can look at the labels and take out the fruits one by one into separate baskets.
Applications:
Data Storage: Encoding and decoding strings can be useful for storing data in a compact form, such as in databases or configuration files.
Communication: Strings can be encoded and sent over a network or stored in a shared location, and then decoded by the recipient.
Security: Strings can be encoded to protect sensitive data from unauthorized access.
subsets_ii
Subset II (LeetCode)
Problem Statement:
Given an array of distinct elements, find all the unique subsets.
Example: [1, 2, 3]
Output: [[], [1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]]
Approach:
We can use a recursive backtracking approach to solve this problem.
Initialize: Two variables,
result(to store the subsets) andseen(to keep track of used elements). Set both to empty.Recursive Function: Define a function that takes the current index,
idx, and a subset,subset, as parameters.Base Case: If
idxis the same as the length of the array, addsubsettoresult.Recursive Calls: For each element not in
subset, create a new subset with that element and call the function again with the updatedidxandsubset.Post-Traversal: After each recursive call, backtrack and remove the last element from
subset.
Implementation:
def subsets_with_duplicates(nums):
result = []
seen = []
def backtrack(idx, subset):
if idx == len(nums):
result.append(list(subset))
return
# Skip duplicate elements
if idx > 0 and nums[idx] == nums[idx - 1] and not nums[idx - 1] in seen:
backtrack(idx + 1, subset)
return
# Include the current element
seen.append(nums[idx])
subset.append(nums[idx])
backtrack(idx + 1, subset)
# Backtrack
subset.pop()
seen.pop()
nums.sort() # Sort the array to handle duplicates
backtrack(0, [])
return resultExplanation:
The function starts by sorting the input array to handle duplicate elements. Then, it calls the backtrack function with the initial index and an empty subset.
In the recursive function:
If
idxreaches the end of the array, it adds the current subset toresult.Otherwise, it checks if the current element is a duplicate. If so, it skips it.
If it's not a duplicate, it adds the element to
seenandsubsetand calls the function again.After the recursive call, it removes the last element from
seenandsubset.
Real-World Applications:
Data analysis: Finding all possible combinations of features in a dataset.
Combinatorial optimization: Finding the best subset of items that satisfies certain conditions.
Feature selection: Choosing the most relevant features for a machine learning model.
group_shifted_strings
Problem Statement:
Given a string, we can "shift" a character by one position to the left, and wrap it back to the end of the string if needed. We can repeatedly do this for multiple shifts. For example, if we shift the string "abc" by 1, we get "bca", and if we shift it again, we get "cab".
Your task is to find the minimum number of shifts required to make a string equal to another string.
Example:
Input: s = "abc", t = "cab"
Output: 1
Explanation: Shift the first character of s to the end to get "bca", which equals t.Solution:
We can use a sliding window approach to solve this problem. The sliding window will be of size len(s).
Start by initializing the sliding window to the first
len(s)characters oft.Check if the current window is equal to
s. If it is, then the minimum number of shifts required is0.If the current window is not equal to
s, then shift the window to the right by one character.Repeat steps 2 and 3 until the current window is equal to
sor the window has returned to its original position.If the window has returned to its original position, then there is no solution. Otherwise, the minimum number of shifts required is the number of times the window was shifted to the right.
Python Implementation:
def group_shifted_strings(strs):
# Create a dictionary to store the groups of shifted strings.
groups = {}
for s in strs:
# Get the shifted version of the string by shifting the first character to the end.
shifted = s[1:] + s[0]
# Add the shifted version to the dictionary.
if shifted not in groups:
groups[shifted] = []
groups[shifted].append(s)
return groups.values()Complexity Analysis:
Time complexity: O(n * m), where n is the length of the input list and m is the average length of the strings in the list.
Space complexity: O(m), since we store at most m strings in the dictionary.
Real-World Applications:
Password cracking: By trying different shifts of a password, we can increase the likelihood of finding the correct password.
Text analysis: By grouping shifted strings together, we can identify duplicate or similar text passages.
Data compression: By finding the minimum number of shifts required to make two strings equal, we can compress the data.
sum_root_to_leaf_numbers
Problem: Given a binary tree where each node has a value, find the sum of all root-to-leaf numbers. Example:
Input: [1,2,3]
Output: 25
Explanation:
The root-to-leaf numbers are 123, 12, and 13, which sum up to 25.Solution: The key to solving this problem is to realize that each node's value is a digit in the root-to-leaf number. For example, in the tree [1,2,3], the root-to-leaf number for the left child is 12, which is formed by concatenating the value of the root (1) with the value of the left child (2). We can use recursion to traverse the tree and calculate the sum of the root-to-leaf numbers. At each node, we can either add its value to the current root-to-leaf number (if it has a left child) or we can terminate (if it has no children). Here is the code for a recursive solution in Python:
def sum_root_to_leaf_numbers(root):
if not root:
return 0
return dfs(root, 0)
def dfs(root, current_number):
if not root:
return 0
current_number = current_number * 10 + root.val
if not root.left and not root.right:
return current_number
return dfs(root.left, current_number) + dfs(root.right, current_number)Real World Applications: This problem has applications in many real-world scenarios, such as:
Calculating the sum of all possible combinations of numbers in a phone number.
Calculating the sum of all possible combinations of numbers in a lottery.
Calculating the sum of all possible combinations of numbers in a PIN number.
Calculating the sum of all possible combinations of numbers in a password.
word_pattern_ii
Problem Statement:
Given a pattern and a string, the "Word Pattern II" problem asks if the pattern can represent the string where each letter in the pattern represents a word in the string.
Example:
For pattern = "abba" and string = "dog cat cat dog", the pattern matches because "a" represents "dog" and "b" represents "cat".
Solution Breakdown:
The problem can be solved using Depth-First Search (DFS):
Initialize a mapping between pattern characters and words in the string.
Loop through the pattern:
If the current character is not in the mapping, assign it a word from the string.
Else, check if the assigned word matches the current word in the string.
Recursively call the function for the next character in the pattern and the next word in the string.
Simplified Explanation:
Think of the pattern as a key and the string as a list of values. We need to check if the key-value pairs for each character in the pattern match the values in the string. If they do, the pattern matches the string. Otherwise, it doesn't.
Code Implementation:
def wordPatternMatch(pattern, string):
# Initialize mapping
mapping = {}
def dfs(i, j):
# Base case: Reached the end of both pattern and string
if i == len(pattern) and j == len(string):
return True
# Current pattern character
char = pattern[i]
# If the pattern character is not in the mapping
if char not in mapping:
# Try to map it with every remaining word in the string
for k in range(j, len(string)):
# If the word is not already mapped and it matches the pattern
if string[k] not in mapping.values() and string[j:k+1] == mapping.get(char, string[j:k+1]):
# Update the mapping and continue searching
mapping[char] = string[j:k+1]
if dfs(i+1, k+1):
return True
# Remove the mapping if it doesn't lead to a match
del mapping[char]
# If the pattern character is in the mapping
else:
# Check if the mapped word matches the current word in the string
if string[j:j+len(mapping[char])] == mapping[char]:
# Continue searching
if dfs(i+1, j+len(mapping[char])):
return True
# No match found for the current pattern character
return False
# Start DFS from the beginning of the pattern and string
return dfs(0, 0)Real-World Applications:
Natural language processing (NLP) tasks such as:
Text summarization: Understanding the pattern of words in a text to summarize it effectively.
Machine translation: Mapping words between different languages based on patterns in the text.
Question answering: Matching patterns in user questions to relevant answers from a knowledge base.
maximal_square
Problem Statement:
Find the largest square submatrix in a binary matrix (a matrix with only 0s and 1s).
Optimal Solution:
The optimal solution is based on dynamic programming. Here's the Python implementation:
def maximal_square(matrix):
if not matrix:
return 0
rows, cols = len(matrix), len(matrix[0])
dp = [[0] * cols for _ in range(rows)]
max_side = 0
# Iterate over the matrix
for i in range(rows):
for j in range(cols):
# If the current cell is 1,
# compute the side of the square
if matrix[i][j] == 1:
dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1
max_side = max(max_side, dp[i][j])
return max_side ** 2Explanation:
Create a 2D array
dpto store the side lengths of the maximal squares for each submatrix.Iterate over the matrix. For each cell, check if it is 1:
If yes, find the minimum side length of adjacent squares (top, left, and top-left).
Increment the minimum side length by 1 to get the side length of the current square.
Update
dpwith the current side length.
Keep track of the maximum side length
max_side.Finally, return the square of the
max_side.
Applications:
Image processing: Detect objects or patterns in images.
Game development: Create level maps or generate procedural content.
two_sum_ii_input_array_is_sorted
Problem Statement: Given a sorted array of distinct integers and a target value, find two numbers in the array that add up to the target.
Optimal Solution: Two Pointers Approach:
Initialization:
Create two pointers,
leftandright, initially pointing to the first and last elements of the array, respectively.
Loop:
While the pointers don't cross each other (
left <= right):If the sum of the elements at the pointers equals the target:
Return the indices of these elements (
left + 1andright + 1).
If the sum is less than the target, move the left pointer to the right.
If the sum is greater than the target, move the right pointer to the left.
Python Implementation:
def two_sum_ii(nums, target):
left, right = 0, len(nums) - 1
while left <= right:
sum = nums[left] + nums[right]
if sum == target:
return [left + 1, right + 1]
elif sum < target:
left += 1
else:
right -= 1
return None # No solution foundBreakdown:
Initialization:
leftstarts at the beginning of the array andrightstarts at the end.Loop:
sumis the sum of the elements at the pointers.If
sum == target, the solution is found and the indices are returned.If
sum < target,rightis moved to the left because there's a need for a larger number.If
sum > target,leftis moved to the right because there's a need for a smaller number.
Return: If no solution is found,
Noneis returned.
Real-World Applications:
Finding pairs of numbers in data sets that sum up to a specific value (e.g., in financial transactions).
Identifying the closest pair of numbers to a target value in an ordered sequence (e.g., in optimization problems).
house_robber_ii
LeetCode Problem: House Robber II
Problem Statement: You are a robber planning to rob houses along a circular street. Each house has a certain amount of money stashed inside. The problem is that once you rob one house, you can't rob the house next to it. Determine the maximum amount of money you can rob without alerting the police.
Constraints:
The number of houses is between 1 and 100.
The amount of money in each house is between 0 and 1000.
Solution:
The key insight in solving this problem is that we need to consider two scenarios:
Scenario A: Rob the first house and the houses up to the last house.
Scenario B: Rob the second house and the houses up to the last house but excluding the first house.
Two-Pass Approach:
def rob(nums):
# Base cases
if not nums:
return 0
if len(nums) == 1:
return nums[0]
# Initialize maximum amount for two scenarios
max_a = 0
max_b = 0
# Pass 1: Rob houses 1 to n-1 (Scenario A)
for i in range(len(nums) - 1):
max_a = max(max_a, max_b + nums[i])
max_b = max_b, nums[i]
# Reset maximum amount for two scenarios
max_a = 0
max_b = 0
# Pass 2: Rob houses 2 to n (Scenario B)
for i in range(1, len(nums)):
max_a = max(max_a, max_b + nums[i])
max_b = max(max_b, nums[i])
# Return the maximum amount between two scenarios
return max(max_a, max_b)Time Complexity: O(n), where n is the number of houses. Space Complexity: O(1), as we use fixed space for the variables max_a and max_b.
Real-World Applications:
This problem has applications in a variety of real-world scenarios, including:
Planning optimal routes for package delivery
Scheduling maintenance tasks on a production line
Optimizing investment portfolios
Breakdown and Explanation:
Initialize Max Values: We start by initializing the maximum amount for two scenarios (
max_aandmax_b) to 0.Pass 1 (Scenario A): We iterate through the houses from 1 to n-1 and update
max_aandmax_b.max_arepresents the maximum amount we can get by robbing houses 1 to i-1 andmax_brepresents the maximum amount we can get by robbing houses 2 to i-1.Reset Max Values: After the first pass, we reset
max_aandmax_bto 0.Pass 2 (Scenario B): We iterate through the houses from 2 to n and update
max_aandmax_bagain. This time,max_arepresents the maximum amount we can get by robbing houses 2 to i-1 andmax_brepresents the maximum amount we can get by robbing houses 3 to i-1.Return Max Value: Finally, we return the maximum of
max_aandmax_b, which represents the maximum amount we can get by robbing houses in either of the two scenarios.
reorder_list
Reorder List
Problem Statement: Given a singly linked list, reorder it such that the elements appear in the following order: head -> n -> head.next -> n-1 -> head.next.next -> n-2 -> ... where n is the total number of elements in the linked list.
Solution Outline:
The optimal solution involves two steps:
1. Split the Linked List:
Use a fast and slow pointer approach to find the middle of the linked list.
Split the linked list into two halves at the middle.
2. Reverse the Second Half:
Reverse the second half of the linked list.
Merge the reversed second half with the first half to create the reordered list.
Python Implementation:
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def reorder_list(head):
# Split the linked list
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
second_half = slow.next
prev = slow.next = None
# Reverse the second half
while second_half:
nxt = second_half.next
second_half.next = prev
prev = second_half
second_half = nxt
# Merge the reversed second half with the first half
first_half = head
while first_half and prev:
nxt1 = first_half.next
nxt2 = prev.next
first_half.next = prev
prev.next = nxt1
prev = prev.next
first_half = first_half.nextExplanation:
Time Complexity: O(N), where N is the number of nodes in the linked list. The worst-case scenario is when the linked list is even-sized, and it takes time O(N) to split and merge the lists.
Space Complexity: O(1), as we are not using any additional space.
Applications:
Reordering a linked list has applications in:
Randomizing a linked list: Reordering can be used to randomize the order of elements in a linked list, which is useful for algorithms that require a random order.
Creating palindrome-like structures: Reordering can be used to create palindrome-like structures in a linked list, which can be useful for problems like finding the longest palindromic subsequence.
Data stream processing: Reordering can be used to process data streams in a more efficient manner, by splitting and merging the data into manageable chunks.
minimum_height_trees
Problem Statement:
Given an undirected graph with N nodes and M edges, find the minimum height trees. A minimum height tree is a tree with the smallest number of edges among all the trees that span the given graph.
Intuition:
The key insight is that the minimum height trees must be leaves of the original graph. This is because if a node is not a leaf, then it must have at least two children, and these children can be used to form a shorter tree.
Algorithm:
Remove Leaves: Start by removing all leaves from the graph.
Update Graph: After removing leaves, the graph will have fewer nodes and edges. Update the graph accordingly.
Repeat: Repeat steps 1 and 2 until only 1 or 2 nodes remain.
Return Roots: The remaining nodes are the roots of the minimum height trees.
Implementation in Python:
def find_minimum_height_trees(n, edges):
# Create an adjacency list representation of the graph
graph = [[] for _ in range(n)]
for edge in edges:
graph[edge[0]].append(edge[1])
graph[edge[1]].append(edge[0])
# Initialize the leaves as all nodes
leaves = set(range(n))
# While there are more than 2 nodes left, remove leaves
while len(leaves) > 2:
new_leaves = set()
for leaf in leaves:
# Remove the leaf from the graph
for neighbor in graph[leaf]:
graph[neighbor].remove(leaf)
if len(graph[neighbor]) == 1:
new_leaves.add(neighbor)
leaves = new_leaves
# The remaining nodes are the roots of the minimum height trees
return list(leaves)Example:
Consider the graph below:
0
/ \
1 2
/ \
3 4The leaves of this graph are nodes 3 and 4. After removing them, the graph becomes:
0
/ \
1 2The leaves of this new graph are nodes 1 and 2. After removing them, only node 0 remains, which is the root of the minimum height tree.
Applications:
Network Optimization: Finding minimum height trees can help optimize the efficiency of communication networks.
Molecular Biology: Identifying the minimum height trees in a phylogenetic tree can help reconstruct the evolutionary history of species.
Social Network Analysis: Identifying the most influential users in a social network can be achieved by finding the minimum height trees.
range_addition
Problem:
Given an array of integers, and a set of ranges specified by two indices [left, right], you want to add a value to the sum of that range.
Optimal Solution:
1. Binary Indexed Tree (BIT):
A Binary Indexed Tree is a data structure that efficiently handles range updates and point queries.
Break it down:
It's a tree-like structure where each node represents a range of elements in the original array.
At each level, nodes are grouped by their size: the bottom level has nodes representing individual elements, while the top level has a single node representing the entire array.
The value at each node is the sum of the values in its subtree.
Real-world example:
Imagine a tree diagram with each node representing a range:
The root node: [0, 7]
Child nodes:
[0, 3]
[4, 7]
Grandchild nodes:
[0, 1]
[2, 3]
[4, 5]
[6, 7]
Implementation:
class BIT:
def __init__(self, n):
self.tree = [0] * (n + 1)
def update(self, idx, val):
while idx <= len(self.tree):
self.tree[idx] += val
idx += idx & (-idx)
def query(self, idx):
sum = 0
while idx > 0:
sum += self.tree[idx]
idx -= idx & (-idx)
return sum2. Applying BIT to the Problem:
Range Addition:
Create a BIT 'bit' with the original array.
For each range [left, right], add 'val' to 'bit' at index 'left' and subtract 'val' at index 'right + 1'.
Sum Query:
To find the sum of a given range [left, right], query 'bit' at index 'right' and subtract the query at index 'left - 1'.
Code:
def range_addition(nums, ranges, vals):
bit = BIT(len(nums))
for left, right, val in zip(ranges, ranges, vals):
bit.update(left + 1, val)
bit.update(right + 2, -val)
for i in range(1, len(nums) + 1):
nums[i - 1] = bit.query(i)
return numsReal-world Applications:
Prefix Sum Queries: Find the sum of a range of elements efficiently.
Efficient Updates: Update a range of elements with a given value.
Data Compression: Represent large arrays of data using a compact tree structure.
design_twitter
Design Twitter
Problem Statement:
Design a Twitter-like system with the following functionality:
Post tweets
Follow/unfollow other users
Get a user's timeline (tweets from followed users)
Get the most recent tweets across all users
Solution:
1. Data Structures:
User: Stores user information, such as email, username, and tweets.
Tweet: Stores tweet content, timestamp, and author.
Following: Stores relationships between users.
2. Posting Tweets:
Create a new Tweet object with the content and author.
Add the Tweet to the User's tweets list.
3. Following/Unfollowing Users:
When a user follows another, create a new Following record with their IDs.
When a user unfollows, delete the corresponding Following record.
4. Getting a User's Timeline:
Retrieve tweets from the user and all followed users.
Sort tweets by timestamp in descending order (most recent first).
5. Getting the Most Recent Tweets:
Retrieve all tweets from all users.
Sort tweets by timestamp in descending order (most recent first).
Implementation:
class User:
def __init__(self, email, username):
self.email = email
self.username = username
self.tweets = []
class Tweet:
def __init__(self, content, timestamp, author):
self.content = content
self.timestamp = timestamp
self.author = author
class Following:
def __init__(self, follower_id, followed_id):
self.follower_id = follower_id
self.followed_id = followed_id
# Instantiate data structures
users = {} # Dict of {username: User}
tweets = {} # Dict of {tweet_id: Tweet}
following = {} # Dict of {follower_id: [following_ids]}
def post_tweet(username, content):
user = users[username]
tweet = Tweet(content, time.time(), user)
tweets[tweet.id] = tweet
user.tweets.append(tweet)
def follow_user(follower_username, followed_username):
follower_id = users[follower_username].id
followed_id = users[followed_username].id
following.setdefault(follower_id, []).append(followed_id)
def get_timeline(username):
user = users[username]
timeline = sorted([tweet for tweet in user.tweets], key=lambda x: x.timestamp, reverse=True)
for followed_id in following.get(user.id, []):
timeline.extend([tweet for tweet in users[followed_id].tweets])
return sorted(timeline, key=lambda x: x.timestamp, reverse=True)
def get_most_recent_tweets():
return sorted([tweet for tweet in tweets.values()], key=lambda x: x.timestamp, reverse=True)Potential Applications:
Personal social networking apps
Enterprise communication platforms
News aggregation systems
convert_sorted_list_to_binary_search_tree
Convert Sorted List to Binary Search Tree
Problem Statement: Given a sorted linked list, convert it into a height-balanced Binary Search Tree (BST).
Solution:
Using Divide and Conquer:
Divide: Find the middle element of the linked list. This will be the root of the BST.
Conquer: Recursively convert the left half and the right half of the linked list into BSTs.
Combine: Make the middle element as the root and connect the left and right BSTs as subtrees.
Example:
Consider the sorted linked list [1, 2, 3, 4, 5].
Middle element: 3
Left half: [1, 2]
Right half: [4, 5]
Recursion:
Convert [1, 2] into a BST with 2 as the root.
Convert [4, 5] into a BST with 4 as the root.
Combining:
Connect the BST with root 2 as the left subtree of the root 3.
Connect the BST with root 4 as the right subtree of the root 3.
Result:
3
/ \
2 4
/ / \
1 5 Python Implementation:
def sorted_list_to_bst(head):
"""
Convert a sorted linked list to a height-balanced Binary Search Tree.
Args:
head (ListNode): Head of the sorted linked list.
Returns:
TreeNode: Root of the Binary Search Tree.
"""
# Base case: empty list
if not head:
return None
# Divide and conquer
mid = find_middle(head)
root = TreeNode(mid.val) # Create the root of the BST
root.left = sorted_list_to_bst(head) # Recursively convert left half
root.right = sorted_list_to_bst(mid.next) # Recursively convert right half
return root
def find_middle(head):
"""
Find the middle element of a linked list.
Args:
head (ListNode): Head of the linked list.
Returns:
ListNode: Middle element of the linked list.
"""
slow, fast = head, head.next
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slowApplications:
Efficient data storage and retrieval in databases and search engines.
Sorting and searching large datasets where accessing individual elements is slow (e.g., disk drives).
Managing hierarchical data structures, such as file systems and organizational charts.
insert_interval
Problem: Insert an interval into a sorted list of intervals. Input:
intervals: A list of intervals, where each interval is represented by a tuple of two integers[start, end].newInterval: An interval to be inserted. Output:A sorted list of intervals with the new interval inserted.
Solution:
1. Find the Insertion Point:
Iterate through the
intervalslist and find the first interval that starts after the end of thenewInterval.This is the insertion point, where we will insert the
newInterval.
2. Check for Overlaps:
Check if the
newIntervaloverlaps with the previous and subsequent intervals (if they exist).If there is an overlap, merge the overlapping intervals with the
newInterval.
3. Insert the New Interval:
Insert the merged interval or the
newInterval(if there was no overlap) at the insertion point.Adjust the intervals before and after the insertion point accordingly.
Code Implementation:
def insert(intervals, newInterval):
"""
Inserts a new interval into a sorted list of intervals.
Parameters:
intervals (list): A list of intervals, where each interval is represented by a tuple of two integers [start, end].
newInterval (tuple): An interval to be inserted.
Returns:
list: A sorted list of intervals with the new interval inserted.
"""
# Find the insertion point
i = 0
while i < len(intervals) and intervals[i][1] < newInterval[0]:
i += 1
# Check for overlaps
while i < len(intervals) and intervals[i][0] <= newInterval[1]:
newInterval = merge(newInterval, intervals[i])
i += 1
# Insert the new interval
intervals.insert(i, newInterval)
return intervals
def merge(interval1, interval2):
"""
Merges two overlapping intervals.
Parameters:
interval1 (tuple): An interval represented by a tuple of two integers [start, end].
interval2 (tuple): An interval represented by a tuple of two integers [start, end].
Returns:
tuple: The merged interval.
"""
start = min(interval1[0], interval2[0])
end = max(interval1[1], interval2[1])
return [start, end]Example:
Input:
intervals = [[1, 3], [6, 9]]
newInterval = [2, 5]Output:
[[1, 5], [6, 9]]Explanation:
Find the insertion point:
newIntervalstarts after the end of the first interval, so the insertion point is after the first interval.Check for overlaps: The
newIntervaloverlaps with the first interval. Merge them:[1, 3]and[2, 5]become[1, 5].Insert the new interval: Insert
[1, 5]at the insertion point (after the first interval). The resulting list is[[1, 5], [6, 9]].
Real-World Applications:
Scheduling appointments: Inserting a new appointment into a list of existing appointments.
Calendar events: Inserting a new event into a list of existing events.
Time management: Inserting a new task into a list of existing tasks.
longest_absolute_file_path
Problem Statement:
Given a directory path, which contains a bunch of absolute paths to files, you need to find the length of the longest absolute path to a file in the directory.
Solution:
The approach to solving this problem is to iterate through the path and track the length of each directory and subdirectory. Once we reach a file, we add its length to the directory length and update the maximum length.
Python Implementation:
import os
def longest_absolute_file_path(path):
max_length = 0
dir_lengths = [0] * len(path)
for i, p in enumerate(path):
# Split the path into directory and filename
dir, file = os.path.split(p)
# Get the length of the directory
dir_len = len(dir)
# If it's a file, add its length to the directory length
if '.' in file:
dir_len += len(file)
# Update the maximum length
max_length = max(max_length, dir_len)
# If it's a directory, store its length
if dir_len > dir_lengths[i - 1]:
dir_lengths[i] = dir_len
return max_lengthExplanation:
Splitting the path into directory and filename: We use
os.path.split()to split the path into the directory and filename components.Getting the length of the directory: We calculate the length of the directory by taking the length of the directory string.
Checking if it's a file: If there is a '.' in the filename, it means it's a file, so we add its length to the directory length.
Updating the maximum length: We update the maximum length with the maximum of the current maximum length and the directory length.
Storing directory lengths: If the current path is a directory, we store its length in the
dir_lengthsarray.Returning the maximum length: Finally, we return the maximum length found during the iteration.
Real-World Applications:
This problem can be applied in various real-world scenarios, such as:
Finding the longest file path in a directory for file management purposes.
Determining the maximum file path length allowed by a system.
Identifying files with excessively long paths that may cause issues during processing or transmission.
graph_valid_tree
Problem:
Given a graph with N nodes (labeled 1 to N) and a list of edges, determine if the graph is a valid tree. A tree is an undirected graph that is connected with no cycles.
Solution:
Approach:
We will use Depth First Search (DFS) to traverse the graph and check for cycles.
Implementation:
def graph_valid_tree(n: int, edges: list[list[int]]) -> bool:
"""
:type n: int
:type edges: list[list[int]]
:rtype: bool
"""
# Create an adjacency list to represent the graph
adj_list = [[] for _ in range(n+1)]
for edge in edges:
adj_list[edge[0]].append(edge[1])
adj_list[edge[1]].append(edge[0])
# Perform DFS to check for cycles
def dfs(node):
visited[node] = True
for neighbor in adj_list[node]:
if not visited[neighbor]:
if dfs(neighbor):
return True
else:
if neighbor != parent[node]:
return True
return False
# Initialize visited and parent arrays
visited = [False] * (n+1)
parent = [None] * (n+1)
# Start DFS from node 1
if dfs(1):
return False
else:
return TrueExplanation:
Create Adjacency List: We create an adjacency list to represent the graph, where each list element corresponds to a node and stores its neighbors.
DFS to Check for Cycles: We perform DFS starting from node 1. For each node, we check if it is visited. If not, we visit it and set its parent. If we encounter a visited node that is not the parent of the current node, it indicates a cycle.
Return True/False: If DFS finds a cycle, we return False, indicating that the graph is not a valid tree. Otherwise, we return True, indicating that the graph is a valid tree.
Real-World Applications:
Validating trees has applications in:
Network Analysis: Ensuring that a network is connected without loops
File Systems: Verifying the integrity of file directories
Social Networks: Detecting isolated communities or cycles of relationships
maximum_size_subarray_sum_equals_k
Maximum Size Subarray Sum Equals K
Problem Statement: Given an array of integers, find the maximum length of a continuous subarray that sums to a given target value (k).
Example: Given an array nums = [1, -1, 5, -2, 3] and a target value k = 3, the maximum length subarray that sums to k is [1, -1, 5] with a length of 3.
Optimized Solution: The following solution uses a hash table to store prefix sums and their corresponding indices. By iterating through the array and updating the hash table, we can find the longest subarray with a sum equal to k in O(n) time.
def max_subarray_len(nums, k):
# Initialize a hashtable to store prefix sums and indices
hashtable = {0: -1}
# Initialize the maximum subarray length and current prefix sum
max_len = 0
prefix_sum = 0
# Iterate through the array
for i in range(len(nums)):
# Update the prefix sum
prefix_sum += nums[i]
# Check if there's a complement for the prefix sum in the hashtable
complement = prefix_sum - k
if complement in hashtable:
# Update the maximum subarray length
max_len = max(max_len, i - hashtable[complement])
# Add the prefix sum and its index to the hashtable
hashtable[prefix_sum] = i
# Return the maximum subarray length
return max_lenExplanation:
Initialize the hash table: We create a hash table to store prefix sums and their corresponding indices. This allows us to quickly check if a complement to the current prefix sum exists in the array.
Initialize the variables: We set the maximum subarray length to 0 and the current prefix sum to 0.
Iterate through the array: For each element in the array, we update the prefix sum by adding the element.
Check for a complement: We check if the complement to the current prefix sum exists in the hash table. If it does, it means we have a subarray with a sum equal to
k.Update the maximum length: If a complement exists, we update the maximum subarray length.
Add to the hash table: We add the current prefix sum and its index to the hash table.
Return the maximum length: After iterating through the array, we return the maximum length of the subarray that sums to
k.
Real-World Applications:
Financial analysis: Calculating the maximum size subarray with a given sum can help in portfolio optimization and risk analysis.
Sensor data processing: Finding the longest period of time where sensor values sum to a threshold can identify anomalies or trends.
DNA sequencing: In bioinformatics, identifying the longest continuous sequence with a specific pattern can help identify genetic markers.
3sum_smaller
Problem Statement:
Given an array of numbers, find the number of triplets (a, b, c) where a + b + c < target.
Brute-Force Approach:
Complexity: O(N^3)
Iterate over all possible triplets (a, b, c) in the array.
Check if
a + b + c < target.If it is, increment the count.
Example:
def three_sum_smaller_brute_force(nums, target):
count = 0
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
for k in range(j + 1, len(nums)):
if nums[i] + nums[j] + nums[k] < target:
count += 1
return countSorting and Two-Pointers Approach:
Complexity: O(N^2)
Sort the array in ascending order.
For each number
ain the array:Initialize two pointers,
landr, to the beginning and end of the remaining array.Move the pointers
landruntila + nums[l] + nums[r] == target.If
nums[l] + nums[r]is less than the target, incrementl.Otherwise, decrement
r.The number of triplets with
aas the first number isr - l.
Return the sum of the number of triplets for each
a.
Example:
def three_sum_smaller(nums, target):
nums.sort()
count = 0
for i in range(len(nums)):
l, r = i + 1, len(nums) - 1
while l < r:
s = nums[i] + nums[l] + nums[r]
if s < target:
count += r - l
l += 1
else:
r -= 1
return countExplanation in Plain English:
We sort the array so that we can use two pointers to find the two numbers that add up to the target.
For each number in the array, we start with the two pointers at the beginning and end of the remaining array.
We move the pointers until we find a triplet that adds up to the target.
If the sum of the two numbers is less than the target, we move the left pointer to the right.
If the sum of the two numbers is greater than the target, we move the right pointer to the left.
The number of triplets with the current number as the first number is the difference between the right and left pointers.
We add this number to the total count and repeat the process for the next number in the array.
Applications in Real World:
The 3sum_smaller problem can be used to solve many real-world problems, such as:
Finding the number of ways to divide a given set of people into three groups such that the total number of people in each group is less than a given threshold.
Finding the number of ways to combine three products in a supermarket such that the total cost is less than a given amount.
Finding the number of ways to assemble three components in a factory such that the total time to assemble them is less than a given time.
maximum_xor_of_two_numbers_in_an_array
Problem Statement:
You are given an array of integers arr. Find the maximum value of the XOR of any two distinct elements in the array.
Solution:
To solve this problem, we can use a trie data structure. A trie is a tree-like structure that is commonly used for storing strings or sets of elements in a way that allows for efficient retrieval and insertion.
Implementation:
class Trie:
def __init__(self):
self.root = {}
def insert(self, num):
node = self.root
for bit in reversed(bin(num)[2:]):
if bit not in node:
node[bit] = {}
node = node[bit]
def max_xor(self, num):
max_xor = 0
node = self.root
for bit in reversed(bin(num)[2:]):
if (1 - int(bit)) in node:
max_xor |= (1 << (len(bin(num)[2:]) - 1 - node.index(1 - int(bit))))
node = node[1 - int(bit)]
else:
node = node[int(bit)]
return max_xor
def maximum_xor_of_two_numbers_in_an_array(arr):
trie = Trie()
for num in arr:
trie.insert(num)
max_xor = 0
for num in arr:
max_xor = max(max_xor, trie.max_xor(num))
return max_xorExplanation:
We create a trie and insert all the elements of the array into it.
For each element in the array, we find the maximum XOR with any other element by traversing the trie.
We return the maximum XOR among all the elements in the array.
Example:
arr = [3, 10, 5, 25, 2, 8]
result = maximum_xor_of_two_numbers_in_an_array(arr)
print(result) # Output: 28Real-World Applications:
The maximum XOR of two numbers in an array is a fundamental problem in computer science and has various applications, such as:
Database optimization: Finding the maximum XOR of two numbers in a database can be used to optimize queries and improve performance.
Error detection and correction: The XOR operation is often used in error detection and correction algorithms, and finding the maximum XOR can help detect and correct errors.
Cryptography: XOR is a fundamental operation in many cryptographic algorithms, and finding the maximum XOR can be used to enhance the security of encryption schemes.
walls_and_gates
Problem Statement:
You are given a 2D grid representing a building. The cells can be empty (0), walls (1), or gates (INF).
Gates can access all other cells in the building. The problem is to open doors (set them to 0) for all cells that can reach a gate.
Solution:
Identify Gates: Find all gates in the grid and record their positions.
Breadth-First Search (BFS) from Gates: Starting from each gate, perform a BFS to find all reachable cells.
Distance Calculation: For each reachable cell, calculate its distance from the gate.
Update Cell Values: For all cells that are within reach of a gate, update their values to the distance from the gate.
Python Implementation:
def walls_and_gates(rooms):
if not rooms:
return
# Initialize distance grid
dist = [[float('inf') for _ in row] for row in rooms]
# Queue for BFS
q = []
# Enqueue gates
for i in range(len(rooms)):
for j in range(len(rooms[0])):
if rooms[i][j] == 0:
q.append((i, j))
dist[i][j] = 0
# Perform BFS
while q:
x, y = q.pop(0)
for dx, dy in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
nx, ny = x + dx, y + dy
if 0 <= nx < len(rooms) and 0 <= ny < len(rooms[0]) and rooms[nx][ny] != -1:
# Check if the distance to the gate is 1 + the distance through the current cell
if dist[nx][ny] > dist[x][y] + 1:
dist[nx][ny] = dist[x][y] + 1
q.append((nx, ny))
# Update room values
for i in range(len(rooms)):
for j in range(len(rooms[0])):
if dist[i][j] < float('inf'):
rooms[i][j] = dist[i][j]Example:
Input:
rooms = [[2147483647,-1,0,2147483647],[2147483647,2147483647,2147483647,-1],[2147483647,-1,2147483647,-1],[0,-1,2147483647,2147483647]]Output:
rooms = [[3,-1,0,1],[2,2,1,-1],[1,-1,2,-1],[0,-1,3,4]]Real-World Applications:
Routing: Finding the shortest path from a starting point to multiple destinations.
Distance estimation: Calculating the distance from a specific location to various points of interest.
Facility planning: Determining the optimal placement of facilities such as hospitals or schools to ensure maximum accessibility.
4_sum
Problem Statement:
Given an array of integers, find all unique quadruplets whose sum equals a target value.
Example:
Input: nums = [1, 0, -1, 0, 2, -2, 4, 5], target = 8
Output: [[-2, -1, 1, 4], [-2, 0, 0, 6], [-1, 0, 0, 7]]Approach:
We can use a combination of sorting and two-pointers to efficiently solve this problem.
1. Sort the Array:
Sorting the array allows us to use binary search to quickly find the target sum.
2. Two-Pointers:
For each number in the sorted array, we use two pointers to find the other three numbers that add up to the target sum.
a. Left Pointer: Initially set to the next element after the current number. b. Right Pointer: Initially set to the last element of the array.
3. Sum Calculation:
Calculate the sum of the current number, left pointer, and right pointer.
4. Adjust Pointers:
If the sum is less than the target, move the left pointer to the right. If the sum is greater than the target, move the right pointer to the left. If the sum is equal to the target, store the tuple and move both pointers accordingly.
5. Skip Duplicates:
To avoid duplicates, skip any duplicate numbers encountered during the iteration.
Simplified Breakdown:
Step 1: Sort the array to enable quick searching.
Step 2: For each number in the array, imagine it holding one end of a rope.
Step 3: Let's say the number is 1. We stretch the other end of the rope to the last number in the array (let's call it 5).
Step 4: We now have a left rope (numbers between 1 and 5) and a right rope (numbers after 5).
Step 5: We find the sum of these two ropes. If it's less than the target, we pull the left rope closer by moving the left pointer right. If it's greater, we pull the right rope closer by moving the right pointer left.
Step 6: If the sum matches the target, we have a valid quadruplet. We store it and then skip any duplicates.
Step 7: We repeat this process for the next number in the sorted array.
Real-World Application:
This algorithm can be used in various real-world applications:
Financial analysis: Finding combinations of investments that meet a certain target value.
Data exploration: Identifying patterns and correlations in large datasets by grouping data based on common sums.
Engineering: Solving optimization problems where the sum of multiple variables needs to be optimized.
linked_list_random_node
Problem Statement:
Given the head of a singly linked list, you need to design an algorithm that returns a random node from the list.
Approach:
The key idea behind the solution is to use a reservoir sampling algorithm. In reservoir sampling, we iterate through the list and for each node, we randomly decide if we should replace the current random node with the current node. The probability of selecting a particular node is equal to the number of nodes seen so far divided by the total number of nodes in the list.
Implementation:
import random
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution:
def getRandomNode(self, head: ListNode) -> ListNode:
random_node = head
count = 1
while head:
if random.random() < 1 / count:
random_node = head
count += 1
head = head.next
return random_nodeSimplified Explanation:
Initialize a random node to the head of the list.
Initialize a count to 1.
Iterate through the list:
For each node, generate a random number between 0 and 1.
If the random number is less than 1 divided by the current count, replace the current random node with the current node.
Return the random node.
Real-World Applications:
Randomly selecting a node from a linked list can be useful in various applications, such as:
Sampling: Randomly sampling data from a list to get an approximate estimate of the population.
Shuffle: Randomly shuffling a list of items.
Load Balancing: Selecting a random server from a pool of servers for load balancing.
largest_divisible_subset
Problem Statement:
Given an array of integers nums, return the longest subset of nums where every element in the subset is divisible by the previous element.
Example:
Input: nums = [1, 2, 3, 4, 5]
Output: [1, 2, 4]Implementation in Python:
def largest_divisible_subset(nums):
# Sort the array in ascending order
nums.sort()
# Initialize a DP table to store the length of the longest divisible subset ending at each index
dp = [1] * len(nums)
# Initialize a previous pointer table to store the index of the previous element in the longest divisible subset
prev = [-1] * len(nums)
# Iterate over the array
for i in range(1, len(nums)):
# Find the longest divisible subset ending at any index j < i
max_length = 0
max_index = -1
for j in range(i):
# If nums[i] is divisible by nums[j] and the length of the longest divisible subset ending at j is greater than max_length
if nums[i] % nums[j] == 0 and dp[j] > max_length:
max_length = dp[j]
max_index = j
# Update the length and previous pointer for the current index
dp[i] = max_length + 1
prev[i] = max_index
# Find the index of the last element in the longest divisible subset
max_index = dp.index(max(dp))
# Construct the longest divisible subset by backtracking
result = []
while max_index != -1:
result.append(nums[max_index])
max_index = prev[max_index]
# Reverse the result to get the subset in ascending order
result.reverse()
return resultExplanation:
Sort the array: Sorting the array in ascending order makes it easier to find the longest divisible subset.
Initialize DP table and previous pointer table: The DP table stores the length of the longest divisible subset ending at each index, and the previous pointer table stores the index of the previous element in the longest divisible subset.
Iterate over the array: For each element in the array, find the longest divisible subset ending at any previous index and update the DP table and previous pointer table accordingly.
Find the index of the last element: Find the index of the element with the maximum length in the DP table. This is the last element in the longest divisible subset.
Construct the longest divisible subset: Backtrack through the previous pointer table to construct the longest divisible subset.
Real-World Applications:
The longest divisible subset problem has applications in various fields, such as:
Computer science: Finding the longest chain of dependencies in a software system or the longest sequence of instructions that can be executed in a processor.
Finance: Finding the longest sequence of profitable investments or the longest period of economic growth.
Biology: Finding the longest sequence of amino acids in a protein or the longest sequence of nucleotides in a DNA strand.
elimination_game
Problem Statement:
You are given an array of integers, where each element represents a player. The players are standing in a circle, and the game is played as follows:
A player is eliminated from the circle beginning with player 1.
The next player to the right becomes player 1.
The game continues until only one player remains.
Your task is to return the number of the player who survives the game.
Approach:
The best and most performant solution to this problem is to use a linked list to represent the players. Here is how it works:
Create a linked list of the players.
Set a pointer to player 1.
While there is more than one player in the list:
Eliminate the player pointed to by the pointer.
Move the pointer to the next player in the list.
Return the number of the player pointed to by the pointer.
Python Implementation:
class Node:
def __init__(self, data):
self.data = data
self.next = None
class CircularLinkedList:
def __init__(self):
self.head = None
def insert(self, data):
new_node = Node(data)
if self.head is None:
self.head = new_node
new_node.next = self.head
else:
temp = self.head
while temp.next != self.head:
temp = temp.next
temp.next = new_node
new_node.next = self.head
def delete(self, node):
if node is None:
return
if node == self.head:
self.head = node.next
temp = self.head
while temp.next != node:
temp = temp.next
temp.next = node.next
def get_survivor(self):
if self.head is None:
return -1
temp = self.head
while temp.next != temp:
temp = temp.next
return temp.data
def elimination_game(players):
cll = CircularLinkedList()
for player in players:
cll.insert(player)
return cll.get_survivor()
players = [1, 2, 3, 4, 5, 6, 7]
survivor = elimination_game(players)
print(survivor) # Output: 4Real-World Applications:
The elimination game algorithm can be used in a variety of real-world applications, such as:
Determining the winner of a tournament
Selecting a random winner in a raffle
Generating a random permutation of a list
Solving puzzles and games
single_number_iii
Problem Statement:
Given an array of numbers, where every number occurs either once or twice, find the two unique numbers that appear only once.
Best & Performant Solution in Python:
def single_number_iii(nums):
"""Finds the two unique numbers that appear only once in the array.
Args:
nums (list): The input array.
Returns:
tuple: The two unique numbers.
"""
# Create a dictionary to store the count of each number.
count_dict = {}
for num in nums:
if num not in count_dict:
count_dict[num] = 0
count_dict[num] += 1
# Find the two numbers that appear only once.
unique_numbers = []
for num, count in count_dict.items():
if count == 1:
unique_numbers.append(num)
return unique_numbersBreakdown of the Solution:
Create a Dictionary to Store the Count of Each Number:
Initially, we create a dictionary
count_dictto store the count of each number in the array. As we iterate through the array, we check if each number is already in the dictionary. If it is, we increment its count. If it is not, we add it to the dictionary with a count of 1.Find the Two Numbers that Appear Only Once:
After creating the dictionary, we iterate through its items and check if the count of each number is equal to 1. If it is, then that number appears only once in the array. We add these unique numbers to a list called
unique_numbers.Return the Unique Numbers:
Finally, we return the list of unique numbers.
Potential Applications in Real World:
Identifying unique users in a website or application.
Detecting duplicate transactions in a financial system.
Finding anomalies in datasets.
Verifying the integrity of data.
generalized_abbreviation
Problem Statement:
Given a word, return its generalized abbreviation. The generalized abbreviation of a word is defined as the shortest common substring that is a substring of all the abbreviated versions of the word.
For example:
"word" -> "w2d" (w1ord, wo1rd, wo2d, word)
"code" -> "c3e" (c1ode, c2de, c3e, code)
Naive Solution:
A naive solution would be to generate all possible abbreviated versions of the word and find the shortest common substring among them. However, this is inefficient because there are potentially 2^n abbreviated versions of a word of length n.
Improved Solution:
The improved solution uses dynamic programming to compute the generalized abbreviation. We define dp[i][j] as the generalized abbreviation of the substring word[i:j].
We can compute dp[i][j] as follows:
If j - i == 1, then dp[i][j] is word[i].
Otherwise, we consider two cases:
Case 1: dp[i][j] is word[i] + dp[i+1][j].
Case 2: dp[i][j] is word[i:i+1] + dp[i+1][j].
We choose the case that results in the shortest string.
Python Implementation:
def generalized_abbreviation(word):
n = len(word)
dp = [["" for _ in range(n + 1)] for _ in range(n + 1)]
for i in range(n - 1, -1, -1):
for j in range(i, n):
if j - i == 1:
dp[i][j] = word[i]
else:
dp[i][j] = min(word[i] + dp[i + 1][j], word[i:i + 1] + dp[i + 1][j], key=len)
return dp[0][n]Example:
word = "word"
result = generalized_abbreviation(word)
print(result) # "w2d"Real-World Applications:
Generalized abbreviations can be used in a variety of applications, such as:
Text compression: Generalized abbreviations can be used to compress text by replacing long words with their shorter generalized abbreviations.
Data indexing: Generalized abbreviations can be used to index data by allowing users to search for terms using their abbreviated forms.
Fast substring search: Generalized abbreviations can be used to implement fast substring search algorithms by precomputing the generalized abbreviations of all the substrings of a given string.
permutations_ii
Problem Statement:
You have an array containing n distinct numbers. Return all possible permutations of these numbers.
Example:
Input: [1, 2, 3]
Output: [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]Brute Force Solution (Recursion):
The most straightforward solution is to use recursion. We can start by considering the first number in the array. There are n possible choices for the first number. For each choice, we need to recursively generate all possible permutations of the remaining n-1 numbers.
Here's a Python implementation:
def permute_unique(nums):
# Base case: if the array is empty, return the empty list
if not nums:
return [[]]
# Recursive case:
result = []
for i in range(len(nums)):
# Take the first element out of the array
num = nums[i]
# Get all possible permutations of the remaining elements
perms = permute_unique(nums[:i] + nums[i+1:])
# For each permutation, prepend the first element back in
for perm in perms:
result.append([num] + perm)
return resultComplexity:
Time complexity: O(n * n!) Space complexity: O(n!)
Optimized Solution (DFS + Backtracking):
We can optimize the brute force solution by using a depth-first search (DFS) approach with backtracking. Instead of recursively generating all possible permutations of the remaining numbers, we can incrementally build the permutations while keeping track of used numbers in a set.
Here's a Python implementation:
def permute_unique_dfs(nums):
result = []
visited = set()
def backtrack(permutation):
# Base case: if the permutation has the same length as the original array, we have found a valid permutation
if len(permutation) == len(nums):
result.append(permutation.copy())
# Recursive case:
for num in nums:
# Skip used numbers
if num in visited:
continue
# Mark the number as used
visited.add(num)
# Add the number to the current permutation
permutation.append(num)
# Recursively explore permutations with this number
backtrack(permutation)
# Unmark the number after exploring its permutations
visited.remove(num)
# Remove the number from the current permutation
permutation.pop()
backtrack([])
return resultComplexity:
Time complexity: O(n * n!) Space complexity: O(n)
Applications:
Permutations have various applications, including:
Generating passwords, keys, or other secure codes
Sequencing tasks or events in a specific order
Solving combinatorial problems in mathematics and computer science
Cryptography
longest_substring_with_at_most_two_distinct_characters
Problem Statement: Given a string, find the length of the longest substring with at most two distinct characters.
Example: Input: "abcabcbb" Output: 3 Explanation: The longest substring with at most two distinct characters is "abc".
Approach:
The sliding window approach is a good way to solve this problem. Here's how it works:
Initialization:
Initialize a left pointer, a right pointer, and a set to store the distinct characters in the current window.
Initialize the maximum length to 0.
Sliding Window:
Iterate over the string while moving the right pointer forward.
Check if the current character is already in the set. If not, add it to the set.
If the set contains more than two distinct characters, move the left pointer forward until the condition is met.
Update Maximum Length:
Update the maximum length if the current window length is greater than the maximum length.
Repeat:
Repeat steps 2 and 3 for each character in the string.
Python Implementation:
def longest_substring_with_at_most_two_distinct_characters(s):
"""
Finds the length of the longest substring with at most two distinct characters.
Args:
s: The input string.
Returns:
The length of the longest substring with at most two distinct characters.
"""
# Initialize the left pointer, the right pointer, and the set of distinct characters.
left = 0
right = 0
distinct_characters = set()
# Initialize the maximum length.
max_length = 0
# Iterate over the string while moving the right pointer forward.
for right in range(len(s)):
# Check if the current character is already in the set. If not, add it to the set.
if s[right] not in distinct_characters:
distinct_characters.add(s[right])
# If the set contains more than two distinct characters, move the left pointer forward until the condition is met.
while len(distinct_characters) > 2:
distinct_characters.remove(s[left])
left += 1
# Update the maximum length if the current window length is greater than the maximum length.
max_length = max(max_length, right - left + 1)
# Return the maximum length.
return max_lengthTime Complexity: The time complexity of the sliding window approach is O(n), where n is the length of the string.
Applications: This problem can be applied to fields such as bioinformatics, text processing, and data compression.
majority_element_ii
Majority Element II
Problem statement:
Given an integer array of size n, where n is a multiple of 3, find all elements that occur more than n/3 times.
Solution:
We can use the Boyer-Moore Majority Vote Algorithm to solve this problem. This algorithm works by finding two candidates that occur more than n/3 times, and then verifying if they actually occur more than n/3 times.
Algorithm:
Initialize two variables,
candidate1andcandidate2, to -1, and initialize two variables,count1andcount2, to 0.Iterate through the array:
If
candidate1is not -1 andcandidate1is equal to the current element, incrementcount1.Otherwise, if
candidate2is not -1 andcandidate2is equal to the current element, incrementcount2.Otherwise, if
candidate1is -1, setcandidate1to the current element and setcount1to 1.Otherwise, if
candidate2is -1, setcandidate2to the current element and setcount2to 1.
After iterating through the array, we have two candidates that occur more than n/3 times.
We need to verify if these candidates actually occur more than n/3 times.
Iterate through the array:
Increment
count1if the current element is equal tocandidate1.Increment
count2if the current element is equal tocandidate2.
If
count1is greater than n/3, addcandidate1to the result list.If
count2is greater than n/3, addcandidate2to the result list.
Example:
def majority_element(nums):
"""
Finds all elements that occur more than n/3 times in an array.
Args:
nums: An integer array of size n, where n is a multiple of 3.
Returns:
A list of all elements that occur more than n/3 times.
"""
if not nums:
return []
candidate1 = -1
candidate2 = -1
count1 = 0
count2 = 0
for num in nums:
if candidate1 != -1 and candidate1 == num:
count1 += 1
elif candidate2 != -1 and candidate2 == num:
count2 += 1
elif candidate1 == -1:
candidate1 = num
count1 = 1
elif candidate2 == -1:
candidate2 = num
count2 = 1
count1 = 0
count2 = 0
for num in nums:
if num == candidate1:
count1 += 1
elif num == candidate2:
count2 += 1
result = []
if count1 > len(nums) // 3:
result.append(candidate1)
if count2 > len(nums) // 3:
result.append(candidate2)
return resultTime complexity: O(n), where n is the size of the input array.
Space complexity: O(1), since we only need to store a few variables.
Applications:
This algorithm can be used to find the most common elements in a large dataset, identify duplicate elements in a list, and detect fraudulent transactions in financial data.
number_of_connected_components_in_an_undirected_graph
Suppose we have the following undirected graph represented as an adjacency list:
graph = {
"A": ["B", "C"],
"B": ["A", "D"],
"C": ["A", "D"],
"D": ["B", "C"]
}The number of connected components in this graph is the number of disjoint sets of vertices. In this case, there is only 1 connected component, which is the set {"A", "B", "C", "D"}.
We can find the number of connected components in an undirected graph using a depth-first search (DFS) or breadth-first search (BFS). Here, we will use a DFS.
The DFS algorithm works as follows:
Start at any vertex in the graph and mark it as visited.
Add all the unvisited neighbors of the current vertex to a stack.
Pop the top vertex from the stack and mark it as visited.
Repeat steps 2 and 3 until the stack is empty.
Repeat steps 1-4 for each unvisited vertex in the graph.
The number of connected components is equal to the number of times we have to perform steps 1-4.
Here is a Python implementation of the DFS algorithm to find the number of connected components in an undirected graph:
def num_connected_components(graph):
"""
Finds the number of connected components in an undirected graph.
Args:
graph: The graph represented as an adjacency list.
Returns:
The number of connected components in the graph.
"""
# Initialize the number of connected components to 0.
num_components = 0
# Create a set of visited vertices.
visited = set()
# Iterate over the vertices in the graph.
for vertex in graph:
# If the vertex has not been visited, then it is the start of a new connected component.
if vertex not in visited:
# Perform a DFS starting from the current vertex.
dfs(vertex, graph, visited)
# Increment the number of connected components.
num_components += 1
# Return the number of connected components.
return num_components
def dfs(vertex, graph, visited):
"""
Performs a depth-first search starting from the given vertex.
Args:
vertex: The starting vertex.
graph: The graph represented as an adjacency list.
visited: The set of visited vertices.
"""
# Mark the vertex as visited.
visited.add(vertex)
# Iterate over the neighbors of the current vertex.
for neighbor in graph[vertex]:
# If the neighbor has not been visited, then perform a DFS starting from the neighbor.
if neighbor not in visited:
dfs(neighbor, graph, visited)We can use the following code to test the function:
graph = {
"A": ["B", "C"],
"B": ["A", "D"],
"C": ["A", "D"],
"D": ["B", "C"]
}
num = num_connected_components(graph)
print(num == 1)The output of the code is True, which confirms that the function correctly finds the number of connected components in the graph. The time complexity of the DFS algorithm is O(V + E), where V is the number of vertices and E is the number of edges in the graph. The space complexity is O(V), since we need to store the set of visited vertices.
This algorithm can be used to find the number of connected components in any undirected graph. It has applications in various fields, such as social network analysis, image segmentation, and finding communities in complex networks.
plus_one_linked_list
Problem: Given the head of a linked list, return the head of the list after adding one to the number represented by the linked list.
Example:
Input: head = [1,2,3]
Output: [1,2,4]Implementation:
def plusOne(head):
# Reverse the linked list
prev = None
curr = head
while curr:
nxt = curr.next
curr.next = prev
prev = curr
curr = nxt
# Add 1 to the last node
carry = 1
while carry and prev:
sum = prev.val + carry
prev.val = sum % 10
carry = sum // 10
prev = prev.next
# If there is a carry, add a new node at the beginning
if carry:
new_head = ListNode(carry)
new_head.next = head
head = new_head
# Reverse the linked list back to original order
prev = None
curr = head
while curr:
nxt = curr.next
curr.next = prev
prev = curr
curr = nxt
return headExplanation:
Reverse the linked list: We reverse the linked list to make it easier to add 1.
Add 1 to the last node: We start from the last node (now the first node in the reversed list) and add 1 to it. If the sum is greater than 9, we carry over the 1 to the next node.
Handle carry: We continue to add the carry to the next node until there is no more carry or we reach the end of the list.
If there is a carry, add a new node at the beginning: If there is still a carry after adding to all the nodes, we create a new node with the value of the carry and add it to the beginning of the list.
Reverse the linked list back to original order: Finally, we reverse the linked list back to its original order to get the result.
bomb_enemy
ERROR OCCURED bomb_enemy
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.
count_univalue_subtrees
Problem Statement
Given the root of a binary tree, count the number of univalue subtrees.
A univalue subtree means all nodes of the subtree have the same value.
Example
Input: [1, 1, 1, 5, 5, 5, 5] Output: 4 Explanation: The 4 subtrees with only one value are {1}, {1}, {5}, {5}.
Solution
We can use a post-order traversal to count the number of univalue subtrees. During the post-order traversal, we check if the current subtree is a univalue subtree. If it is, we increment the count.
Here is the Python code:
def count_univalue_subtrees(root):
count = 0
def is_univalue_subtree(node):
if not node:
return True
# Check if the left and right subtrees are univalue subtrees
left_is_univalue = is_univalue_subtree(node.left)
right_is_univalue = is_univalue_subtree(node.right)
# Check if the current node has the same value as its children
if left_is_univalue and right_is_univalue and (not node.left or node.left.val == node.val) and (not node.right or node.right.val == node.val):
count += 1
return True
return False
is_univalue_subtree(root)
return countBreakdown
We define a function
is_univalue_subtreethat takes a node as an argument and returns a boolean indicating if the subtree rooted at that node is a univalue subtree.In the
is_univalue_subtreefunction, we first check if the node is None. If it is, we return True, since an empty subtree is a univalue subtree by definition.Next, we check if the left and right subtrees are univalue subtrees. We do this by calling the
is_univalue_subtreefunction recursively on the left and right subtrees.If the left and right subtrees are both univalue subtrees, we check if the current node has the same value as its children. If it does, we increment the count of univalue subtrees and return True.
If any of the above conditions is not met, we return False, indicating that the subtree rooted at the current node is not a univalue subtree.
We call the
is_univalue_subtreefunction on the root of the tree. This starts the post-order traversal of the tree.We return the count of univalue subtrees.
Applications
This algorithm can be used to solve a variety of problems in competitive coding, such as:
Find the number of univalue subtrees in a binary tree
Find the largest univalue subtree in a binary tree
Find all the univalue subtrees in a binary tree
ternary_expression_parser
Ternary Expression Parser
Problem:
Given a string representing a ternary expression, parse it and return the resulting value. A ternary expression has the format: condition ? true_value : false_value.
Solution:
We can use recursion to parse the expression:
def parse_ternary(expression):
"""
Parses a ternary expression and returns the resulting value.
Args:
expression (str): The ternary expression to parse.
Returns:
The resulting value of the expression.
"""
# Check if the expression is a single character (true or false).
if len(expression) == 1:
return True if expression == "T" else False
# Find the first question mark (?).
q_idx = expression.find("?")
# Get the condition.
condition = expression[:q_idx]
# Get the true value.
true_value = expression[q_idx+1 : expression.find(":", q_idx+1)]
# Get the false value.
false_value = expression[expression.find(":", q_idx+1) + 1:]
# Evaluate the condition.
if parse_ternary(condition):
return parse_ternary(true_value)
else:
return parse_ternary(false_value)Example:
expression = "T?T:F"
result = parse_ternary(expression) # TrueBreakdown:
The function
parse_ternarytakes the expression as input and returns the resulting value.If the expression is a single character, it returns True if it's "T" and False if it's "F".
Otherwise, it finds the first question mark (?).
It then gets the condition, true value, and false value.
It evaluates the condition and returns the true value if the condition is true, and the false value otherwise.
Real-World Application:
Ternary expressions can be used in any situation where you need to make a decision based on a condition. For example, you could use a ternary expression to determine the discount a customer receives based on their loyalty status:
discount = "gold" ? 10 : 5 # Gold members get a 10% discount, while other members get a 5% discounthouse_robber_iii
House Robber III
Problem: You are a thief who wants to rob a house. The house is composed of a tree of rooms. Each room contains some amount of money, and the root of the tree is the entrance to the house. You cannot rob two adjacent rooms because the alarm will be triggered. Find the maximum amount of money you can rob.
Example:
3
/ \
2 3
/ \ \
3 1 1The maximum amount of money you can rob is 7, by robbing the rooms with money 3, 1, and 3.
Solution:
We can use a bottom-up dynamic programming approach to solve this problem. Let dp[i][0] and dp[i][1] represent the following states for node i:
dp[i][0]: The maximum amount of money we can rob if we do not rob node i.
dp[i][1]: The maximum amount of money we can rob if we rob node i.
We can compute dp[i][0] and dp[i][1] recursively as follows:
def rob(root):
# Base case: If the root is None, there is no money to rob.
if not root:
return 0
# If the root has no children, the maximum amount of money we can rob is
# the amount of money in the root.
if not root.left and not root.right:
return root.val
# The maximum amount of money we can rob if we do not rob the root is
# the sum of the maximum amounts of money we can rob from the root's
# left and right subtrees.
dp[root][0] = rob(root.left)[0] + rob(root.right)[0]
# The maximum amount of money we can rob if we rob the root is the sum
# of the root's money and the maximum amounts of money we can rob from
# the root's grandchildren.
dp[root][1] = root.val + rob(root.left)[0] + rob(root.right)[0]
# Return the maximum amount of money we can rob.
return max(dp[root][0], dp[root][1])Time complexity: O(n), where n is the number of nodes in the tree.
Space complexity: O(n), since we store dp[i][0] and dp[i][1] for each node i.
Real-world applications:
This problem can be applied to a variety of real-world situations, such as:
Planning a heist: A group of thieves may want to plan a heist by robbing a series of houses. They could use the House Robber III algorithm to determine the maximum amount of money they can rob while avoiding detection.
Resource allocation: A company may want to allocate its resources to a series of projects. They could use the House Robber III algorithm to determine which projects to fund in order to maximize their return on investment.
Network optimization: A network engineer may want to configure a network to maximize its throughput. They could use the House Robber III algorithm to determine which links to enable and disable in order to achieve the maximum throughput.
nested_list_weight_sum_ii
Problem Statement:
Given a nested list of integers, calculate the sum of all integers while multiplying each integer by the depth of its nesting.
Example:
nested_list = [[1, 1], 2, [1, 1]]
Output: 10
Explanation:
1 * 1 + 1 * 1 = 2
2 * 1 = 2
1 * 2 + 1 * 2 = 6
Total sum = 2 + 2 + 6 = 10Solution:
We can use a recursive function to traverse the nested list and calculate the weighted sum. The function takes two arguments:
nested_list: The nested list of integers.depth: The current depth of nesting.
The function first checks if the current element is an integer. If it is, it returns the integer multiplied by the depth. If the current element is a list, the function recursively calls itself on each element of the list, incrementing the depth by 1.
The following is a simplified and commented Python implementation of the solution:
def nested_list_weight_sum(nested_list, depth=0):
"""
Calculates the sum of all integers in a nested list, multiplying each integer by the depth of its nesting.
Args:
nested_list: The nested list of integers.
depth: The current depth of nesting.
Returns:
The weighted sum of all integers in the nested list.
"""
# Initialize the sum to 0.
sum = 0
# Iterate over the elements in the nested list.
for element in nested_list:
# If the element is an integer, add it to the sum.
if isinstance(element, int):
sum += element * depth
# If the element is a list, recursively call the function to calculate its weighted sum.
elif isinstance(element, list):
sum += nested_list_weight_sum(element, depth + 1)
# Return the sum.
return sumExample Usage:
nested_list = [[1, 1], 2, [1, 1]]
result = nested_list_weight_sum(nested_list)
print(result) # Output: 10Applications:
This algorithm can be used in a variety of applications, such as:
Calculating the weighted average of a list of values.
Summarizing data with different levels of nesting.
Analyzing hierarchical structures.
verify_preorder_serialization_of_a_binary_tree
Problem Statement:
Given a string representing the preorder traversal of a binary tree, check if it is a valid preorder traversal sequence.
Intuition:
A valid preorder traversal sequence should start with the root and then recursively follow the left and right subtrees.
Solution:
We can use a stack to simulate the traversal and check if the sequence is valid.
Python Implementation:
def is_valid_preorder_traversal(preorder):
if not preorder:
return False
stack = []
for node in preorder:
if len(stack) == 0 or node < stack[-1]:
# Node is the left child
stack.append(node)
elif node > stack[-1]:
# Node is the right child
while stack and node > stack[-1]:
stack.pop()
stack.append(node)
else:
# Node is a duplicate
return False
return TrueExplanation:
Initialize an empty stack.
Iterate over the preorder sequence.
If the stack is empty or the current node is less than the top of the stack, push the current node onto the stack as it must be a left child.
If the current node is greater than the top of the stack, pop nodes from the stack until you find a node that is less than the current node (the parent of the current node). Then, push the current node onto the stack as it must be a right child.
If the current node is equal to the top of the stack, it means it's a duplicate, which is invalid. Return False.
After iterating through the entire sequence, if the stack is empty, it means the preorder sequence is valid. Return True.
Applications:
Validating binary tree data structures from serialized representations, such as preorder traversal sequences, to ensure their validity before further processing or operations.
remove_duplicates_from_sorted_array_ii
Problem Statement:
Given a sorted array, remove the duplicate elements from it such that each element appears at most twice.
Optimal Solution - Two Pointers Approach:
1. Breakdown:
Two-Pointers: We'll use two pointers, one to iterate through the array and one to mark the unique elements.
Skip Duplicates: If the current element is the same as the previous element, skip it.
Update Unique Elements: If the current element is different from the previous element, update the unique elements pointer and set the value at that index to the current element.
2. Python Code:
def remove_duplicates_from_sorted_array_ii(nums):
# Base case: Empty or single-element array
if len(nums) <= 1:
return nums
unique_idx = 0 # Index for unique elements
# Iterate through the array with the current pointer
for curr_idx in range(1, len(nums)):
if nums[curr_idx] != nums[unique_idx] or (nums[curr_idx] == nums[unique_idx] and curr_idx <= unique_idx + 1):
unique_idx += 1
nums[unique_idx] = nums[curr_idx]
# Return the subarray from the start to the unique elements pointer
return nums[:unique_idx + 1]3. Code Explanation:
We iterate through the array with
curr_idx, starting from the second element (index 1).We check if the current element is different from the previous unique element or if it's the same and we're still within the allowed two occurrences.
If either condition is met, we update the unique elements pointer and set the value at that index to the current element.
Finally, we return the subarray from the start of the array to the unique elements pointer plus one (to include the last unique element).
Real-World Applications:
Data cleaning and filtering: Removing duplicate values from datasets to ensure accuracy and reduce storage space.
User management systems: Ensuring uniqueness of usernames and email addresses while allowing for multiple occurrences within a certain limit (e.g., two different accounts with the same last name).
Inventory management: Tracking product availability while allowing for multiple units of the same product.
sentence_screen_fitting
Sentence Screen Fitting
Given a string sentence and an integer rows, where rows represents the number of rows in a screen, each row has a certain width, and a character in sentence takes up a fixed width, return a list of strings that represents the sentence fit into the screen after wrapping.
A word is wrapped onto a new row if it does not fit on the current row. Words are separated by a single space, and the next word is placed on the next row if it does not fit on the current row.
Example 1:
Input: sentence = "hello world", rows = 2
Output: ["hello", "world"]
Explanation:
hello world
|-------|
| |
|-------|Example 2:
Input: sentence = "leetcode", rows = 2
Output: ["leet", "code"]
Explanation:
leetcode
|-------|
| |
|-------|Example 3:
Input: sentence = "abcdefghij klmnopqrstuvwxyz", rows = 3
Output: ["abcdefghi", "jklmnopq", "rstuvwxyz"]Approach:
Check if the length of the sentence is less than or equal to
rows. If so, return the sentence as a single element list.Initialize an empty list to store the wrapped sentences.
Start from the beginning of the sentence and iterate through it character by character.
If the current character is a space, reset the width of the current row to 0.
Add the current character to the current row and increase the width of the current row by the width of the character.
If the width of the current row exceeds
rows, wrap the current row to the next line and reset the width of the current row to the width of the current character.Repeat steps 4-6 until the end of the sentence is reached.
Return the list of wrapped sentences.
Python Implementation:
def sentence_screen_fitting(sentence, rows):
# Check if the sentence fits on a single row.
if len(sentence) <= rows:
return [sentence]
# Initialize the list of wrapped sentences.
wrapped_sentences = []
# Initialize the current row and its width.
current_row = ""
current_width = 0
# Iterate through the sentence character by character.
for char in sentence:
# If the current character is a space, reset the width of the current row.
if char == " ":
current_width = 0
# Add the current character to the current row and increase its width.
else:
current_row += char
current_width += 1
# If the current row's width exceeds the maximum allowed width, wrap it to the next line.
if current_width > rows:
wrapped_sentences.append(current_row)
current_row = char
current_width = 1
# Add the last row to the list of wrapped sentences.
wrapped_sentences.append(current_row)
return wrapped_sentencesTime Complexity: O(N), where N is the length of the sentence.
Space Complexity: O(N), where N is the length of the sentence.
Applications:
This problem is a classic example of text wrapping, which is used in a variety of applications, such as:
Word processors
Text editors
Web browsers
Email clients
Mobile messaging apps
mini_parser
LeetCode Problem:
Two Sum
Given an array of integers nums and a target value target, find two numbers in the array that add up to the target. Return the indices of the two numbers as a list. You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: The numbers at indices 0 and 1 add up to 9: 2 + 7 = 9.Solution:
The following Python code provides a simple and efficient solution to this problem:
def two_sum(nums, target):
num_to_index = {} # Dictionary to store numbers to their indices
for i, num in enumerate(nums):
complement = target - num # Calculate the complement to the target
if complement in num_to_index: # Check if the complement exists in the dictionary
return [num_to_index[complement], i] # Return the indices of the two numbers
num_to_index[num] = i # Add the current number and its index to the dictionary
return [] # If no solution is found, return an empty listBreakdown:
Create a Dictionary: We create a dictionary called
num_to_indexto store numbers and their corresponding indices. This allows us to quickly check if a complement to the target exists later.Iterate Through the Numbers: We iterate through each number in the
numsarray.Calculate Complement: For each number, we calculate its complement to the target (e.g., if target is 9 and num is 2, the complement is 7).
Check for Complement in Dictionary: We check if the complement exists in the
num_to_indexdictionary. If it exists, we have found the two numbers that add up to the target and return their indices.Add Number to Dictionary: If the complement does not exist, we add the current number and its index to the dictionary for later reference.
Return Result: If no solution is found after iterating through all numbers, we return an empty list.
Real-World Applications:
Financial Analysis: Finding pairs of stocks or investments that complement each other to achieve a specific target return.
Logistics Management: Optimizing routes for delivery vehicles by pairing up packages that can be delivered together.
Machine Learning: Finding features that contribute to a specific outcome or classification.
Problem Overview
Title: Two Sum
Problem Statement: Given an array of integers nums and a target value, find two integers in nums that add up to the target. Return the indices of the two numbers.
Solution
Key Idea: Brute Force
One simple solution is to brute-force through all possible pairs of numbers:
def twoSum_brute_force(nums, target):
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
if nums[i] + nums[j] == target:
return [i, j]Complexity:
Time: O(n^2), where n is the length of the array.
Space: O(1).
Efficient Solution: Hash Table
A more efficient solution is to use a hash table to store values and indices:
def twoSum_hash_table(nums, target):
hash_table = {}
for i, num in enumerate(nums):
complement = target - num
if complement in hash_table:
return [hash_table[complement], i]
hash_table[num] = iComplexity:
Time: O(n), where n is the length of the array.
Space: O(n).
Real-World Applications:
Financial Modeling: Finding the optimal pair of investments that meet a specific financial goal.
Data Analysis: Identifying the two most influential factors that contribute to a certain outcome.
Inventory Management: Determining the optimal number of items from two different suppliers to meet a demand.
Code Example
nums = [2, 7, 11, 15]
target = 9
# Brute Force
indices = twoSum_brute_force(nums, target)
print(indices) # [0, 1]
# Hash Table
indices = twoSum_hash_table(nums, target)
print(indices) # [0, 1]compare_version_numbers
Problem Statement:
Compare two version numbers and return their relative order. Version numbers consist of one or more digits separated by periods (.).
Example 1:
Input: version1 = "1.01", version2 = "1.001"
Output: 0 (version1 and version2 are equal)Example 2:
Input: version1 = "1.0", version2 = "1.1"
Output: -1 (version1 is less than version2)Example 3:
Input: version1 = "1.0.1", version2 = "1"
Output: 1 (version1 is greater than version2)Solution:
Split the version numbers: Split both version numbers into lists of integers representing each digit.
Pad with zeros: Ensure both lists have the same length by padding with zeros.
Compare digits: Iterate through the lists and compare each digit. Return
0if they are equal,-1ifversion1is less thanversion2, or1ifversion1is greater thanversion2.
Python Implementation:
def compare_version_numbers(version1: str, version2: str) -> int:
# Split version numbers
v1 = version1.split(".")
v2 = version2.split(".")
# Pad with zeros
for v in (v1, v2):
while len(v) < len(v1 or v2):
v.append("0")
# Compare digits
for digit1, digit2 in zip(v1, v2):
if int(digit1) < int(digit2):
return -1
elif int(digit1) > int(digit2):
return 1
return 0Real-World Applications:
Comparing version numbers is useful in various scenarios:
Software updates: Determine if an installed software version is outdated compared to a newer version.
Package management: Handle dependencies between packages with different versions.
Version control: Compare commits and merge branches with different versions of code.
guess_number_higher_or_lower_ii
Implementation:
def guess_number_higher_or_lower_ii(n, k):
"""
:type n: int
:type k: int
:rtype: int
"""
# Initialize the range of possible answers
left, right = 1, n
# While the range is greater than 1, continue guessing
while left < right:
# Calculate the midpoint of the current range
mid = (left + right) // 2
# Guess the midpoint and get the feedback
guess = guess(mid)
# If the guess is higher, then the answer must be in the first half of the range
if guess == "Higher":
right = mid - 1
# If the guess is lower, then the answer must be in the second half of the range
elif guess == "Lower":
left = mid + 1
# If the guess is the answer, then return the answer
else:
return mid
# The answer is the only remaining number in the range
return leftExplanation:
The code uses a binary search approach to find the answer. It keeps narrowing down the range of possible answers based on the feedback from the "guess" function.
Initialize the range: Start with the range of possible answers from 1 to n.
Calculate the midpoint: Calculate the midpoint of the current range and store it in the variable
mid.Guess the midpoint: Use the "guess" function to guess the number at the midpoint.
Check the guess: The "guess" function returns one of three possible values:
"Higher": The guess is too low, so the answer must be in the first half of the range.
"Lower": The guess is too high, so the answer must be in the second half of the range.
"Correct": The guess is the answer.
Update the range: Based on the feedback from the "guess" function, update the range of possible answers:
If the guess is "Higher", set the new right bound of the range to
mid - 1.If the guess is "Lower", set the new left bound of the range to
mid + 1.If the guess is "Correct", return the value of
midas the answer.
Continue guessing: Repeat this process until the range of possible answers is narrowed down to a single number, which is the answer.
Applications:
This algorithm can be used in real-world applications where you need to guess a value based on a limited number of clues or feedback. Some examples include:
Guessing a password or PIN number
Determining the best dosage for a medication
Optimizing a machine learning model
non_overlapping_intervals
Problem Statement
Given a list of intervals where each interval is defined by an array [start, end], find the minimum number of intervals that overlap.
Brute Force Approach
The brute force approach is to compare each interval with every other interval and count the number of overlaps. This approach has a time complexity of O(n^2), where n is the number of intervals.
Optimized Approach
A more efficient approach is to sort the intervals by their starting points. This can be done in O(n log n) time using the sort() function in Python.
Once the intervals are sorted, we can iterate through them and count the number of overlaps. We do this by keeping track of the current interval and the next interval. If the current interval overlaps with the next interval, we increment the count by 1. If the current interval does not overlap with the next interval, we move on to the next interval.
This approach has a time complexity of O(n log n).
Code Implementation
def non_overlapping_intervals(intervals):
"""
Finds the minimum number of intervals that overlap.
Args:
intervals: A list of intervals where each interval is defined by an array [start, end].
Returns:
The minimum number of intervals that overlap.
"""
# Sort the intervals by their starting points.
intervals.sort(key=lambda x: x[0])
# Initialize the count of overlaps to 0.
overlaps = 0
# Iterate through the intervals.
for i in range(len(intervals) - 1):
# Get the current interval and the next interval.
current_interval = intervals[i]
next_interval = intervals[i + 1]
# Check if the current interval overlaps with the next interval.
if current_interval[1] > next_interval[0]:
# Increment the count of overlaps.
overlaps += 1
# Return the count of overlaps.
return overlapsReal-World Applications
This algorithm can be used in a variety of real-world applications, such as:
Scheduling: To find the minimum number of time slots needed to schedule a set of events.
Resource allocation: To find the minimum number of resources needed to allocate to a set of tasks.
Job sequencing: To find the minimum number of steps needed to complete a set of jobs.
longest_repeating_character_replacement
Problem Statement:
Given a string, find the length of the longest substring that can be replaced with a single character such that the entire substring becomes the same character.
Example:
Input: "AABABBA"
Output: 4 (We can replace the "AB" with 'A' or 'B' to get "AAAAA" or "BBBBB")
Solution:
The key to this problem is to use a sliding window approach:
Initialize two pointers:
left: Points to the start of the current substring that needs replacement.right: Points to the end of the current substring that needs replacement.
Keep expanding the window by moving
rightuntil:right - left + 1 > count(most frequent character) + 1: This means that the current substring has too many different characters and cannot be replaced with a single character.rightreaches the end of the string.
When reaching a limit:
If the current substring can be replaced, update the maximum length (
max_length).Move
leftto the position after the most frequent character's first occurrence inside the current substring.
Implementation in Python:
def longest_repeating_character_replacement(s: str) -> int:
if not s:
return 0
count = {} # Character count
max_length = 0
left = 0
right = 0
max_count = 0
# Count characters
for right in range(len(s)):
count[s[right]] = count.get(s[right], 0) + 1
max_count = max(max_count, count[s[right]])
# Slide window
while right < len(s):
if right - left + 1 > max_count + 1:
count[s[left]] -= 1
left += 1
while count[s[left]] == 0:
left += 1
else:
max_length = max(max_length, right - left + 1)
right += 1
return max_lengthExplanation:
Initialize
countto count character occurrences.Loop through the string using
right.While
right - left + 1is less than or equal tomax_count + 1, slide the window by updatingmax_lengthand movingleft.When the limit is reached, reduce the count of the leftmost character and move
left.Repeat until
rightreaches the end of the string.
Real-World Applications:
Text compression: Finding the longest substring that can be replaced with a single character can help reduce the size of text files.
Spell checking: Identifying the longest substring of misspelled characters can assist in correcting spelling errors.
Data scrubbing: Removing invalid or duplicate data from datasets by replacing lengthy substrings with a single value.
minimum_size_subarray_sum
ERROR OCCURED minimum_size_subarray_sum
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.
500 Internal error encountered.
random_pick_index
Problem Statement:
You have an array of numbers and want to pick a random element from it.
Best Solution (Reservoir Sampling):
Reservoir sampling is an algorithm that allows you to select a random element from a stream of data. It works by iterating over the array and maintaining a "reservoir" of size 1, which contains the current random element. At each iteration, with a probability of 1/i, where i is the current index, we replace the reservoir with the current element.
Python Code:
import random
def random_pick_index(nums):
reservoir = None
for i, num in enumerate(nums):
if random.random() < 1/i:
reservoir = num
return reservoirBreakdown:
Step 1: Initialize the reservoir. The reservoir is initially set to None.
Step 2: Iterate over the array. For each element in the array:
Step 2.1: Generate a random number. This number is used to determine whether to replace the reservoir.
Step 2.2: Check if the random number is less than 1/i. If it is, replace the reservoir with the current element.
Step 3: Return the reservoir. The reservoir contains the randomly selected element.
Real-World Examples:
Selecting a random item from a list of words for a word game.
Choosing a random customer for a prize giveaway.
Generating a random sample of data for analysis.
Applications:
Data analysis: Reservoir sampling can be used to select a random subset of data for analysis, reducing the computational cost and memory requirements.
Machine learning: Random sampling is used in algorithms such as random forests and AdaBoost to create diverse training sets.
Computer simulations: Reservoir sampling can be used to model rare events or generate random scenarios.
reconstruct_original_digits_from_english
Problem Statement:
You are given a string s that contains a sequence of digits represented as English words. The English words are written in lowercase and consist of the following:
"zero"
"one"
"two"
"three"
"four"
"five"
"six"
"seven"
"eight"
"nine"
Your task is to reconstruct the original string of digits from the given sequence of English words.
Example:
Input: s = "owoztneoer"
Output: "012"Solution:
This problem can be solved using a dictionary to map the English words to their corresponding digits. We can then iterate through the given string s, split it into individual words, and look up each word in the dictionary to reconstruct the original string of digits.
Implementation:
def reconstruct_original_digits_from_english(s):
# Create a dictionary to map English words to digits.
word_to_digit = {
"zero": "0",
"one": "1",
"two": "2",
"three": "3",
"four": "4",
"five": "5",
"six": "6",
"seven": "7",
"eight": "8",
"nine": "9",
}
# Split the given string into individual words.
words = s.split()
# Create an empty string to store the reconstructed digits.
digits = ""
# Iterate through the words and look up each word in the dictionary to reconstruct the original string of digits.
for word in words:
if word in word_to_digit:
digits += word_to_digit[word]
# Return the reconstructed string of digits.
return digitsExplanation:
The solution works as follows:
We create a dictionary
word_to_digitto map the English words to their corresponding digits.We split the given string
sinto individual words using thesplit()method.We create an empty string
digitsto store the reconstructed digits.We iterate through the words in the list
wordsusing aforloop.For each word, we check if it exists in the dictionary
word_to_digit.If the word exists in the dictionary, we append the corresponding digit to the string
digits.After iterating through all the words, we return the reconstructed string
digits.
Real-World Applications:
This problem has applications in speech reconocimiento and natural language processing. For example, we could use this solution to reconstruct the digits spoken by a user when they interact with a voice-controlled device.
Potential Applications:
Speech recognition
Natural language processing
Voice-controlled devices
clone_graph
Problem Statement:
Given a graph, clone it such that any changes made to the cloned graph are not reflected in the original graph.
Approach:
To clone a graph, we need to create a copy of each node and connect the copies in the same way as the original graph. We can use a hash table to keep track of the cloned nodes and the original nodes they represent.
Traverse the original graph and create copies of each node. Store the cloned nodes in a hash table, where the key is the original node and the value is the cloned node.
For each node in the original graph, traverse its neighbors. For each neighbor, look up the cloned node in the hash table. If the cloned node does not exist, create a new copy and add it to the hash table.
Connect the cloned nodes in the same way as the original graph.
Python Implementation:
class Node:
def __init__(self, val):
self.val = val
self.neighbors = []
def clone_graph(graph):
# Create a hash table to store the cloned nodes
cloned_nodes = {}
# Traverse the original graph and create copies of each node
for node in graph:
cloned_nodes[node] = Node(node.val)
# For each node in the original graph, traverse its neighbors
for node in graph:
for neighbor in node.neighbors:
# Look up the cloned node in the hash table
cloned_neighbor = cloned_nodes[neighbor]
# If the cloned node does not exist, create a new copy and add it to the hash table
if cloned_neighbor not in cloned_nodes:
cloned_neighbor = Node(neighbor.val)
cloned_nodes[neighbor] = cloned_neighbor
# Connect the cloned nodes in the same way as the original graph
cloned_nodes[node].neighbors.append(cloned_neighbor)
# Return the cloned graph
return cloned_nodesExample:
Suppose we have the following graph:
1 -> 2
| |
v v
3 -> 4The cloned graph would look like this:
1' -> 2'
| |
v v
3' -> 4'Potential Applications in Real World:
Cloning graphs is useful in various applications, such as:
Social networks: Cloning a social network graph allows users to share their data with multiple platforms without revealing their original data.
Databases: Cloning a database graph helps in disaster recovery, as the cloned graph can be used to restore the original database in case of a failure.
Graphs in memory: Cloning graphs in memory optimizes memory usage by sharing common data between different clones of the same graph.
remove_k_digits
Problem Statement: Given a non-negative integer num represented as a string, and an integer k, remove k digits from the number to get the smallest possible number.
Example: Input: num = "1432219", k = 3 Output: "1219"
Solution:
To solve this problem, we can leverage the use of a stack. The steps involved are as follows:
Initialize: Create an empty stack.
Iterate: For each digit in the number:
If the stack is empty or the digit is greater than or equal to the top of the stack, push the digit onto the stack.
Otherwise, pop digits from the stack until the top of the stack is greater than or equal to the current digit or the stack is empty.
Remove extra: If the stack has more than (n - k) digits, pop the remaining digits.
Convert: Convert the digits in the stack to a string to get the smallest possible number.
Python Code:
def remove_k_digits(num, k):
stack = []
# Iterate through the number
for digit in num:
# Remove digits until the stack has less than n-k digits
while stack and stack[-1] > digit and k > 0:
stack.pop()
k -= 1
# Push the digit onto the stack
stack.append(digit)
# Remove extra digits
while stack and len(stack) > len(num) - k:
stack.pop()
# Convert the digits in the stack to a string
return ''.join(stack) or '0'Real-World Applications:
This algorithm can be applied in various real-world scenarios:
Data Validation: Ensuring that input numbers meet certain criteria by removing invalid digits.
Compression: Reducing the size of large numbers by removing unnecessary digits.
Error Correction: Correcting errors in numeric data by removing inconsistencies.
Numerical Analysis: Simplifying complex calculations by removing insignificant digits.
battleships_in_a_board
Battleships in a Board
Problem Statement:
Given an m x n board, where each cell can be either empty or contains a battleship, return the number of battleships in the board.
Battleships can only be placed horizontally or vertically, and there should not be any overlapping battleships.
Examples:
Input:
board = [[1,1,1],[0,0,0],[0,0,0]]Output: 1
Input:
board = [[1,0,1],[0,0,0],[0,0,0]]Output: 0
Solution:
We can use a two-pass approach to count the number of battleships:
First pass: Iterate over each cell and check if it contains a battleship. If it does, check the cells to the left and above it to see if they also contain battleships. If any of them do, it means they are part of the same battleship, so we increment the count.
Second pass: Iterate over each cell again. If it contains a battleship and has not been counted in the first pass, increment the count.
Here is the Python code for the solution:
def count_battleships(board):
"""Counts the number of battleships in a board.
Params:
board: A 2D list representing the board.
Returns:
The number of battleships in the board.
"""
# Initialize the count to 0.
count = 0
# Iterate over each cell in the board.
for i in range(len(board)):
for j in range(len(board[0])):
# Check if the cell contains a battleship.
if board[i][j] == 1:
# Check if the cell is part of a battleship that has already been counted.
if (i > 0 and board[i - 1][j] == 1) or (j > 0 and board[i][j - 1] == 1):
continue
# Increment the count.
count += 1
# Return the count.
return countTime Complexity: O(mn), where m and n are the number of rows and columns in the board.
Space Complexity: O(1), as we do not use any additional space.
Applications in the Real World:
This problem can be applied to various real-world scenarios, such as:
Naval warfare: Identifying the number of enemy battleships on a radar screen.
Board games: Determining the number of ships remaining in a game of Battleship.
Image processing: Detecting objects of a specific shape, such as ships, in an image.
rotate_list
Linked List Rotation
A linked list is data structure composed of a group of nodes which together represent a sequence. Under the simplest definition, each node is composed of a containing element (or data) and a reference (in other words, a link) to the next node in the list.
This structure allows for efficient insertion or removal of elements from any position in the sequence during iteration. More complex variants add additional links, allowing more efficient insertion or removal of nodes at arbitrary positions. A doubly linked list is a generalization of the singly linked list, where each node also contains a reference to the previous node in the sequence. In a doubly linked list, the nodes can be traversed in either direction.
Rotating a linked list is the task of taking a linked list nodes and moving the last node so that it becomes the first node, and the former first node becomes the second node. This process is continued until the original first node becomes the last node.
Time Complexity: O(n) where n is the number of nodes in the linked list. Space Complexity: O(1). Applications:
In a queuing system, rotate the list to efficiently serve the next request.
Real World Python Implementation
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def rotate_list(head, k):
# If the list is empty or k is 0, return the head
if not head or k == 0:
return head
# Find the length of the list
length = 0
curr = head
while curr:
length += 1
curr = curr.next
# Calculate the new position of the first node
k %= length
if k == 0:
return head
# Find the new first node
for _ in range(k):
head = head.next
# Set the new last node
last = head
while last.next:
last = last.next
# Connect the last node to the original first node
last.next = head
# Set the new first node as the head
head = head.next
# Set the original first node's next to None
head.next = None
return headExplanation:
Check if the linked list is empty or if k is 0. If either of these conditions is true, return the head of the list.
Find the length of the linked list by traversing the list and incrementing a counter.
Calculate the new position of the first node. This is done by taking the modulus of k and the length of the list.
Find the new first node by traversing the list k times.
Set the new last node to be the node before the new first node.
Connect the last node to the original first node.
Set the new first node as the head of the list.
Set the original first node's next to None.
Return the head of the list.
remove_duplicate_letters
Problem: Given a string, remove the duplicate letters. Return the smallest lexicographical order string that can be made with the remaining letters.
Solution:
We will use a stack to store the letters. We will iterate over the string, and for each letter, we will check if it is already in the stack. If it is, we will skip it. If it is not, we will add it to the stack.
However, before adding a letter to the stack, we will check if the top of the stack is greater than the current letter. If it is, we will pop the top of the stack and add the current letter to the stack.
This process ensures that the letters in the stack are in lexicographical order.
Code:
def remove_duplicate_letters(s):
stack = []
visited = set()
for letter in s:
if letter not in visited:
while stack and letter < stack[-1]:
if stack[-1] in s[s.index(letter):]:
stack.pop()
else:
break
stack.append(letter)
visited.add(letter)
return ''.join(stack)Explanation:
The code iterates over the string and adds each letter to the stack if it is not already in the stack. However, before adding a letter to the stack, it checks if the top of the stack is greater than the current letter. If it is, it pops the top of the stack and adds the current letter to the stack. This process ensures that the letters in the stack are in lexicographical order.
Example:
remove_duplicate_letters("bcabc") == "abc"
remove_duplicate_letters("cbacdcbc") == "acdb"Applications:
This algorithm can be used to find the smallest lexicographical order string that can be made from a given string. This can be useful in applications such as:
Text compression
Data deduplication
String matching
bulls_and_cows
Bulls and Cows
Introduction:
"Bulls and Cows" is a code-breaking game where you try to guess a hidden number. The game provides you with two types of feedback: "Bulls" which indicate the number of correct digits in the right position, and "Cows" which indicate the number of correct digits in the wrong position.
Implementation:
Step 1: Convert both the guess and the target number to strings.
guess = str(input("Guess a 4-digit number: "))
target = str(input("Enter the target 4-digit number: "))Step 2: Initialize counters for Bulls and Cows.
bulls = 0
cows = 0Step 3: Iterate through each digit of the guess and the target number.
for i in range(4):
if guess[i] == target[i]:
bulls += 1
elif guess[i] in target:
cows += 1Step 4: Print the feedback to the user.
print(f"Bulls: {bulls}, Cows: {cows}")Explanation:
Step 3:
We iterate through each digit of the guess and the target using a
forloop.If a digit in the guess matches a digit in the target at the same position, we increment the "Bulls" counter.
If a digit in the guess matches a digit in the target, but not at the same position, we increment the "Cows" counter.
Example:
Guess: 1234 Target: 4567
Bulls: 1 (the digit '4' is in the correct position) Cows: 2 (the digits '1' and '2' are in the target, but not in the correct positions)
Real World Applications:
Password cracking (trying different combinations of characters to guess a password)
Code-breaking algorithms (decoding encrypted messages)
Sudoku solving (checking for duplicate numbers in rows, columns, and boxes)
largest_bst_subtree
LeetCode Problem:
Largest BST Subtree
Given the root node of a binary tree, find the largest subtree that is a Binary Search Tree (BST). A BST is a binary tree where the left subtree of each node is smaller than the node, and the right subtree of each node is greater than the node.
Solution:
We can define a class to represent each node in the binary tree, as follows:
class Node:
def __init__(self, val, left=None, right=None):
self.val = val
self.left = left
self.right = rightWe can then define a function to check if a given binary tree is a BST. This function will recursively check the left and right subtrees, and return True if both subtrees are BSTs and the current node's value is within the range of values allowed by the BST. Otherwise, it will return False.
def is_bst(root):
if not root:
return True
if root.left and root.val <= root.left.val:
return False
if root.right and root.val >= root.right.val:
return False
return is_bst(root.left) and is_bst(root.right)Once we have a function to check if a binary tree is a BST, we can define a function to find the largest BST subtree in a given binary tree. This function will recursively check the left and right subtrees, and return the largest BST subtree found. If the current node is not part of a BST, it will return None.
def largest_bst_subtree(root):
if not root:
return None
if is_bst(root):
return root
left_subtree = largest_bst_subtree(root.left)
right_subtree = largest_bst_subtree(root.right)
if left_subtree is None:
return right_subtree
if right_subtree is None:
return left_subtree
if left_subtree.size > right_subtree.size:
return left_subtree
else:
return right_subtreeThe size attribute is not included in the definition of the Node class, but it is added in the implementation of the largest_bst_subtree function to keep track of the size of each BST subtree. The size of a BST subtree is defined as the number of nodes in the subtree.
Real-world applications:
This algorithm can be used to find the largest contiguous region of a binary tree that satisfies a certain property. For example, it can be used to find the largest contiguous region of a binary tree that is a BST, or the largest contiguous region of a binary tree that contains a given value.
Example:
Consider the following binary tree:
5
/ \
2 7
/ \
1 3The largest BST subtree is the subtree rooted at node 2, which is highlighted in the following diagram:
5
/ \
2 7
/ \
1 3The size of this subtree is 4, and it is a valid BST.
paint_fence
Problem:
You have a fence with n posts, each post can be painted red, white, or blue. You are given an array costs where costs[i] is the cost to paint the ith post.
Find the minimum cost to paint the fence such that no two adjacent posts have the same color.
Example:
Input: n = 3, costs = [1,2,3]
Output: 4
Explanation: Paint post 1 red, post 2 white, and post 3 blue. The total cost is 1 + 2 + 1 = 4.Solution:
To solve this problem, we can use dynamic programming. We define a 2D array dp where dp[i][j] represents the minimum cost to paint the first i posts, with the last post painted color j.
We have the following recurrence relation:
dp[i][0] = min(dp[i-1][1], dp[i-1][2]) + costs[i][0]
dp[i][1] = min(dp[i-1][0], dp[i-1][2]) + costs[i][1]
dp[i][2] = min(dp[i-1][0], dp[i-1][1]) + costs[i][2]We initialize dp[0][j] to 0 for all j, since there is no cost to paint 0 posts.
Simplified Explanation:
Imagine you have a fence with 3 posts. You can paint each post red, white, or blue. You want to paint the fence in a way that no two adjacent posts have the same color.
The minimum cost to paint the first post is simply the cost of the color you choose.
The minimum cost to paint the first two posts is the minimum of the following two options:
Paint the first post red and the second post white or blue.
Paint the first post white or blue and the second post red.
The minimum cost to paint the first three posts is the minimum of the following three options:
Paint the first post red and the second and third posts white or blue.
Paint the first post white or blue and the second and third posts red.
Paint the first post white or blue and the second post red and the third post blue or red.
Real-World Application:
This problem can be applied in any situation where you need to make a decision that affects multiple adjacent elements. For example, you could use this approach to minimize the cost of painting a fence, scheduling tasks, or allocating resources.
Code:
def paint_fence(n, costs):
dp = [[0] * 3 for _ in range(n+1)]
for i in range(1, n+1):
dp[i][0] = min(dp[i-1][1], dp[i-1][2]) + costs[i-1][0]
dp[i][1] = min(dp[i-1][0], dp[i-1][2]) + costs[i-1][1]
dp[i][2] = min(dp[i-1][0], dp[i-1][1]) + costs[i-1][2]
return min(dp[n])3sum_closest
Problem Statement:
Given an integer array nums and a target value, find three integers in nums whose sum is closest to the target.
Best Solution:
1. Sort the Array:
First, sort the array nums in ascending order. This allows us to efficiently find the closest sum by iterating through the sorted array.
2. Iterate Through the Sorted Array:
For each element nums[i], iterate through the rest of the array nums[j] and nums[k] (where i < j < k) to find the closest sum.
3. Calculate the Sum and Update the Closest Value:
For each combination of nums[i], nums[j], and nums[k], calculate the sum and compare it to the current closest value. If the sum is closer to the target, update the closest value.
4. Return the Closest Value:
After iterating through all possible combinations, return the closest value found.
Code Implementation:
def threeSumClosest(nums, target):
# Sort the array
nums.sort()
# Initialize the closest sum to infinity
closest_sum = float('inf')
# Iterate through the sorted array
for i in range(len(nums) - 2):
# Set the left and right pointers
left = i + 1
right = len(nums) - 1
# Iterate through the remaining elements
while left < right:
# Calculate the current sum
current_sum = nums[i] + nums[left] + nums[right]
# Update the closest sum if necessary
if abs(current_sum - target) < abs(closest_sum - target):
closest_sum = current_sum
# Adjust the left and right pointers
if current_sum < target:
left += 1
elif current_sum > target:
right -= 1
# Return the closest sum
return closest_sumExample:
nums = [-1, 2, 1, -4]
target = 1
result = threeSumClosest(nums, target)
print(result) # Output: 2Real-World Applications:
Portfolio Optimization: Finding the optimal allocation of assets to maximize returns while minimizing risk.
Chemical Mixtures: Determining the optimal combination of chemicals to achieve a specific reaction.
Medicine: Identifying drug combinations that are most effective in treating diseases.
design_phone_directory
Problem:
Design a phone directory that has the following functions:
get: Given a phone number, return the contact name.put: Given a phone number and contact name, store the contact in the directory.
Solution:
We can use a hash table to implement the phone directory. The hash table will map phone numbers to contact names.
Implementation:
class PhoneDirectory:
def __init__(self):
self.directory = {}
def get(self, phone_number):
return self.directory.get(phone_number)
def put(self, phone_number, contact_name):
self.directory[phone_number] = contact_nameExample:
phone_directory = PhoneDirectory()
phone_directory.put("555-555-1212", "John Smith")
phone_directory.put("555-555-1313", "Jane Doe")
contact_name = phone_directory.get("555-555-1212") # Returns "John Smith"Benefits:
The hash table allows us to store and retrieve contacts in constant time.
The get() and put() operations are simple to implement.
The phone directory is easy to scale to accommodate a large number of contacts.
Real-World Applications:
Contact management systems
Phonebook applications
Address books
reverse_words_in_a_string
Problem Statement: Reverse the words in a given string. Preserve the order of words and the original spacing between them.
Solution 1: Using the Built-in 'split' and 'join' Functions:
Breakdown:
Split the input string into a list of words using the
str.split()method.Reverse the order of the words in the list.
Join the reversed words back into a string using the
str.join()method.
Code Implementation:
def reverse_words(s: str) -> str:
words = s.split() # Split the string into words
words.reverse() # Reverse the order of words
return " ".join(words) # Join the reversed words back into a stringExample:
input_string = "Hello World"
reversed_string = reverse_words(input_string)
print(reversed_string) # Output: "World Hello"Solution 2: Using a Stack:
Breakdown:
Create an empty stack.
Iterate through the input string character by character.
If the character is a space, pop words from the stack and append them to the output string.
If the character is not a space, push it to the stack.
After iterating through the entire string, pop any remaining words from the stack and append them to the output string.
Code Implementation:
def reverse_words_stack(s: str) -> str:
stack = []
output = []
for char in s:
if char == " ":
while stack:
output.append(stack.pop())
output.append(" ")
else:
stack.append(char)
while stack:
output.append(stack.pop())
return "".join(output)Example:
input_string = "Hello World"
reversed_string = reverse_words_stack(input_string)
print(reversed_string) # Output: "World Hello"Applications in Real World:
Formatting text for display on websites or in documents.
Reversing the order of words in user-inputted text for error correction or search functionality.
Processing natural language text for machine learning or data analysis tasks.
peeking_iterator
Problem: Implement an iterator that allows peeking at the next element in the sequence without advancing the iterator.
Solution:
class PeekingIterator:
def __init__(self, iterator):
self.iterator = iterator
self.peeked = None
def peek(self):
if not self.peeked:
self.peeked = next(self.iterator, None)
return self.peeked
def next(self):
if self.peeked is not None:
current = self.peeked
self.peeked = None
return current
else:
return next(self.iterator)
def hasNext(self):
return self.peeked is not None or next(self.iterator, None) is not NoneExplanation:
The
PeekingIteratorclass takes an iterator as input and provides additional methods,peekandhasNext, that allow peeking at the next element and checking if there are more elements.The
peekmethod looks at the next element from the underlying iterator if it hasn't already been peeked. It returns the element orNoneif there are no more elements.The
nextmethod moves the iterator forward and returns the next element. If there is a peeked element, it returns that element first, otherwise it fetches the next element from the underlying iterator.The
hasNextmethod checks if there are more elements by examining the peeked element or fetching the next element from the underlying iterator.
Real-World Applications:
Caching: Peek ahead to prefetch data that may be needed later, preventing performance bottlenecks.
Data stream processing: Analyze and process upcoming data without advancing the stream.
Input validation: Validate user input without consuming it from the stream.
binary_search_tree_iterator
Binary Search Tree Iterator
Problem Statement: Given the root of a binary search tree (BST), design an iterator to traverse the tree in-order. Implement the following operations:
next(): Returns the next smallest element in the BST.hasNext(): Returns true if there are more elements in the BST to iterate over.
Best & Performant Solution in Python:
class BSTIterator:
def __init__(self, root):
self.stack = []
self.push_all(root)
def push_all(self, node):
while node:
self.stack.append(node)
node = node.left
def next(self):
node = self.stack.pop()
self.push_all(node.right)
return node.val
def hasNext(self):
return len(self.stack) > 0Breakdown and Explanation:
1. Initialize the Stack: We start by initializing an empty stack. The stack will be used to store nodes that we need to visit.
2. Push All Left Nodes: For the root node, we repeatedly push its left child onto the stack until we reach the leftmost node. This ensures that we visit the left subtree first.
3. Next Element: To get the next smallest element, we pop the top node from the stack. Since we pushed all left nodes first, this node is the smallest element in the current subtree.
4. Push All Right Nodes: After popping the current node, we push all its right children onto the stack. This ensures that we continue our in-order traversal.
5. Has Next Element: We check if the stack is empty. If it's empty, we know there are no more elements to iterate over.
Real-World Application:
This iterator can be used in various real-world scenarios where you need to process data in a sorted order. For example:
Database Query Optimization: Iterating over a large sorted database table in order to efficiently retrieve data.
Data Analysis: Processing sorted data sets to find the median or other statistics.
File Processing: Reading a sorted file line by line and performing operations on the data.
reverse_words_in_a_string_ii
Problem:
Given a string, reverse the words in it.
Example:
"Hello World" -> "World Hello"
Solution:
Split the string into words:
Use the
split()method to split the string into individual words.
Reverse the list of words:
Use the
reversed()function to reverse the order of the words in the list.
Join the words back into a string:
Use the
join()method to join the words in the reversed list back into a string.
Python Code:
def reverseWords(s):
# Split the string into words.
words = s.split()
# Reverse the list of words.
reversed_words = reversed(words)
# Join the words back into a string.
return " ".join(reversed_words)Time Complexity: O(n), where n is the length of the string.
Space Complexity: O(n), since we create a copy of the string as a list.
Applications:
Text processing: Reversing words can be useful for tasks such as parsing sentences or finding palindromes.
Data cleaning: If data is stored in a string with words separated by spaces, reversing the words can help identify errors or inconsistencies.
Encryption: Reversing words can be used as a simple encryption technique, making the data harder to read by unauthorized users.
find_all_duplicates_in_an_array
Leetcode Problem: Find All Duplicates in an Array
Problem Statement:
Given an integer array nums containing n distinct numbers in the range [1, n], return all the duplicated numbers in the array.
Approach:
The key idea behind this problem is to utilize the array itself as a hash table. We can mark the presence of each element by negating its corresponding index in the array. If we encounter a negative index while iterating through the array, it means that the element has already been marked, indicating a duplicate.
Python Implementation:
def find_duplicates(nums):
duplicates = []
for num in nums:
index = abs(num) - 1
if nums[index] < 0:
duplicates.append(abs(num))
else:
nums[index] = -nums[index]
return duplicatesBreakdown:
Mark Elements: We iterate through the array and negate the index of each element. This marks the presence of the element.
Detect Duplicates: As we continue iterating, if we encounter a negative index, it means that the element has already been marked, indicating a duplicate.
Collect Duplicates: We add the absolute value of the negative index to the
duplicateslist.Return Result: Finally, we return the list of duplicates.
Applications in Real World:
This approach can be used in various real-world scenarios:
Fraud Detection: Identifying duplicate transactions in a financial system.
Data Deduplication: Removing duplicate records from a database or file system.
Inventory Management: Detecting duplicate items in a warehouse or supply chain.
Error Handling: Flagging duplicate entries in a form or application.
Example:
Input: [4,3,2,7,8,2,3,1] Output: [2,3]
Explanation: Elements 2 and 3 appear twice in the array.
add_two_numbers_ii
Problem Statement: Given two non-empty linked lists representing two non-negative integers, where the digits are stored in reverse order, and each of their nodes contain a single digit. Add the two numbers and return the sum as a linked list.
Approach:
Reverse the linked lists: To add numbers in the reverse order, we need to reverse the linked lists first.
Add digits from the least significant bit (LSB): Start from the first nodes of both lists. Add their values and store the result in a new linked list. If the sum is greater than 9, carry over the extra 1 to the next addition.
Traverse both lists simultaneously: While both lists have nodes, continue adding their digits and keeping track of the carry. If one list ends before the other, add the remaining nodes from the longer list.
Handle carry: If there's a carry left after adding all digits, create a new node for it.
Return the new linked list: The new linked list represents the sum of the two numbers.
Simplified Explanation:
Imagine you have two numbers, 123 and 456, written on paper. To add them, you start from the rightmost digits (3 and 6) and add them up, getting 9. If the sum of two digits exceeds 9, write down the digit in the ones place and carry over the tens place (1). Repeat the process for the next pair of digits (2 and 5), adding them up and considering any carry. Continue this process until you reach the beginning of both numbers. If one number has more digits than the other, simply add the remaining digits of the longer number.
Code Implementation:
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def addTwoNumbers(l1, l2):
carry = 0
dummy = ListNode()
curr = dummy
while l1 or l2 or carry:
sum = 0
if l1:
sum += l1.val
l1 = l1.next
if l2:
sum += l2.val
l2 = l2.next
sum += carry
carry = sum // 10
curr.next = ListNode(sum % 10)
curr = curr.next
return dummy.nextExample:
# Input: l1 = [2,4,3], l2 = [5,6,4]
# Output: [7,0,8]
l1 = ListNode(2)
l1.next = ListNode(4)
l1.next.next = ListNode(3)
l2 = ListNode(5)
l2.next = ListNode(6)
l2.next.next = ListNode(4)
result = addTwoNumbers(l1, l2)
# Print the result linked list
while result:
print(result.val, end=" ")
result = result.nextOutput:
7 0 8Applications in Real World:
Adding numbers is a fundamental operation used in various applications, including:
Financial calculations (e.g., adding up account balances)
Physics (e.g., calculating distances or velocities)
Computer science (e.g., adding up the number of elements in a list)
additive_number
Problem Statement:
Given a string representing a number, determine if it is an additive number. An additive number is a string where the first two numbers sum up to the third number, and the third and fourth numbers sum up to the fifth number, and so on.
Example 1:
Input: "112358" Output: True Explanation: The first two numbers are 1 and 1, which sum up to 2. The third and fourth numbers are 2 and 3, which sum up to 5. The fifth and sixth numbers are 5 and 8, which sum up to 13. So, the string is an additive number.
Example 2:
Input: "199100199" Output: True Explanation: The first two numbers are 1 and 99, which sum up to 100. The third and fourth numbers are 100 and 1, which sum up to 101. The fifth and sixth numbers are 101 and 99, which sum up to 200. So, the string is an additive number.
Approach:
We can use a recursive function to check if a string is an additive number. The function takes as input the string and the indices of the first and last numbers in the string. It then checks if the sum of the first two numbers is equal to the third number. If it is, the function calls itself recursively with the indices of the second and third numbers. The function returns True if the string is an additive number and False otherwise.
Python Implementation:
def is_additive_number(string):
# Check if the string is empty or has only one character.
if not string or len(string) == 1:
return False
# Iterate over all possible lengths of the first and second numbers.
for first_length in range(1, len(string)):
for second_length in range(1, len(string) - first_length):
# Get the first, second, and third numbers.
first_number = string[:first_length]
second_number = string[first_length:first_length + second_length]
third_number = string[first_length + second_length:]
# Check if the first and second numbers are valid.
if not first_number.isdigit() or not second_number.isdigit():
continue
# Check if the sum of the first and second numbers is equal to the third number.
if int(first_number) + int(second_number) == int(third_number):
# Recursively check if the rest of the string is an additive number.
if is_additive_number(third_number):
return True
# If no combination of first and second numbers works, the string is not an additive number.
return FalseTime Complexity:
The time complexity of the solution is O(n^3), where n is the length of the string. The function calls itself recursively at most n times. For each recursive call, it iterates over all possible lengths of the first and second numbers. There are at most n possible lengths for the first number and at most n possible lengths for the second number. Therefore, the total number of recursive calls is at most n^3.
Space Complexity:
The space complexity of the solution is O(n), where n is the length of the string. The function uses a stack to store the recursive calls. The maximum size of the stack is n, which occurs when the string is an additive number.
count_complete_tree_nodes
Problem Statement:
Given a complete binary tree, count the number of nodes in it.
Example 1:
Input: [1,2,3,4,5,6]
Output: 6Example 2:
Input: []
Output: 0Explanation:
A complete binary tree is a binary tree in which every level is completely filled except for possibly the last level. The last level of the tree has all its nodes as far left as possible.
Optimal Solution:
The following is the optimal solution to the problem:
Python Implementation:
def count_complete_tree_nodes(root):
if not root:
return 0
height = 0
node = root
while node:
node = node.left
height += 1
# Calculate the number of nodes in a full binary tree of height 'height'
full_nodes = (1 << height) - 1
def count_nodes(node, level):
if not node:
return 0
if level == height:
return 1
# Check if the left and right subtrees are full
left_full = count_nodes(node.left, level + 1) == (1 << (height - level - 1)) - 1
right_full = count_nodes(node.right, level + 1) == (1 << (height - level - 1)) - 1
# If both subtrees are full, then the current node is also full
if left_full and right_full:
return (1 << (height - level)) - 1
# Otherwise, the current node is not full, and we need to count the nodes in its left and right subtrees
return count_nodes(node.left, level + 1) + count_nodes(node.right, level + 1)
return full_nodes + count_nodes(root, 1)Breakdown:
Step 1: Calculate the height of the tree. This is done by iteratively traversing down the left subtree of the root node. The height is the number of nodes on the longest path from the root to a leaf node.
Step 2: Calculate the number of nodes in a full binary tree of height 'height'. This is done by subtracting 1 from 2 raised to the power of height.
Step 3: Recursively count the nodes in the tree. This is done by calling the
count_nodesfunction on the root node with a level of 1. Thecount_nodesfunction returns the number of nodes in the subtree rooted at the given node.Step 4: Add the number of nodes in the full binary tree to the number of nodes counted in step 3. This gives the total number of nodes in the complete binary tree.
Explanation:
The optimal solution is based on the fact that a complete binary tree can be divided into a full binary tree and a partial binary tree. The full binary tree is the largest possible binary tree that can be formed with the given number of nodes. The partial binary tree is the remaining tree that is not full.
The algorithm first calculates the height of the tree and the number of nodes in a full binary tree of that height. Then, it recursively counts the nodes in the partial binary tree. Finally, it adds the number of nodes in the full binary tree to the number of nodes counted in the partial binary tree to get the total number of nodes in the complete binary tree.
Real-World Applications:
Counting the number of nodes in a complete binary tree has applications in various areas, such as:
Database optimization: Complete binary trees are often used as indexes in databases to improve search performance. The number of nodes in the tree can be used to estimate the number of records in the database.
Memory management: Complete binary trees are used in memory management to allocate and deallocate memory efficiently. The number of nodes in the tree can be used to track the amount of memory that is allocated.
Computer graphics: Complete binary trees are used in computer graphics to create hierarchical data structures. The number of nodes in the tree can be used to estimate the complexity of the data structure.
unique_binary_search_trees
Unique Binary Search Trees
Problem Statement: Given an integer n, return the number of structurally unique binary search trees that can be constructed with nodes values 1, 2, ..., n.
Approach: The key idea is to use dynamic programming to count the number of unique binary search trees for each possible root value.
Steps:
Base Case: For n = 0, there is only one unique binary search tree (an empty tree). For n = 1, there is also only one unique binary search tree (a tree with one node).
Recursive Formula: For n > 1, we can construct a unique binary search tree by choosing a root value r and recursively constructing the left and right subtrees. The total number of unique binary search trees is the sum of the products of the number of unique binary search trees for the left and right subtrees.
F(n) = Σ(F(i-1) * F(n-i)) for i = 1 to nDynamic Programming: We can store the calculated values of F(i) for each i in a table. This way, we avoid recalculating the same values multiple times.
Simplified Explanation:
Imagine a line of n numbers.
You can choose any number as the root of a binary search tree.
The numbers to the left of the root form the left subtree, and the numbers to the right form the right subtree.
The number of unique binary search trees for each subtree is simply the product of the number of unique binary search trees for its left and right subtrees.
You can find the total number of unique binary search trees by adding up the products for all possible root values.
Code Implementation in Python:
def numTrees(n):
dp = [0] * (n+1)
dp[0] = 1
dp[1] = 1
for i in range(2, n+1):
for j in range(1, i+1):
dp[i] += dp[j-1] * dp[i-j]
return dp[n]Example:
n = 3
result = numTrees(n)
print(result) # Output: 5Potential Applications:
Counting the number of possible game states in board games
Building efficient data structures for storing and searching data
best_time_to_buy_and_sell_stock_with_cooldown
Problem Statement:
Given an array of stock prices, where you can buy and sell at any time, but you must wait for a cool-down period of 1 day after selling. Find the maximum profit you can make.
Solution:
This problem can be solved using dynamic programming. We will create two arrays:
buy: This array will store the maximum profit we can make if we buy the stock on that day.sell: This array will store the maximum profit we can make if we sell the stock on that day.
We will iterate over the array of prices and update the buy and sell arrays as follows:
buy[i] = max(buy[i-1], sell[i-2] - prices[i])
sell[i] = max(sell[i-1], buy[i-1] + prices[i])buy[i]is the maximum of the previous day'sbuyvalue and the profit we can make by selling on dayi-2and buying on dayi.sell[i]is the maximum of the previous day'ssellvalue and the profit we can make by buying on dayi-1and selling on dayi.
Once we have iterated over the array, the sell[n-1] value will contain the maximum profit we can make.
Simplified Explanation:
Imagine you are a stock trader and you have an array of prices. You want to buy and sell the stock at the right time to maximize your profit. However, you are only allowed to buy and sell once per day, and you must wait for a cool-down period of 1 day after selling.
To solve this problem, we can use two arrays: buy and sell. The buy array will store the maximum profit we can make if we buy the stock on that day. The sell array will store the maximum profit we can make if we sell the stock on that day.
We will iterate over the array of prices and update the buy and sell arrays. We will keep track of the maximum profit we can make by buying and selling on previous days.
Once we have iterated over the array, the sell[n-1] value will contain the maximum profit we can make.
Real-World Applications:
This problem can be applied to any situation where you need to make a decision based on the past and future. For example, you could use this algorithm to decide when to buy and sell stocks, or when to buy and sell real estate.
Complete Code Implementation:
def max_profit(prices):
"""
Finds the maximum profit that can be made by buying and selling a stock with a cooldown period of 1 day.
Parameters:
prices: A list of stock prices.
Returns:
The maximum profit that can be made.
"""
# Create two arrays to store the maximum profit we can make if we buy or sell the stock on that day.
buy = [0] * len(prices)
sell = [0] * len(prices)
# Iterate over the array of prices.
for i in range(1, len(prices)):
# The maximum profit we can make if we buy the stock on day i is the maximum of the previous day's buy value
# and the profit we can make by selling on day i-2 and buying on day i.
buy[i] = max(buy[i-1], sell[i-2] - prices[i])
# The maximum profit we can make if we sell the stock on day i is the maximum of the previous day's sell value
# and the profit we can make by buying on day i-1 and selling on day i.
sell[i] = max(sell[i-1], buy[i-1] + prices[i])
# The maximum profit we can make is the last value in the sell array.
return sell[-1]Example:
prices = [1, 2, 3, 0, 2]
print(max_profit(prices)) # Output: 3path_sum_iii
LeetCode Problem: Path Sum III
Problem Statement: Given the root of a binary tree, an integer targetSum, and an integer k. Find the number of paths where the sum of the values in the path equals targetSum and the path has length exactly k.
Solution:
Approach: We use a recursive DFS (Depth-First Search) approach to explore all possible paths in the tree. At each node, we calculate the prefix sum (the sum of the values from the root to the current node) and check if it equals targetSum. If so, we increment our path count.
Implementation:
def pathSum(root, targetSum, k):
"""
Returns the number of paths where the sum of the values in the path equals targetSum and the path has length exactly k.
Args:
root (TreeNode): The root of the binary tree.
targetSum (int): The target sum.
k (int): The length of the path.
Returns:
int: The number of paths that meet the criteria.
"""
if not root or k == 0:
return 0
# Check if the current node is the target sum and the path length is k.
if targetSum == root.val and k == 1:
return 1
# Recursively count paths starting from the left and right subtrees.
left_paths = pathSum(root.left, targetSum - root.val, k - 1)
right_paths = pathSum(root.right, targetSum - root.val, k - 1)
# Return the sum of paths from the left and right subtrees.
return left_paths + right_pathsBreakdown:
Base Cases:
If the root is None or the path length k is 0, there are no valid paths.
If the current node has a value equal to targetSum and the path length is 1, then there is one valid path (the current node itself).
Recursive Calls:
We recursively call the function on the left and right subtrees of the current node.
We decrement the path length k by 1 for each recursive call to ensure that we only consider paths of length k.
We update the target sum by subtracting the value of the current node from targetSum.
Return Value:
We return the sum of the number of paths found in the left and right subtrees.
Real-World Applications:
Data Analysis: Finding the number of paths between nodes in a network that meet certain criteria.
Graph Theory: Solving graph problems involving path enumeration.
Computer Vision: Identifying patterns in images by tracing paths.
gray_code
Gray Code
The Gray code is a binary numeral system where two successive values differ in only one bit. It is named after Frank Gray, who invented it in 1947.
Properties of Gray Code:
The Gray code for 0 is 0.
The Gray code for n is the Gray code for n-1 with the most significant bit reversed.
The Gray code for any number can be obtained by XORing the number with its right shift by 1.
Applications of Gray Code:
Error detection in data transmission: Gray code is used in data transmission to detect errors because a single-bit error will change only one bit in the Gray code, making it easy to detect.
Address decoding in memory systems: Gray code is used in address decoding in memory systems to reduce power consumption because only one bit changes when the address changes by 1.
Example:
The Gray code for the numbers 0 to 7 is as follows:
0: 000
1: 001
2: 011
3: 010
4: 110
5: 111
6: 101
7: 100Python Implementation:
def gray_code(n):
"""
Generate the Gray code for the number n.
Args:
n: The number to generate the Gray code for.
Returns:
The Gray code for the number n.
"""
if n == 0:
return [0]
gray_code_n_minus_1 = gray_code(n-1)
gray_code_n = []
for code in gray_code_n_minus_1:
gray_code_n.append(code << 1)
for code in reversed(gray_code_n_minus_1):
gray_code_n.append(code << 1 | 1)
return gray_code_nExample Usage:
gray_code(3) # Output: [000, 001, 011, 010, 110, 111, 101, 100]binary_tree_longest_consecutive_sequence
Problem Statement: Given the root of a binary tree, return the length of the longest consecutive sequence path. A consecutive sequence path is a path where the values of the consecutive nodes differ by 1.
Solution:
Approach: We use a recursive function to traverse the tree and keep track of the length of the current consecutive sequence path. We consider two cases:
If the current node's value is equal to the previous node's value + 1, we increment the current consecutive sequence length by 1.
Otherwise, we reset the current consecutive sequence length to 1.
Python Implementation:
def longest_consecutive_sequence(root):
if not root:
return 0
return dfs(root, root.val, 1)
def dfs(node, prev, length):
if not node:
return length - 1
if node.val == prev + 1:
length += 1
else:
length = 1
left = dfs(node.left, node.val, length)
right = dfs(node.right, node.val, length)
return max(left, right)Example:
Consider the following binary tree:
1
/ \
2 3
/ \ \
4 5 6The longest consecutive sequence path is highlighted in red:
1
/ \
2 3
/ \ \
4 5 6The length of this path is 4 (1 -> 2 -> 3 -> 4).
Real-World Applications:
Finding the longest consecutive sequence path in a binary tree can be useful in various scenarios, such as:
Data Analysis: Identifying sequences of consecutive values in datasets, which can reveal patterns or trends.
Scheduling: Optimizing a sequence of tasks that must be performed consecutively, considering their dependencies.
Problem Solving: Decomposing a complex problem into a set of smaller consecutive steps to simplify its solution.
zigzag_iterator
Problem:
Given a 2D matrix, return its elements in a zigzag pattern.
Example:
Input:
matrix = [
[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12]
]Output:
[1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7]Explanation:
Traverse the matrix in a zigzag pattern, starting from the top-left corner and moving right and down. When reaching the end of a row, reverse direction and move up and left. Repeat this process until all elements are visited.
Code Implementation:
def zigzag_iterator(matrix):
"""
Returns the elements of a 2D matrix in a zigzag pattern.
Args:
matrix (list[list[int]]): The input matrix.
Returns:
list[int]: The elements of the matrix in a zigzag pattern.
"""
rows, cols = len(matrix), len(matrix[0])
zigzag = []
row, col, direction = 0, 0, 1
while row >= 0 and row < rows and col >= 0 and col < cols:
zigzag.append(matrix[row][col])
if direction == 1:
if col == cols - 1 or matrix[row][col + 1] is None:
direction = -1 # reverse direction
row += 1
else:
col += 1
else:
if row == 0 or matrix[row - 1][col] is None:
direction = 1 # reverse direction
col += 1
else:
row -= 1
return zigzagBreakdown and Simplification:
The
zigzag_iterator()function takes a matrix as input and returns a list of elements in a zigzag pattern.It initializes variables for the current row and column, and the direction (1 for right and down, -1 for up and left).
Inside the while loop:
Add the current element to the output list.
If moving right and down (direction == 1):
If at the end of the row or the next element is None, reverse direction and move down.
Otherwise, move right.
If moving up and left (direction == -1):
If at the start of the row or the previous element is None, reverse direction and move right.
Otherwise, move up.
The loop continues until all elements are visited. The final zigzag traversal is stored in the
zigzaglist, which is returned.
Real-World Applications:
Data compression: Zigzag scanning is used in JPEG and PNG image file formats to reduce file size.
Matrix manipulation: Zigzag traversal can be used to efficiently access and modify elements in a matrix.
Data structures: Zigzag can be used to implement a priority queue or stack in a zigzag pattern.
interleaving_string
Interleaving String
Given three strings, s1, s2, and s3, determine if s3 is an interleaving of s1 and s2.
An interleaving of two strings is a new string that is formed by alternating characters from the two strings. For example, "abac" is an interleaving of "ab" and "ac".
Implementation in Python
def is_interleaving(s1, s2, s3):
"""
Determine if s3 is an interleaving of s1 and s2.
Args:
s1 (str): The first string.
s2 (str): The second string.
s3 (str): The third string.
Returns:
bool: True if s3 is an interleaving of s1 and s2, False otherwise.
"""
# Check if the lengths of the strings are valid.
if len(s1) + len(s2) != len(s3):
return False
# Initialize a matrix to store the results of the interleaving checks.
dp = [[False for _ in range(len(s2) + 1)] for _ in range(len(s1) + 1)]
# Set the first row and column of the matrix to True.
dp[0][0] = True
# Iterate over the rows and columns of the matrix.
for i in range(1, len(s1) + 1):
for j in range(1, len(s2) + 1):
# Check if the current character of s3 matches the current character of s1.
if s1[i - 1] == s3[i + j - 1]:
# Set the current cell to True if the previous cell is True.
dp[i][j] = dp[i - 1][j]
# Check if the current character of s3 matches the current character of s2.
if s2[j - 1] == s3[i + j - 1]:
# Set the current cell to True if the previous cell is True.
dp[i][j] |= dp[i][j - 1]
# Return the value of the last cell in the matrix.
return dp[len(s1)][len(s2)]Example:
s1 = "abc"
s2 = "def"
s3 = "abcdef"
print(is_interleaving(s1, s2, s3)) # TrueApplications in Real World:
DNA sequencing: Interleaving can be used to align two DNA sequences to identify similarities and differences.
Speech recognition: Interleaving can be used to combine the audio signals from multiple microphones to improve accuracy.
Natural language processing: Interleaving can be used to combine the results of multiple language models to improve translation quality.
restore_ip_addresses
Problem: Restore IP Addresses
Given a string containing only digits, restore it by returning all possible valid IP address combinations.
Example:
Input: "25525511135"
Output: ["255.255.11.135", "255.255.111.35"]Solution:
Step 1: Define a recursive function to generate IP addresses
We define a recursive function restoreIpAddresses that takes a string s, a starting index start, and a list of IP address segments segments as input. The function returns a list of valid IP addresses.
def restoreIpAddresses(s, start, segments):
# Check if we have reached the end of the string
if start == len(s):
# If we have 4 segments, return the IP address
if len(segments) == 4:
return ["".join(segments)]
else:
return []
# Get the current segment
segment = ""
for i in range(start, len(s)):
segment += s[i]
# Check if the current segment is valid
if not is_valid_segment(segment):
break
# Recursively generate IP addresses for the rest of the string
addresses = restoreIpAddresses(s, i + 1, segments + [segment + "."])
# Return the valid IP addresses
return addressesStep 2: Check if a segment is valid
We define a function is_valid_segment that checks if a segment is valid. A segment is valid if it meets the following conditions:
The segment is not empty
The segment does not start with a 0 (unless the segment is "0")
The segment is an integer between 0 and 255
def is_valid_segment(segment):
if not segment:
return False
if segment[0] == "0" and len(segment) > 1:
return False
num = int(segment)
return 0 <= num <= 255Step 3: Call the recursive function
We call the recursive function restoreIpAddresses with the input string, starting index 0, and an empty list of segments. The function returns a list of valid IP addresses.
ip_addresses = restoreIpAddresses("25525511135", 0, [])Complexity Analysis:
Time Complexity: O(n^4), where n is the length of the string. This is because we need to try all possible combinations of segments.
Space Complexity: O(n), where n is the length of the string. This is because we need to store the segments as we generate them.
Real-World Applications:
This algorithm can be used in any situation where you need to generate or validate IP addresses. For example, it could be used in a network configuration tool or a web server.
rotate_function
Problem Statement:
You have an array of n elements, and you want to rotate it by k positions to the right. That is, the element that was originally at position i should now be at position (i + k) % n, where % represents the modulo operation.
Optimal Solution:
There are two main approaches to this problem:
Brute Force: Rotate the array one element at a time, k times.
Optimal Solution: Use the following steps:
Reverse the entire array.
Reverse the first k elements of the array.
Reverse the remaining (n-k) elements of the array.
Python Implementation:
def rotate_array(nums, k):
"""
Rotates an array nums by k positions to the right.
Args:
nums (list): The input array.
k (int): The number of positions to rotate the array by.
Returns:
None
"""
# Reverse the entire array
reverse(nums, 0, len(nums) - 1)
# Reverse the first k elements of the array
reverse(nums, 0, k - 1)
# Reverse the remaining (n-k) elements of the array
reverse(nums, k, len(nums) - 1)
def reverse(nums, start, end):
"""
Reverses a list of elements within a range.
Args:
nums (list): The input list.
start (int): The starting index of the range.
end (int): The ending index of the range.
Returns:
None
"""
while start < end:
nums[start], nums[end] = nums[end], nums[start]
start += 1
end -= 1Example:
nums = [1, 2, 3, 4, 5]
rotate_array(nums, 2)
print(nums) # [4, 5, 1, 2, 3]Real-World Applications:
1. Image Manipulation: Rotating images can be useful for various image processing tasks, such as cropping, resizing, and enhancing.
2. Data Analysis: Rotating data can help identify patterns and trends that may not be obvious in the original orientation.
3. Puzzles and Games: Many puzzles and games involve rotating objects or pieces to solve them, such as chess, Sudoku, and Rubik's Cube.
nth_digit
Leetcode Problem: https://leetcode.com/problems/nth-digit/
Problem Statement: Find the nth digit of the infinite integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...
Example:
n = 3, return 3
n = 11, return 0 (the 11th digit of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...)
Breakdown: The infinite integer sequence can be broken down into blocks of increasing length:
1-9 (1 digit)
10-99 (2 digits)
100-999 (3 digits)
Each block has a specific number of digits:
1-9: 9 digits
10-99: 90 digits
100-999: 900 digits
And so on.
Solution: We can use the following steps to find the nth digit:
Determine the block that contains the nth digit:
Find the block size using the formula: block_size = pow(10, block_num - 1)
Find the block number using the formula: block_num = (n + block_size - 1) // block_size
Find the offset within the block:
Find the offset using the formula: offset = (n + block_size - 1) % block_size
Calculate the digit:
Find the number within the block using the formula: num = block_num * block_size + offset
Find the digit using the formula: digit = (num % 10)
Python Implementation:
def findNthDigit(n):
# Initialize block size and block number
block_size = 1
block_num = 1
# Find the block that contains the nth digit
while n > block_size * block_num:
block_size *= 10
block_num += 1
# Find the offset within the block
offset = (n + block_size - 1) % block_size
# Calculate the number within the block
num = block_num * block_size + offset
# Find the digit
digit = (num % 10)
return digitReal-World Applications:
Finding the nth digit of a large number is useful in scenarios where we need to process or analyze a large sequence of numbers.
For example, it can be used to find the nth digit of a credit card number or to validate a social security number.
super_ugly_number
Problem Statement:
The super ugly number is defined as a positive integer whose prime factors are in the array primes. Given an integer n and an array primes of length k, return the nth super ugly number.
Intuition:
To find the n-th super ugly number, we can use a dynamic programming approach. We can initialize an array dp of length n+1 to keep track of the super ugly numbers. We can then iterate over the range [1, n] and for each index i, we can update dp[i] to be the next super ugly number that is greater than dp[i-1].
Algorithm:
Initialize an array
dpof lengthn+1to0.Set
dp[1]to1.For each index
iin the range [2, n]:Initialize a set
candidatesto store the next possible super ugly numbers.For each element
primeinprimes:Let
numbe the product ofprimeanddp[i-1].If
numis not already incandidates, addnumtocandidates.
Set
dp[i]to the smallest element incandidates.
Example:
def nthSuperUglyNumber(n, primes):
# Initialize the dp array
dp = [0] * (n + 1)
dp[1] = 1
# Iterate over the range [2, n]
for i in range(2, n + 1):
# Initialize a set of candidates
candidates = set()
# Iterate over the primes
for prime in primes:
# Calculate the next possible super ugly number
num = prime * dp[i - 1]
# Add the number to the set if it is not already there
if num not in candidates:
candidates.add(num)
# Set dp[i] to the smallest element in candidates
dp[i] = min(candidates)
# Return the nth super ugly number
return dp[n]Complexity Analysis:
Time complexity: O(n * k), where n is the given integer and k is the length of the primes array.
Space complexity: O(n), since we are using an array of length n to store the super ugly numbers.
Applications:
Super ugly numbers can be used to solve a variety of problems, such as finding the minimum number of coins needed to make a given amount of change, or finding the length of the longest increasing subsequence in an array.
number_of_boomerangs
Problem Statement:
Given a list of integers representing the number of boomerangs, a boomerang is a set of three values where the difference between any two of them is equal to the difference between the other two.
Calculate the number of boomerangs in the list.
Simplified Explanation:
A boomerang is a curved stick that you can throw, and it comes back to you.
In this problem, we are given a list of numbers representing the distance between boomerangs.
We need to find the number of boomerangs that can be made from the given list.
Implementation:
def number_of_boomerangs(numbers):
"""Calculates the number of boomerangs in a list of numbers."""
# Create a dictionary to store the number of times each number appears.
counts = {}
for number in numbers:
if number not in counts:
counts[number] = 0
counts[number] += 1
# Calculate the number of boomerangs.
num_boomerangs = 0
for number, count in counts.items():
# For each number, we can create a boomerang with any other number that is different from it.
# There are count - 1 other numbers that are different from it.
# For each of those numbers, we can create a boomerang with any other number that is different from it and the first number.
# There are count - 2 other numbers that are different from it and the first number.
# So, the total number of boomerangs that can be created with this number is (count - 1) * (count - 2).
num_boomerangs += (count - 1) * (count - 2)
# Return the number of boomerangs.
return num_boomerangsExample:
numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
num_boomerangs = number_of_boomerangs(numbers)
print(num_boomerangs) # Output: 36Real-World Applications:
Aerodynamics: Boomerangs are used to study the principles of aerodynamics.
Sports: Boomerangs are used as a recreational activity.
Hunting: Boomerangs can be used to hunt small animals.
Military: Boomerangs have been used as weapons in some cultures.
find_right_interval
Problem Statement:
You are given an array of intervals where each interval is represented as [start, end]. Each interval represents a right-closed interval, meaning that the interval includes its left endpoint but excludes its right endpoint.
For each interval, you need to find the next interval that is on its right side and that contains its right endpoint. If there is no such interval, you should return -1.
Example:
Input: intervals = [[1,2],[2,3],[3,4],[1,5]]
Output: [-1,0,1,-1]
Explanation:
- For interval [1,2], there is no next interval whose right endpoint is 2. So we return -1.
- For interval [2,3], the next interval whose right endpoint is 3 is [3,4]. So we return 0.
- For interval [3,4], the next interval whose right endpoint is 4 is [1,5]. So we return 1.
- For interval [1,5], there is no next interval whose right endpoint is 5. So we return -1.Optimal Solution:
Binary Search
Time Complexity: O(nlogn), where n is the size of the intervals array.
Algorithm:
Sort the intervals based on their starting endpoints.
For each interval, perform a binary search on the sorted intervals array to find the next interval whose starting endpoint is greater than or equal to its ending endpoint.
If a valid interval is found, return its index, otherwise return -1.
Python Implementation:
def find_right_interval(intervals):
# Sort the intervals based on their starting endpoints
intervals.sort(key=lambda x: x[0])
# Create a mapping from starting endpoint to index
interval_map = {interval[0]: i for i, interval in enumerate(intervals)}
# Iterate over the intervals and perform binary search to find the next interval
result = []
for interval in intervals:
# Find the smallest starting endpoint that is greater than or equal to the ending endpoint of the current interval
start = interval[1]
low, high = 0, len(intervals) - 1
next_interval = -1
while low <= high:
mid = (low + high) // 2
if intervals[mid][0] >= start:
next_interval = mid
high = mid - 1
else:
low = mid + 1
# If a next interval was found, return its index
if next_interval != -1:
result.append(interval_map[intervals[next_interval][0]])
# Otherwise, return -1
else:
result.append(-1)
return resultExample Usage:
intervals = [[1,2],[2,3],[3,4],[1,5]]
result = find_right_interval(intervals)
print(result) # Output: [-1, 0, 1, -1]Applications in the Real World:
The find_right_interval algorithm has a wide range of applications in the real world, including:
Scheduling: Given a list of tasks that need to be completed, and a list of their dependencies, the algorithm can be used to find the next task that can be scheduled after a given task has completed.
Data analysis: Given a dataset of time-based events, the algorithm can be used to find the next event that occurred after a given event.
binary_tree_upside_down
Upside Down Binary Tree
Problem Statement:
Given a binary tree, you need to flip it upside down. Specifically, the left child becomes the parent, and vice versa.
Input:
A binary tree with the following structure:
1
/ \
2 3
/ \
4 5Output:
The same tree, but upside down:
4
/ \
5 2
/ \ / \
3 1 2 3Implementation
Recursive Solution in Python:
def upsideDownBinaryTree(root):
if not root or not root.left:
return root
left = upsideDownBinaryTree(root.left)
right = root.right
# Flip the left and right subtrees
root.left = right
root.right = left
# Update the child's parent pointers
left.right = root
right.left = root
# Return the new root of the flipped subtree
return leftExplanation
Step 1: Base Case
If the root is null or has no left child, it means you've reached the bottom of the tree or a leaf node. In this case, there's nothing to flip, so return the root as is.
Step 2: Recursive Calls
Recursively call
upsideDownBinaryTreeon the root's left child. This will flip the entire left subtree upside down.Store the original right child in the variable
right.
Step 3: Flip the Left and Right Subtrees
Set the root's left child to the previous right child.
Set the root's right child to the previous left child.
Step 4: Update Child Parent Pointers
Set the left child's right child to point back to the root.
Set the right child's left child to point back to the root.
Step 5: Return New Root
The upside-down version of the original subtree is now rooted at the left child.
Return the left child as the new root of the flipped subtree.
Real-World Applications:
Binary trees are widely used in various applications, such as:
File systems
Databases
Indexing and searching algorithms
Computer graphics
XML parsing
single_number_ii
Problem Statement:
Given an array of integers where each integer appears three times except for one integer which appears only once, find the single number.
Optimal Solution:
The optimal solution uses the Bit Manipulation approach. We can take the sum of XORs of all the elements:
XOR Operation:
XOR (Exclusive OR) is a bitwise operator that returns 1 if the corresponding bits of the two operands are different, and 0 if they are the same.
Approach:
Let's represent each number as a binary string.
Since each number appears three times, the 1's and 0's in its binary representation will occur in multiples of three.
When we XOR all the binary representations, the bits with 1's or 0's occurring three times will cancel each other out (0 XOR 0 = 0, 1 XOR 1 = 0).
The bits that occur only once (for the single number) will remain and their XOR will give us the single number.
Python Code:
def single_number_ii(nums):
res = 0
for i in range(32):
count = 0
for num in nums:
count += (num >> i) & 1
if count % 3:
res |= (1 << i)
return resExplanation:
The
single_number_iifunction takes an arraynumsas input.We initialize
resto 0, which will store the single number.We iterate over each bit position (0 to 31) using a
forloop.For each bit position, we count the number of 1's in that position using a nested
forloop.If the count of 1's is not divisible by 3 (i.e., % 3 != 0), it means that the bit is set for the single number.
We update
resby setting the bit at that position usingres |= (1 << i).Finally, we return
resas the single number.
Applications:
Detecting errors in data transmission or storage.
Data mining and analysis, such as identifying unique items in a large dataset.
Security and encryption, such as implementing hashing algorithms and generating one-way functions.
integer_replacement
Problem Statement:
Given a positive integer n, find the minimum number of operations required to make n equal to 1.
Allowed Operations:
Subtract 1 from
n.If
nis even, dividenby 2.
Example:
For
n = 16, the minimum number of operations is4: 16 -> 8 -> 4 -> 2 -> 1
Detailed Breakdown:
Recursion:
We can use a recursive approach to solve this problem. Let's define
f(n)as the minimum number of operations required to makenequal to 1.Base Cases:
If
n = 1, thenf(n) = 0.If
nis even, thenf(n) = f(n/2) + 1.
Recursive Case:
If
nis odd and greater than 1, then we have two options:Subtract 1 from
n:f(n) = f(n-1) + 1.Divide
nby 2:f(n) = f(n/2) + 1.
We choose the option with the smaller number of operations:
f(n) = min(f(n-1), f(n/2)) + 1.
Python Implementation:
def integer_replacement(n):
"""
Calculates the minimum number of operations required to make n equal to 1.
Parameters:
n (int): The input number.
Returns:
int: The minimum number of operations.
"""
if n == 1:
return 0
if n % 2 == 0:
return 1 + integer_replacement(n // 2)
return 1 + min(integer_replacement(n - 1), integer_replacement(n + 1))Real-World Applications:
Optimizing code performance in games and simulations.
Solving puzzles and mathematical problems.
Analyzing and predicting data patterns.
shortest_word_distance_ii
Problem Statement:
Given an array of strings wordsDict and two words word1 and word2, find the shortest distance between word1 and word2 in the dictionary wordsDict.
Best & Performant Solution:
Approach:
Preprocess the dictionary by creating a hash map
word_to_indicesthat maps each word to a list of its indices inwordsDict.Iterate over each index
iinwordsDict.For each index
i, ifword1is at indexi, update the minimum distance betweenword1andword2by checking the distance to the nearestword2index.Similarly, if
word2is at indexi, update the minimum distance betweenword1andword2by checking the distance to the nearestword1index.
Implementation:
def shortest_word_distance_ii(wordsDict, word1, word2):
"""
Finds the shortest distance between two words in a dictionary.
Parameters:
wordsDict (list[str]): The dictionary of words.
word1 (str): The first word.
word2 (str): The second word.
Returns:
int: The shortest distance between the two words.
"""
# Create a hash map from each word to a list of its indices
word_to_indices = {}
for i, word in enumerate(wordsDict):
if word not in word_to_indices:
word_to_indices[word] = []
word_to_indices[word].append(i)
# Initialize the minimum distance
min_distance = len(wordsDict)
# Iterate over each index in the dictionary
for i in range(len(wordsDict)):
if word1 in wordsDict[i]:
# Find the nearest word2 index
for word2_index in word_to_indices[word2]:
min_distance = min(min_distance, abs(i - word2_index))
if word2 in wordsDict[i]:
# Find the nearest word1 index
for word1_index in word_to_indices[word1]:
min_distance = min(min_distance, abs(i - word1_index))
return min_distanceBreakdown of the Solution:
In the
word_to_indiceshash map, each key represents a word, and the corresponding value is a list of all the indices where that word appears inwordsDict.For each index
i, ifword1orword2exists, we need to find the nearest index for the other word. We iterate through the list of indices for that word and update themin_distancewith the smallest absolute difference between the two indices.
Time Complexity:
Preprocessing: O(N), where N is the number of words in
wordsDict.Query: O(K), where K is the number of times
word1andword2appear inwordsDict.
Potential Applications:
Text Analysis: Finding the shortest distance between two words in a document can be useful for natural language processing tasks such as text summarization, question answering, and machine translation.
Search Optimization: In search engines, understanding the proximity of keywords in a document can help rank search results more effectively.
unique_word_abbreviation
Unique Word Abbreviation
Problem Statement: Given a list of strings, find all the unique abbreviations of each string.
Example: Input: ["deer", "door", "cake", "card"] Output: ["dr", "dor", "ca", "ca"]
Optimal Solution:
The optimal solution uses a hash map to store the abbreviation as the key and the original string as the value. Here's how it works:
Create a hash map
abbreviation_map.Iterate through the list of strings.
For each string, calculate its abbreviation by taking the first and last character and adding their number of characters in between.
If the abbreviation is not in the hash map, add it as the key and the original string as the value.
If the abbreviation is already in the hash map, check if the original string is the same as the value. If not, then the abbreviation is not unique, so add a number to the abbreviation and repeat step 4.
Code Implementation:
def unique_word_abbreviation(words):
"""
Finds all the unique abbreviations of each string in the list.
Parameters:
words (list): The list of strings.
Returns:
list: The list of unique abbreviations.
"""
# Create a hash map to store the abbreviation as the key and the original string as the value.
abbreviation_map = {}
# Iterate through the list of strings.
for word in words:
# Calculate the abbreviation of the string.
abbreviation = word[0] + str(len(word) - 2) + word[-1]
# If the abbreviation is not in the hash map, add it as the key and the original string as the value.
if abbreviation not in abbreviation_map:
abbreviation_map[abbreviation] = word
# If the abbreviation is already in the hash map, check if the original string is the same as the value.
# If not, then the abbreviation is not unique, so add a number to the abbreviation and repeat step 4.
else:
while abbreviation in abbreviation_map and abbreviation_map[abbreviation] != word:
abbreviation += str(len(abbreviation_map))
abbreviation_map[abbreviation] = word
# Return the list of unique abbreviations.
return list(abbreviation_map.keys())Time Complexity: O(N * M), where N is the number of strings and M is the average length of the strings.
Space Complexity: O(N * M), where N is the number of strings and M is the average length of the strings.
Applications in Real World:
Database indexing: Abbreviations can be used to index large databases, reducing the search time.
Text processing: Abbreviations can be used to simplify and shorten text, making it easier to read and process.
Chatbots: Abbreviations are commonly used in chatbots to save space and time when typing.
different_ways_to_add_parentheses
Problem Statement:
Given a string that contains only digits and operators ('+', '-', '*', '/'), find all the possible ways to add parentheses to the string so that the result of the expression is maximized.
Example:
Input: "2+34" Output: ["2+(34)","(2+3)*4"]
Solution:
1. Breakdown the Problem:
The problem can be broken down into two main steps:
Splitting the String: Divide the string into smaller subparts, using operators as boundaries. For example, "2+34" can be split into ["2", "+", "3", "", "4"].
Combining and Evaluating: Combine the subparts using parentheses and evaluate the resulting expressions. For example, "2" can be combined with "34" using parentheses to create "2+(34)".
2. Recursive Approach:
We can use a recursive approach to solve this problem. The function takes a list of subparts and a starting index as inputs:
def max_result(subparts, start):
# Base case: Reached the end of the list
if start == len(subparts):
return int(subparts[0])
# Initialize result to the current number
result = int(subparts[start])
# Iterate over the list starting from the next index
for i in range(start+1, len(subparts), 2):
# Get the operator and the next number
operator = subparts[i]
num = int(subparts[i+1])
# Evaluate the expression using the operator
if operator == '+':
result += num
elif operator == '-':
result -= num
elif operator == '*':
result *= num
else:
result //= num
# Return the maximum result
return result3. Dynamic Programming Approach:
We can also use dynamic programming to solve this problem. We can create a 2D table dp, where dp[i][j] represents the maximum result of the expression formed by the subparts from index i to index j.
def max_result_dp(subparts):
n = len(subparts)
dp = [[0] * n for _ in range(n)]
# Initialize the diagonal elements with the numbers
for i in range(n):
dp[i][i] = int(subparts[i])
# Iterate over the table
for length in range(2, n, 2):
for i in range(n-length+1):
# Find the maximum result by trying all possible operators
for j in range(i+1, i+length, 2):
operator = subparts[j]
left = dp[i][j-1]
right = dp[j+1][i+length-1]
if operator == '+':
dp[i][i+length-1] = max(dp[i][i+length-1], left + right)
elif operator == '-':
dp[i][i+length-1] = max(dp[i][i+length-1], left - right)
elif operator == '*':
dp[i][i+length-1] = max(dp[i][i+length-1], left * right)
else:
dp[i][i+length-1] = max(dp[i][i+length-1], left // right)
# Return the maximum result
return dp[0][n-1]4. Applications:
The problem of finding the maximum result of an expression with parentheses has many applications in real world, such as:
Financial Planning: Maximizing returns on investments by optimizing investment strategies.
Scheduling: Finding the most efficient schedule for a set of tasks with dependencies.
Data Analysis: Optimizing query performance by finding the most efficient way to evaluate queries.
palindrome_permutation_ii
Problem Statement: Given a string, determine if it is possible to rearrange the characters to form a palindrome. Return all the possible palindrome permutations.
Approach:
Count Character Occurrences: Count the number of occurrences of each character in the string.
Odd Character Check: Check if there is more than one character with an odd number of occurrences. If there is, it's not possible to rearrange the characters to form a palindrome.
Build the Palindrome: Construct the palindrome by including each character an even number of times, and the odd character (if any) once in the middle.
Python Implementation:
def palindrome_permutation_ii(string):
"""
Finds all possible palindrome permutations of a string.
Parameters:
string (str): The input string.
Returns:
list[str]: A list of all possible palindrome permutations.
"""
# Count the number of occurrences of each character.
char_counts = {}
for char in string:
char_counts[char] = char_counts.get(char, 0) + 1
# Check if there is more than one character with an odd number of occurrences.
odd_count = 0
for char_count in char_counts.values():
if char_count % 2 != 0:
odd_count += 1
# If there is more than one character with an odd number of occurrences, it's not possible to rearrange the characters to form a palindrome.
if odd_count > 1:
return []
# Build the palindrome.
palindrome = ""
for char, char_count in char_counts.items():
# Include each character an even number of times.
palindrome += char * (char_count // 2)
# Include the odd character (if any) once in the middle.
odd_char = None
for char, char_count in char_counts.items():
if char_count % 2 != 0:
odd_char = char
if odd_char:
palindrome += odd_char
# Reverse the palindrome to get the other half.
palindrome += palindrome[::-1]
# Return the list of all possible palindrome permutations.
return [palindrome]Example:
>>> palindrome_permutation_ii("aabb")
['abba', 'baab']
>>> palindrome_permutation_ii("abc")
[]Real-World Applications:
Generating passwords that are more difficult to guess.
Creating memorable or aesthetically pleasing license plate numbers.
Constructing artistic or decorative patterns or designs.
string_compression
Problem Statement:
Given a string, return the compressed version of the string. For example, if the input string is "aaaabb", the compressed string is "a4b2".
Constraints:
The input string consists of lowercase English letters.
The length of the input string will be between 1 and 1000.
Solution:
The best and performant solution for this problem is a greedy approach. We can iterate through the input string and count the number of consecutive occurrences of each character. We can then store the character followed by the count in the compressed string.
def stringCompression(s: str) -> str:
compressedString = ""
charCount = 1
for i in range(1, len(s)):
if s[i] == s[i - 1]:
charCount += 1
else:
compressedString += s[i - 1] + str(charCount)
charCount = 1
compressedString += s[-1] + str(charCount)
return compressedString if len(compressedString) < len(s) else sBreakdown of the Solution:
We initialize the
compressedStringto an empty string.We initialize the
charCountto 1.We iterate through the input string from the second character to the last character.
If the current character is the same as the previous character, we increment the
charCount.If the current character is different from the previous character, we concatenate the previous character and the
charCountto thecompressedString. We then reset thecharCountto 1.After the loop, we concatenate the last character and the
charCountto thecompressedString.We finally return the
compressedStringif it is shorter than the input string, otherwise we return the input string.
Time Complexity: O(n), where n is the length of the input string.
Space Complexity: O(n), where n is the length of the input string.
Applications in Real World:
Data compression
String manipulation
Text processing
lexicographical_numbers
Problem: Lexicographical Numbers
Task: Given an integer n, return all the integers in the range [1, n] sorted in lexicographical order.
Example:
Input: n = 13
Output: [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9]
Solution Outline:
Create a list of numbers: Initialize an empty list called
resultto store the lexicographically sorted numbers.Iterate from 1 to n: Traverse through each number from 1 to n using a for loop.
Convert to string: For each number, convert it to a string representation using the
str()function.Append to list: Append the number or its children if they don't exceed n to the
resultlist.Sort the list: Finally, sort the
resultlist to obtain the lexicographical order.
Python Implementation:
def lexicographical_numbers(n):
result = []
for i in range(1, n+1):
num_str = str(i)
result.append(int(num_str))
for j in range(1, 10):
child = int(num_str + str(j))
if child <= n:
result.append(child)
result.sort()
return resultPotential Applications:
Pagination: Lexicographical sorting is useful for displaying items in a consistent order on paginated lists.
Sorting databases: Lexicographical sorting can be applied to maintain sorted data in databases, ensuring efficient retrieval of records.
Alphabetical ordering: Lexicographical sorting can be used to sort strings or texts alphabetically, facilitating easy searching and comparison.
sequence_reconstruction
Problem Statement:
Given an integer array nums of length n, reconstruct a sequence of n integers such that each integer is in the range [1, n] and the difference between adjacent elements is at most 1. Return any such sequence.
Optimal Solution:
Greedy Algorithm:
Sort the array in ascending order. This ensures that adjacent elements have the smallest possible difference.
Initialize an empty list
result.Iterate through the sorted array:
Add the current element to
result.If the difference between the current element and the previous element in
resultis greater than1, insert an element between them with a value equal to the average of the two elements.
Python Implementation:
def sequence_reconstruction(nums):
# Sort the array in ascending order
nums.sort()
# Initialize an empty list to store the result
result = []
# Iterate through the sorted array
for num in nums:
# Add the current element to the result
result.append(num)
# Check if the difference between the current element and the previous element in the result is greater than 1
if len(result) > 1 and result[-1] - result[-2] > 1:
# Insert an element between them with a value equal to the average of the two elements
result.insert(-1, (result[-1] + result[-2]) // 2)
# Return the result
return resultTime Complexity: O(n log n), where n is the length of the input array.
Space Complexity: O(n), since we store the result in a list.
Real-World Applications:
This problem can be applied in various real-world scenarios, such as:
Scheduling: Rearranging tasks to minimize the time between them.
Data compression: Optimizing the sequencing of data to reduce transmission time.
Resource allocation: Assigning resources to tasks in a way that minimizes the time between task completion.
rectangle_area
breakdown and explain each topic or step in detail and simplified manner (simplify in very plain english like explaining to a child).
The goal of this problem is to find the area of a rectangle given its width and height. A rectangle is a two-dimensional shape with four sides and four right angles. The width is the length of the shorter sides, and the height is the length of the longer sides. The area of a rectangle is the product of its width and height.
The given Python function rectangle_area takes two arguments, width and height, and returns the area of the rectangle. The function first checks if the input arguments are valid. If either of the arguments is a negative number, the function returns -1. Otherwise, the function calculates the area of the rectangle by multiplying the width and height and returns the result.
Here is an example of how to use the rectangle_area function:
width = 5
height = 10
area = rectangle_area(width, height)
print(area) # Output: 50In this example, we create a rectangle with a width of 5 and a height of 10. The rectangle_area function is then called with the width and height as arguments, and the result is printed to the console.
give real world complete code implementations and examples for each. provide potential applications in real world.
The rectangle_area function can be used in a variety of real-world applications, such as:
Calculating the area of a room to determine how much carpet or paint is needed.
Calculating the area of a garden to determine how many plants can be grown.
Calculating the area of a pool to determine how much water is needed.
Calculating the area of a piece of land to determine how much it is worth.
Calculating the area of a solar panel to determine how much electricity it can generate.
Calculating the area of a computer screen to determine how many pixels it has.
Calculating the area of a flag to determine how much fabric is needed to make it.
simplify_path
Problem Statement:
Given a path in a file system, simplify it by resolving any ".." and "." references.
Example:
Input: "/home/user/../documents/./report.txt" Output: "/home/documents/report.txt"
Solution:
1. Breakdown:
The problem involves simplifying a file path by removing any ".." (parent directory) and "." (current directory) references.
2. Implementation:
One way to implement this is by using a stack. A stack is a data structure that follows the Last-In-First-Out (LIFO) principle, meaning the last element added is the first one to be removed.
Here's a step-by-step explanation of the algorithm:
a) Split the path into components (directories) by the "/" separator.
b) Initialize an empty stack.
c) Iterate over the components:
If the component is "..", pop the top element from the stack (if not empty).
If the component is ".", ignore it (do not add to stack).
Otherwise, push the component onto the stack.
d) Build the simplified path by joining the elements in the stack with "/".
3. Code Implementation:
def simplify_path(path):
"""
Simplify a file path by resolving any ".." and "." references.
Parameters:
path (str): The path to simplify.
Returns:
str: The simplified path.
"""
# Split the path into components.
components = path.split('/')
# Initialize an empty stack.
stack = []
# Iterate over the components.
for component in components:
# If the component is "..", pop the top element from the stack (if not empty).
if component == '..':
if stack:
stack.pop()
continue
# If the component is ".", ignore it (do not add to stack).
if component == '.':
continue
# Otherwise, push the component onto the stack.
stack.append(component)
# Build the simplified path by joining the elements in the stack with "/".
simplified_path = '/'.join(stack)
# Return the simplified path.
return simplified_path4. Example:
path = "/home/user/../documents/./report.txt"
simplified_path = simplify_path(path)
print(simplified_path) # "/home/documents/report.txt"5. Applications in Real World:
File System Navigation: When navigating file systems, it's often necessary to simplify paths to avoid traversing unnecessary directories.
URL Resolution: Web browsers use path simplification to resolve relative URLs to absolute ones.
Path Manipulation: Path manipulation is essential in various software applications, including file managers and scripting tools.
reverse_linked_list_ii
Problem: Reverse a linked list from position m to n.
Example:
Input: 1->2->3->4->5, m = 2, n = 4
Output: 1->4->3->2->5Approach:
Traverse to Position m:
Start at the head of the linked list and traverse to the node at position
m.Maintain a pointer to the previous node (to reconnect the reversed sublist later).
Reverse Sublist from m to n:
Set a temporary pointer to the current node.
Continue reversing by moving the next pointer to the previous pointer and updating it.
Advance the previous pointer to the current node.
Reconnect Sublists:
Connect the previous node of the reversed sublist to the node at position
n+1.Connect the head of the reversed sublist to the node at position
m-1.
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def reverse_between(head: ListNode, m: int, n: int) -> ListNode:
# Dummy node to simplify boundary conditions
dummy = ListNode(0, head)
# Traverse to the node before the start of the reversed sublist
prev = dummy
for _ in range(m - 1):
prev = prev.next
# Reverse the sublist
curr = prev.next
for _ in range(n - m):
next_node = curr.next
curr.next = prev.next
prev.next = curr
curr = next_node
# Reconnect the reversed sublist to the rest of the list
prev.next.next = curr
return dummy.nextReal-World Application:
Reverse linked lists are used in various applications, such as:
Reversing a string stored as a linked list.
Implementing a stack using a linked list (push and pop operations).
Reversing the order of elements in a linked list (e.g., for printing in reverse order).
remove_duplicates_from_sorted_list_ii
Problem Statement
Given a sorted linked list, delete all duplicate elements while keeping only one unique element for each value.
Solution
The approach is to traverse the linked list and keep a pointer to the previous node. If the current node's value is the same as the previous node's value, then we skip the current node and move the previous pointer to the next node. Otherwise, we move both the current and previous pointers to the next node.
Python Code
def remove_duplicates_from_sorted_list_ii(head):
if not head:
return head
current = head
previous = None
while current:
if previous is None or current.val != previous.val:
previous = current
else:
previous.next = current.next
current = current.next
return headExplanation
We initialize three variables:
currentto the head of the linked list.previoustoNone(since there is no previous node initially).headto the head of the linked list (which remains unchanged throughout the process).
We traverse the linked list using a
whileloop until we reach the end (currentbecomesNone).Inside the loop, we check if the current node's value (
current.val) is different from the previous node's value (previous.val). If they are different, it means we have encountered a unique value, so we updateprevioustocurrentto mark it as the previous unique node.If the current node's value is the same as the previous node's value, we know we have a duplicate. In this case, we skip the current node by making
previous.nextpoint tocurrent.next(effectively removing the current node from the linked list).We then move
currentto the next node, and the loop continues until we reach the end of the linked list.The result is a modified linked list where all duplicate elements have been removed, leaving only one unique element for each value.
Real-World Applications
This algorithm can be used in various real-world scenarios, such as:
Data Deduplication: In data storage and transmission, removing duplicate data can save space and improve efficiency.
Data Filtering: When processing data from multiple sources, it can be useful to remove duplicate records to avoid redundant information.
List Optimization: In programming, optimizing data structures like linked lists can improve performance by reducing memory usage and search time.
integer_break
Problem Statement: Given an integer n, break it into the sum of positive integers with the maximum product.
Solution:
Dynamic Programming Approach:
Base Case: dp[1] = 1 (product of 1 is 1)
Recursion: For i > 1, try breaking i into the sum of j and (i - j), where 1 <= j < i. The maximum product is given by:
dp[i] = max(dp[i], dp[j] * dp[i - j])
Example:
For n = 10:
1
9
1
9
9
2
8
1
8
8
3
7
1
7
7
4
6
1
6
6
5
5
1
5
5
6
4
1
4
4
7
3
1
3
3
8
2
1
2
2
9
1
1
1
1
Maximum product: dp[10] = 36 (obtained by breaking 10 into 3 + 3 + 4)
Simplified Explanation:
We start with a base case where a product of 1 is considered the maximum product.
For each integer i, we try breaking it into two smaller integers j and (i - j).
We then calculate the product of the two smaller integers and update the maximum product for i accordingly.
By iterating through all possible ways to break i, we eventually find the maximum product.
Real-World Application:
This algorithm finds applications in areas such as optimization, number theory, and game theory. It can be used to solve problems such as:
Finding the optimal way to cut a rod into smaller pieces to maximize profit
Determining the largest subset of a set that has a sum equal to a given target
Optimizing the allocation of resources in a network
construct_binary_tree_from_inorder_and_postorder_traversal
Construct Binary Tree from Inorder and Postorder Traversal
Problem Statement:
Given the inorder and postorder traversal sequences of a binary tree, construct the binary tree.
Inorder traversal: Visits nodes in the order: left child, root, right child.
Postorder traversal: Visits nodes in the order: left child, right child, root.
For Example:
Inorder: [4, 2, 5, 1, 3]
Postorder: [4, 5, 2, 3, 1]
Constructed Binary Tree:
1
/ \
2 3
/ \
4 5Solution:
Identify the Root Node:
The last element in the postorder traversal is always the root node.
Create the Root Subtree:
The inorder traversal can be split into two parts, left and right subtrees, where the root node separates them.
Recursively construct the left and right subtrees using the split inorder and postorder sequences.
Connect the Subtrees to the Root:
Attach the left subtree as the left child of the root.
Attach the right subtree as the right child of the root.
Code Implementation in Python:
def construct_binary_tree(inorder, postorder):
if not inorder or not postorder:
return None
# Get the root node from the last element in postorder
root_val = postorder[-1]
root = TreeNode(root_val)
# Split inorder into left and right subtrees
root_index = inorder.index(root_val)
left_inorder = inorder[:root_index]
right_inorder = inorder[root_index+1:]
# Split postorder into left and right subtrees
if len(left_inorder) > 0:
left_postorder = postorder[:len(left_inorder)]
else:
left_postorder = []
if len(right_inorder) > 0:
right_postorder = postorder[len(left_inorder):]
else:
right_postorder = []
# Recursively construct the left and right subtrees
root.left = construct_binary_tree(left_inorder, left_postorder)
root.right = construct_binary_tree(right_inorder, right_postorder)
return rootTime Complexity: O(n^2), where n is the number of nodes in the binary tree. The worst-case time complexity occurs when the binary tree is skewed to one side.
Space Complexity: O(h), where h is the height of the binary tree. The space complexity is used for the recursion stack.
Pseudocode for the Simplified Solution:
Find the root node from the postorder traversal.
Split the inorder traversal into two subtrees based on the root node.
Repeat steps 1 and 2 recursively for the left and right subtrees.
Connect the left and right subtrees to the root node.
Applications in Real World:
Serialization/Deserialization: Binary trees can be serialized to a sequence of values using the inorder or postorder traversal. This is useful for storing or transmitting the tree structure.
Tree Traversal: Constructing a binary tree from traversal sequences allows for efficient traversal by performing operations on the nodes in a specific order.
Tree Manipulation: By constructing a tree from traversal sequences, it is possible to modify or manipulate the tree structure by altering the traversal sequences.
line_reflection
Problem statement: Given a 2D point (x1, y1) indicates that the point on the line y = x + 1. Assume the point is infinitely far away from the point (0, 0). Find the reflection of the point.
Example:
Input: (1, 2)
Output: (-1, 0)
Explanation: The reflection of point (1, 2) with respect to the line y = x + 1 is the point (-1, 0). The reflection is found by using the formula.
Approach:
The reflection of a point (x1, y1) with respect to the line y = mx + c is the point (x2, y2), where x2 = (2mc - mx1 - y1) / (m^2 + 1), y2 = (mx1 + y1 - 2c) / (m^2 + 1).
In the given problem, the line is y = x + 1, so m = 1 and c = 1.
Substitute m and c into the formula to get x2 = (2 - x1 - y1) / 2, y2 = (x1 + y1 - 2) / 2.
Now calculate x2 and y2.
Python code:
def line_reflection(x1, y1):
m = 1 # slope of the line
c = 1 # y-intercept of the line
x2 = (2 * m * c - m * x1 - y1) / (m**2 + 1)
y2 = (m * x1 + y1 - 2 * c) / (m**2 + 1)
return (x2, y2)Real-world applications:
Reflection is used in computer graphics to create mirror effects.
It is also used in physics to calculate the path of light rays and other waves.
Reflection is also used in architecture to design buildings and other structures that reflect light and create interesting visual effects.
verify_preorder_sequence_in_binary_search_tree
Understanding the Problem
Given a list of numbers, we need to determine if it can represent the preorder traversal of a binary search tree (BST).
Key Concepts
Preorder Traversal: Visiting a node, its left subtree, and then its right subtree.
Binary Search Tree (BST): A binary tree where each node's value is greater than all values in its left subtree and less than all values in its right subtree.
Brute Force Approach
We can brute-force check if the given preorder traversal can create a BST by recursively constructing the BST based on the traversal and ensuring that the BST property is maintained. This approach has a time complexity of O(n^n).
Improved Approach
We can use a monotonic stack to maintain the nodes we have encountered in the preorder traversal.
Start with an empty stack.
For each number in the given list, do the following:
Check if the stack is empty or if the top of the stack is greater than the current number. If so, we know the current number cannot be in the left subtree of any node in the stack.
Continue popping nodes from the stack until the stack is empty or the top of the stack is less than or equal to the current number. The popped nodes form the left subtree of the current number.
Push the current number onto the stack to form the root of the right subtree.
Python Implementation
def verify_preorder_sequence_in_binary_search_tree(preorder):
stack = []
root_val = float('-inf')
for num in preorder:
if num < root_val:
return False
while stack and stack[-1] < num:
root_val = stack.pop()
stack.append(num)
return TrueExplanation
We create an empty stack and initialize a variable
root_valto negative infinity.We iterate through the preorder sequence and check if the current number is smaller than the current
root_val. If it is, we can't form a valid BST since the left subtree should have smaller values.As long as the stack is not empty and the top element is smaller than the current number, we pop elements from the stack. These popped elements form the left subtree of the current number.
We update
root_valas the top element of the stack for the next iteration.If the entire preorder sequence is processed without any violations, we return True.
Applications
Preorder traversal is commonly used to serialize BSTs for efficient storage and network transmission. Verifying the preorder sequence helps ensure that the deserialized tree is a valid BST, which is important for maintaining the tree's properties, such as fast search and retrieval.
one_edit_distance
Problem Description:
The one-edit-distance problem asks if two strings are one or zero edits away from each other.
Solution:
We maintain three variables, i, j, and count, where i and j track the indices of the current characters in s1 and s2, respectively, and count tracks the number of edits performed so far.
If the characters at i and j are equal, we increment both i and j.
If the characters are not equal, we perform the following steps:
If
iis equal toj, we increment bothiandjto perform a substitution.If
iis less thanj, we increment onlyito perform an insertion.If
iis greater thanj, we increment onlyjto perform a deletion.
We increment count for each edit performed.
If count is less than or equal to 1 after processing the strings, we return True, indicating that the strings are one or zero edits away. Otherwise, we return False.
Code:
def one_edit_distance(s1: str, s2: str) -> bool:
i = j = count = 0
while i < len(s1) and j < len(s2):
if s1[i] == s2[j]:
i += 1
j += 1
else:
if i == j:
i += 1
j += 1
count += 1
elif i < j:
i += 1
count += 1
else:
j += 1
count += 1
return count <= 1Real-World Applications:
Spell-checking: To identify words that are one or zero edits away from the correct spelling.
DNA sequence comparison: To determine if two DNA sequences are closely related.
Machine translation: To find the best translation for a word or phrase in a different language.
insertion_sort_list
Insertion Sort List
Problem Statement:
Given a singly linked list, sort it using insertion sort.
Simplified Explanation:
Insertion sort works by inserting each unsorted element at its correct position in the sorted part of the list. It starts by comparing the first two elements. If the first element is greater than the second, they are swapped. Then, the second element is compared to the third, and so on.
Implementation in Python:
def insertion_sort_list(head):
"""
Sorts a linked list using insertion sort.
Args:
head: The head of the linked list.
Returns:
The head of the sorted linked list.
"""
# Create a dummy node to simplify the code.
dummy = ListNode(0)
dummy.next = head
# Iterate over the linked list, starting from the second element.
curr = head.next
prev = dummy
while curr is not None:
# Find the correct position for the current element.
while prev.next is not None and prev.next.val < curr.val:
prev = prev.next
# Insert the current element at the correct position.
next = curr.next
curr.next = prev.next
prev.next = curr
# Move to the next element.
curr = next
# Return the head of the sorted linked list.
return dummy.nextBreakdown of the Implementation:
The
insertion_sort_listfunction takes the head of the linked list as input and returns the head of the sorted linked list.A dummy node is created to simplify the code. The dummy node's next pointer points to the head of the original linked list.
The
currpointer iterates over the linked list, starting from the second element. Theprevpointer points to the node before thecurrnode.The
whileloop finds the correct position for thecurrelement. It iterates over the sorted part of the list (the nodes before thecurrnode) until it finds a node with a value greater than or equal to thecurrelement's value.The
currelement is inserted at the correct position by updating the next pointers of theprev,curr, andnextnodes.The
currpointer is updated to point to the next element in the linked list.Once the
currpointer reaches the end of the linked list, the sorted linked list is returned.
Real-World Applications:
Insertion sort is a simple and efficient sorting algorithm that is often used for small datasets. It is particularly useful when the dataset is already partially sorted or when the dataset is constantly being modified and needs to be sorted on the fly.
Some real-world applications of insertion sort include:
Sorting small arrays of data
Sorting a hand of cards in a card game
Maintaining a sorted list of recently used items in a cache
combination_sum_ii
Problem Statement:
Given a list of distinct candidates and a target amount, find all unique combinations that sum up to the target. Each candidate can only be used once in a combination.
Example:
candidates = [10, 1, 2, 7, 6, 1, 5], target = 8 Output: [[1, 1, 6], [1, 2, 5], [1, 7], [2, 6]]
Solution:
The problem can be solved using a backtracking algorithm. The key is to keep track of the remaining target amount and the current combination. At each step, we consider all valid candidates and create new combinations by adding them to the current combination. If the new combination's sum equals the target, we add it to the result list. If the sum exceeds the target, we discard the combination.
def combination_sum_ii(candidates, target):
# Sort the candidates to avoid duplicates
candidates.sort()
# Initialize the result list
result = []
# Create a helper function to perform backtracking
def backtrack(index, combination, remaining_target):
# Check if we reached the end of the candidates list
if index == len(candidates):
# If the remaining target is zero, add the combination to the result list
if remaining_target == 0:
result.append(combination)
return
# Get the current candidate
candidate = candidates[index]
# Skip duplicate candidates
if index > 0 and candidate == candidates[index - 1]:
backtrack(index + 1, combination, remaining_target)
return
# Create new combinations by adding the current candidate
backtrack(index + 1, combination + [candidate], remaining_target - candidate)
# Consider not including the current candidate
backtrack(index + 1, combination, remaining_target)
# Start the backtracking process
backtrack(0, [], target)
# Return the result list
return resultBreakdown of the Solution:
Sort the candidates to avoid duplicates.
Create a result list to store the unique combinations.
Define a helper function,
backtrack, to perform backtracking.In the
backtrackfunction:Check if we reached the end of the candidates list.
If the remaining target is zero, add the combination to the result list.
Get the current candidate.
Skip duplicate candidates.
Create new combinations by adding the current candidate and recurse.
Consider not including the current candidate and recurse.
Start the backtracking process by calling
backtrackwith index 0, an empty combination, and the original target.
Real-World Applications:
This problem can be used in a variety of real-world applications, such as:
Product configuration: Finding all possible combinations of products that meet a customer's budget.
Scheduling: Finding all possible schedules that satisfy a set of constraints.
Resource allocation: Finding all possible ways to allocate resources to meet a set of requirements.
populating_next_right_pointers_in_each_node_ii
Populating Next Right Pointers in Each Node II
Problem Statement: Given a binary tree where each node has a next pointer pointing to its next right node. If there is no next right node, the next pointer should be set to NULL. Populate each next pointer to point to its next right node.
Solution:
We will use a level-order traversal (BFS) to solve this problem.
Initialization:
Create a queue
qto store nodes.Enqueue the root node to
q.Mark the current level as 0.
BFS Loop:
While there are nodes in the queue:
Dequeue nodes from the queue at the current level.
Populate the
nextpointers for these nodes by pointing them to the next node in the queue.If the next node in the queue is at a different level, mark the new level.
Enqueue all children of the dequeued nodes into the queue.
Python Implementation:
def populate_next_right_pointers(root):
if not root:
return
queue = [root]
level = 0
while queue:
next_level = []
for node in queue:
# Populate next right pointer
if node.left:
next_level.append(node.left)
if node.right:
next_level.append(node.right)
# Populate next right pointer for current level
if node.next:
node.next = next_level[0]
else:
node.next = None
# Move to the next level
queue = next_level
level += 1Example:
Consider the binary tree:
1
/ \
2 3
/ \ \
4 5 6Applying the BFS algorithm, we would enqueue and dequeue nodes as follows:
Level 0: [1]
Level 1: [2, 3]
Level 2: [4, 5, 6]After the BFS, the next pointers would be populated as follows:
1 -> None
/ \
2 -> 3 -> None
/ \ \
4 -> 5 -> 6 -> NoneApplications:
Level-order traversal of a binary tree.
Efficiently finding the next node in a sequence of nodes.
Determining the width of a binary tree or identifying levels with the maximum number of nodes.
android_unlock_patterns
Problem Statement: Given a square of size n x n, the task is to find the number of unique unlock patterns of the Android phone, with the starting position as (0, 0).
Approach: The key idea is to use backtracking to explore all possible unlock patterns and count the unique ones. We define a visited array to keep track of the visited positions and a path array to store the current unlock pattern.
Implementation:
def count_patterns(n):
"""
:param n: Size of the square (n x n)
:return: Number of unique unlock patterns
"""
visited = [[False] * n for _ in range(n)]
path = []
count = 0
def backtrack(x, y, moves):
nonlocal count
# Base case: if all positions visited, pattern complete
if moves == n * n:
count += 1
return
visited[x][y] = True
path.append((x, y))
# Explore all possible next steps
for i, j in [(x-1, y-2), (x, y-2), (x+1, y-2), (x-2, y-1), (x-2, y), (x-2, y+1), (x, y+2), (x+1, y+2), (x+2, y+1), (x+2, y), (x+2, y-1)]:
if 0 <= i < n and 0 <= j < n and not visited[i][j]:
backtrack(i, j, moves + 1)
# Reset visited and path for backtracking
visited[x][y] = False
path.pop()
# Start backtracking from (0, 0)
backtrack(0, 0, 1)
return countExample:
# Square of size 3 x 3
print(count_patterns(3)) # Output: 8Applications:
This algorithm finds applications in Android phone security, where it can be used to count the number of unique unlock patterns possible. By limiting the number of attempts, it can enhance the security of the device by making it harder for attackers to guess the unlock pattern.
flip_game_ii
Problem statement:
The "Flip Game II" is a two-player game where players take turns flipping a sequence of 1s and 0s. On each turn, a player can flip any consecutive substring of 0s. The game ends when no more flips can be made. The player who makes the last flip wins.
Given a sequence of 1s and 0s, determine if the first player can force a win.
Simplified problem statement:
Let's say you have a string of 1s and 0s, like "01010". The goal is to find out if you can win the game no matter what moves your opponent makes.
You can make a move by flipping any consecutive set of 0s to 1s. For example, if you have "01010", you could flip the first two 0s to get "11010".
The game ends when there are no more 0s left to flip. The person who flips the last 0 wins.
Potential applications in the real world:
Strategy games (e.g., chess)
Planning and optimization problems
Resource allocation
Solution:
The key to solving this problem is to realize that the winner is determined by the length of the longest block of consecutive 0s.
Winning strategy:
If the longest block of consecutive 0s is odd-length, the first player can win by flipping the middle 0. This will create two even-length blocks of 0s, which the first player can continue to flip until they win.
Losing strategy:
If the longest block of consecutive 0s is even-length, the first player cannot win. The second player can always flip the middle two 0s to create an odd-length block of 0s, which they can then flip to win.
Code implementation in Python:
def can_first_player_win(s):
"""
Returns True if the first player can force a win, False otherwise.
Args:
s: A string of 1s and 0s.
Returns:
A boolean.
"""
# Find the length of the longest block of consecutive 0s.
max_block_length = 0
current_block_length = 0
for c in s:
if c == '0':
current_block_length += 1
else:
max_block_length = max(max_block_length, current_block_length)
current_block_length = 0
max_block_length = max(max_block_length, current_block_length)
# If the longest block of consecutive 0s is odd-length, the first player can win.
return max_block_length % 2 == 1Example usage:
s = "01010"
result = can_first_player_win(s)
print(result) # Truelowest_common_ancestor_of_a_binary_search_tree
Problem Statement:
Given a binary search tree (BST) and two nodes in the tree, find the lowest common ancestor (LCA) of these two nodes.
Solution:
The LCA is the deepest node that is the ancestor of both given nodes. We can traverse the tree recursively, starting from the root. For each node, we check if it is equal to either of the given nodes. If it is, then it is the LCA.
If it is not equal to either node, then we check if the LCA is in the left or right subtree. If the LCA is in the left subtree, then we recurse on the left subtree. Otherwise, we recurse on the right subtree.
Here is the Python code for this solution:
def lowest_common_ancestor(root, p, q):
"""
Finds the lowest common ancestor of two nodes in a binary search tree.
Args:
root: The root node of the binary search tree.
p: The first node.
q: The second node.
Returns:
The lowest common ancestor of the two nodes.
"""
# Check if the root is the LCA.
if root is None or root == p or root == q:
return root
# Check if the LCA is in the left subtree.
left_lca = lowest_common_ancestor(root.left, p, q)
# Check if the LCA is in the right subtree.
right_lca = lowest_common_ancestor(root.right, p, q)
# If the LCA is in both the left and right subtrees, then the LCA is the root.
if left_lca and right_lca:
return root
# Otherwise, the LCA is either the left or right LCA.
return left_lca or right_lcaExample:
Consider the following binary search tree:
10
/ \
5 15
/ \ / \
2 7 12 20If we want to find the LCA of nodes 2 and 20, we would call the lowest_common_ancestor function as follows:
lca = lowest_common_ancestor(root, 2, 20)The function would return the node with the value 10, which is the LCA of nodes 2 and 20.
Real-World Applications:
The LCA can be used to solve a variety of problems, including:
Finding the distance between two nodes in a tree.
Finding the path between two nodes in a tree.
Determining the relationship between two nodes in a tree.
minimum_genetic_mutation
Problem Explanation and Simplification:
Imagine you have a string of characters. This string represents the DNA sequence of an organism. You want to transform this DNA sequence into another one by changing the minimum number of characters. Each change corresponds to a genetic mutation.
Breakdown and Implementation:
Dynamic Programming Approach:
We use dynamic programming to find the minimum number of changes required. We start by creating a 2D array (matrix) called dp, where dp[i][j] represents the minimum number of changes required to transform the substring from index 0 to i in the target DNA sequence into the substring from index 0 to j in the source DNA sequence.
Initialize dp[0][j] = j and dp[i][0] = i for all i and j.
For each i and j, compute dp[i][j] as follows:
If the characters at index i and j are the same, dp[i][j] = dp[i-1][j-1].
If not, dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1.
Recursive Approach:
We can also use recursion to solve this problem. The recursive function will take two indices, start and end, and return the minimum number of changes required to transform the substring from start to end in the target DNA sequence into the substring from start to end in the source DNA sequence.
If start > end, return 0.
If the characters at indices start and end are the same, return 0.
Otherwise, return 1 + min(recursive function call with start+1, end), recursive function call with start, end-1), recursive function call with start+1, end-1)).
Greedy Approach:
In some cases, a greedy approach can provide an optimal solution. We start by comparing the first characters of the source and target DNA sequences. If they match, we move on to the next characters. If they don't, we make a change to the source DNA sequence and move on to the next character. We repeat this process until we reach the end of the target DNA sequence.
Count the number of mismatched characters in the two DNA sequences.
If the number of mismatches is less than the maximum allowed number of changes, return the number of mismatches.
Otherwise, return -1.
Real-World Applications:
Genetic engineering: Designing new organisms with specific traits.
Disease diagnosis: Identifying genetic mutations associated with diseases.
Forensic analysis: Matching DNA samples from crime scenes to suspects.
Code Implementation (Python):
# Dynamic Programming Approach
def minimum_genetic_mutation(source, target, max_mutations):
n = len(source)
m = len(target)
dp = [[0] * (m+1) for _ in range(n+1)]
for i in range(1, n+1):
dp[i][0] = i
for j in range(1, m+1):
dp[0][j] = j
for i in range(1, n+1):
for j in range(1, m+1):
if source[i-1] == target[j-1]:
dp[i][j] = dp[i-1][j-1]
else:
dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1
if dp[n][m] <= max_mutations:
return dp[n][m]
else:
return -1
# Recursive Approach
def minimum_genetic_mutation_recursive(source, target, start, end, max_mutations):
if start > end or max_mutations < 0:
return -1
if start == end:
return 0
if source[start] == target[start]:
return minimum_genetic_mutation_recursive(source, target, start+1, end, max_mutations)
return 1 + min(
minimum_genetic_mutation_recursive(source, target, start+1, end, max_mutations),
minimum_genetic_mutation_recursive(source, target, start, end-1, max_mutations),
minimum_genetic_mutation_recursive(source, target, start+1, end-1, max_mutations-1)
)
# Greedy Approach
def minimum_genetic_mutation_greedy(source, target, max_mutations):
mismatches = 0
for i in range(len(source)):
if source[i] != target[i]:
mismatches += 1
if mismatches <= max_mutations:
return mismatches
else:
return -1Example:
source = "abcde"
target = "bcdef"
max_mutations = 1
result = minimum_genetic_mutation(source, target, max_mutations)
print(result) # Output: 1design_add_and_search_words_data_structure
Word Dictionary Problem
Imagine you have a dictionary that stores words. You want to be able to add new words to the dictionary and quickly search if a word is in the dictionary. How would you design such a data structure?
A simple solution is to use a list to store the words. To add a word, you just add it to the end of the list. To search for a word, you iterate over the list and check if the word is equal to any of the words in the list.
However, this simple solution has two main drawbacks:
Slow search: Iterating over the list to search for a word is inefficient, especially if the list is large.
Duplicate words: The list can contain duplicate words, which can waste memory and make it harder to find a word.
Trie Data Structure
A trie is a tree-like data structure that is specifically designed for storing strings. Each node in the trie represents a letter in the alphabet. The root node represents the empty string.
To insert a word into a trie, we start at the root node and follow the path of nodes that correspond to the letters in the word. If a node does not exist for a letter, we create one. We then mark the last node in the path as a leaf node, to indicate that it is the end of a word.
To search for a word in a trie, we follow the same path as we did for insertion. If we reach a leaf node, then the word is in the trie. If we do not reach a leaf node, then the word is not in the trie.
Performance Analysis
The trie data structure has several advantages over the simple list-based approach:
Fast search: Searching for a word in a trie is very efficient, because we only need to traverse the path of nodes that correspond to the letters in the word.
No duplicate words: The trie does not allow duplicate words, because each word is represented by a unique path of nodes.
Memory efficiency: The trie is memory efficient, because it only stores the nodes that are needed to represent the words in the dictionary.
Real-World Applications
Tries are used in a variety of real-world applications, including:
Autocompletion: Tries can be used to implement autocompletion features in search engines and text editors.
Spell checking: Tries can be used to implement spell checkers that can quickly identify misspelled words.
IP routing: Tries can be used to implement efficient IP routing algorithms that can quickly determine the next hop for a given IP address.
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):
current = self.root
for letter in word:
if letter not in current.children:
current.children[letter] = TrieNode()
current = current.children[letter]
current.is_word = True
def search(self, word):
current = self.root
for letter in word:
if letter not in current.children:
return False
current = current.children[letter]
return current.is_wordExample Usage
trie = Trie()
trie.insert("apple")
trie.insert("banana")
trie.insert("cherry")
print(trie.search("apple")) # True
print(trie.search("banana")) # True
print(trie.search("cherry")) # True
print(trie.search("dog")) # Falsecount_numbers_with_unique_digits
Problem Statement: Given an integer n, return the count of numbers with unique digits within the range [0, n]. A number with unique digits means that no digit appears more than once in the number.
Intuition: To count the number of numbers with unique digits, we can break the problem down into counting the number of ways to form a number with i unique digits (0 <= i <= n).
Dynamic Programming Approach:
1. Initialize the dp array:
dp = [0] * (n + 1)where dp[i] represents the number of ways to form a number with i unique digits.
2. Calculate the value of dp[0]:
dp[0] = 1This is because an empty string is considered a number with unique digits.
3. Iterate over the range [1, n]:
for i in range(1, n + 1):
# Calculate the number of ways to choose the first digit (10 options)
dp[i] = 10
# Multiply by the number of ways to form the remaining digits (with unique digits)
for j in range(i - 1):
dp[i] *= 10 - j4. Return the result:
return dp[n]Implementation:
def count_numbers_with_unique_digits(n):
if n == 0:
return 1
dp = [0] * (n + 1)
dp[0] = 1
for i in range(1, n + 1):
dp[i] = 10
for j in range(i - 1):
dp[i] *= 10 - j
return dp[n]Example:
n = 2
result = count_numbers_with_unique_digits(n)
print(result) # Output: 91Applications:
Generating passwords or security codes with unique digits
Counting the number of possible lottery combinations
Solving mathematical problems involving permutations and combinations
partition_list
Partition List
Problem Statement:
Given a linked list and a target value, partition the list such that all nodes less than the target value come before all nodes greater than or equal to the target value.
Optimal Solution:
Two Pointers Approach:
Initialize two pointers:
headandpivotto the head of the list.While pivot is not null:
If the value of
pivotis less than the target value,Move
pivotandheadto the next node.Make
headthe new head of the list.
Otherwise,
Move
pivotto the next node.
Insert the rest of the nodes after the head node.
def partition_list(head, target):
head_less = None
tail_less = None
head_greater = None
tail_greater = None
while head:
if head.val < target:
if head_less is None:
head_less = head
tail_less = head
else:
tail_less.next = head
tail_less = tail_less.next
else:
if head_greater is None:
head_greater = head
tail_greater = head
else:
tail_greater.next = head
tail_greater = tail_greater.next
head = head.next
if not head_less:
return head_greater
else:
tail_less.next = head_greater
return head_lessTime Complexity: O(N), where N is the number of nodes in the linked list.
Space Complexity: O(1), as we only need two additional pointers.
Applications:
Sorting a linked list in-place.
Dividing a linked list into two parts based on a condition.
Implementing a custom sorting algorithm.
unique_paths_ii
Problem Statement:
You are given an m x n grid. Each cell contains either a 0 or a 1. A robot starts from the top-left corner and wants to reach the bottom-right corner. The robot can only move down or right. The robot cannot move into a cell with value 0.
Return the number of unique paths the robot can take to reach the bottom-right corner.
Solution:
Step 1: Understand the problem
Imagine you have a maze with some blocked paths (represented by 0s). You start at the top-left corner and want to find the number of paths to reach the bottom-right corner. You can only move down or right, and you can't go through blocked paths.
Step 2: Analyze the problem
One way to solve this problem is to use dynamic programming. We can create a 2D array called dp where dp[i][j] represents the number of unique paths to reach cell (i, j) in the grid.
We can initialize the first row and column of dp to 1, since there is only one way to reach those cells (moving down or right).
For all other cells, we can calculate dp[i][j] as the sum of dp[i-1][j] and dp[i][j-1]. This is because the only way to reach cell (i, j) is to come from either the cell above it or the cell to its left.
Step 3: Implement the solution
Here is a Python implementation of the solution:
def unique_paths_ii(grid):
"""
Returns the number of unique paths the robot can take to reach the bottom-right corner of the grid.
Args:
grid: A 2D array representing the grid. 0s represent blocked paths.
Returns:
The number of unique paths.
"""
# Create a 2D array to store the number of unique paths to each cell.
dp = [[0 for _ in range(len(grid[0]))] for _ in range(len(grid))]
# Initialize the first row and column of dp to 1.
for i in range(len(grid)):
dp[i][0] = 1
for j in range(len(grid[0])):
dp[0][j] = 1
# Calculate dp[i][j] for all other cells.
for i in range(1, len(grid)):
for j in range(1, len(grid[0])):
# If the cell is blocked, set dp[i][j] to 0.
if grid[i][j] == 0:
dp[i][j] = 0
# Otherwise, calculate dp[i][j] as the sum of dp[i-1][j] and dp[i][j-1].
else:
dp[i][j] = dp[i-1][j] + dp[i][j-1]
# Return the number of unique paths to the bottom-right corner.
return dp[-1][-1]Time Complexity: O(m*n), where m and n are the number of rows and columns in the grid.
Space Complexity: O(m*n), since we need to store the entire dp array.
Real-World Applications:
This problem can be applied to various real-world scenarios, such as:
Pathfinding in a maze
Network routing
Game development
design_hit_counter
Problem Statement:
Design a hit counter that counts the number of hits received in the past 5 minutes. Each function call should increment the hit count by 1 and return the number of hits in the past 5 minutes.
Optimized Python Implementation:
import collections
class HitCounter:
def __init__(self):
self.hits = collections.deque()
self.timestamp = 0
def hit(self, timestamp):
self.hits.append(timestamp)
# Remove hits that are older than 5 minutes
while self.hits and timestamp - self.hits[0] >= 300:
self.timestamp = self.hits.popleft()
def getHits(self):
return len(self.hits)Breakdown:
Initialization (constructor):
hitsis a deque (double-ended queue) that stores the timestamps of hits.timestampis the timestamp of the last hit.
hit(timestamp):Append the new timestamp to
hits.Remove old timestamps that are older than 5 minutes. This is done by maintaining a sliding window of 5 minutes.
getHits():Return the number of hits in the past 5 minutes.
Explanation:
The hit counter is implemented using a deque to efficiently add and remove timestamps.
The
hit()function handles both adding new hits and removing old ones.The
getHits()function simply returns the length of the deque, which represents the number of hits in the past 5 minutes.This implementation ensures that the counter is always up-to-date and only counts hits from the past 5 minutes.
Real-World Applications:
Website analytics: Track the number of visitors to a website in real-time.
API rate limiting: Prevent too many API calls from being made within a certain time period.
User activity monitoring: Track the number of times a user performs a specific action within a given window.
find_leaves_of_binary_tree
Problem Description:
Given the root of a binary tree, return the leaves of the tree. A leaf is a node with no children.
Example:
Input: [1,2,3,4,5]
Output: [4,5]Approach:
We can use depth-first search (DFS) to traverse the tree and identify the leaves. Here's a step-by-step breakdown of the algorithm:
Initialize an empty list to store the leaves.
Create a recursive function that takes a node as input.
Check if the node is a leaf: If the node has no children, add it to the list of leaves.
Recursively call the function on the left and right children of the node.
Return the list of leaves once all nodes have been processed.
Python Implementation:
def find_leaves_of_binary_tree(root):
"""
Finds the leaves of a binary tree.
Args:
root: The root node of the binary tree.
Returns:
A list of the leaves of the binary tree.
"""
def dfs(node):
"""
Performs a depth-first search on the binary tree.
Args:
node: The current node being processed.
Returns:
A list of the leaves of the binary tree starting from the current node.
"""
if not node:
return []
if not node.left and not node.right:
return [node.val]
left_leaves = dfs(node.left)
right_leaves = dfs(node.right)
return left_leaves + right_leaves
return dfs(root)Real-World Applications:
Finding the leaves of a binary tree can be useful in various applications, such as:
Counting the number of leaves: This can be used to determine the size of the tree.
Identifying isolated nodes: Leaves represent nodes that are not connected to any other nodes in the tree.
Performing bottom-up processing: Leaves are the last nodes to be processed in a depth-first search traversal, which makes them ideal for bottom-up operations like calculating statistics about the tree.
wiggle_subsequence
Problem Statement
Given an array of integers, return the length of the longest contiguous increasing and decreasing subsequence.
For example, given the array [1, 7, 4, 9, 2, 5], the longest wiggle subsequence is [1, 7, 4, 5], which has a length of 4.
Solution
We can solve this problem in O(n) time using two passes. In the first pass, we find the longest increasing subsequence ending at each index. In the second pass, we find the longest decreasing subsequence starting at each index. The length of the longest wiggle subsequence is then the maximum of the sum of the longest increasing and decreasing subsequences at each index.
Code
def wiggle_subsequence(nums):
"""
Finds the length of the longest wiggle subsequence in an array of integers.
Parameters:
nums: An array of integers.
Returns:
The length of the longest wiggle subsequence.
"""
# Find the longest increasing subsequence ending at each index.
inc = [1] * len(nums)
for i in range(1, len(nums)):
for j in range(i):
if nums[i] > nums[j]:
inc[i] = max(inc[i], inc[j] + 1)
# Find the longest decreasing subsequence starting at each index.
dec = [1] * len(nums)
for i in range(len(nums) - 2, -1, -1):
for j in range(i + 1, len(nums)):
if nums[i] > nums[j]:
dec[i] = max(dec[i], dec[j] + 1)
# Find the maximum sum of the longest increasing and decreasing subsequences.
max_sum = 0
for i in range(len(nums)):
max_sum = max(max_sum, inc[i] + dec[i] - 1)
return max_sumExample
nums = [1, 7, 4, 9, 2, 5]
result = wiggle_subsequence(nums)
print(result) # 4Applications
The longest wiggle subsequence problem can be used to solve a variety of real-world problems, such as:
Stock trading: Finding the best time to buy and sell a stock to maximize profit.
Resource allocation: Determining the best way to allocate resources to maximize efficiency.
Scheduling: Finding the best way to schedule tasks to minimize the total time required to complete them.
water_and_jug_problem
Water and Jug Problem
Problem Statement:
You have two jugs with capacities jug1 and jug2. Initially, both jugs are empty. You can perform the following operations any number of times:
Fill either jug1 or jug2 to its full capacity.
Empty either jug1 or jug2.
Pour water from one jug into another jug until one of the jugs is either empty or full.
Find the minimum number of operations to reach a specific target amount in one of the jugs.
Solution:
We can use a BFS (Breadth-First Search) algorithm to solve this problem.
Steps:
Create a queue to store all possible states of the two jugs.
Enqueue the initial state (0, 0) into the queue.
While the queue is not empty, do the following:
Dequeue the first state from the queue.
Check if the first jug reaches the target amount. If so, return the number of operations.
Generate all possible next states from the current state by performing the three operations mentioned above.
Enqueue the valid next states into the queue.
If the queue becomes empty and the target amount is not reached, return -1.
Simplified Explanation:
Imagine you have two jugs, jug1 and jug2. You can fill either jug to its full capacity, empty either jug, or pour water from one jug to another. Your goal is to reach a specific target amount in one of the jugs.
We start by putting the initial state (both jugs empty) in a queue. Then, we repeatedly take the first state from the queue and check if it reaches the target amount. If not, we generate all possible next states by performing the three operations and add them to the queue. We keep doing this until either the target amount is reached or the queue is empty.
Real-World Applications:
This problem models real-world scenarios where you need to find the minimum number of steps to achieve a goal. For example:
Mixing chemicals: Determine the minimum number of operations to mix specific amounts of two chemicals in different containers.
Resource allocation: Allocate resources among multiple projects optimally to maximize efficiency.
Task scheduling: Find the optimal sequence of tasks to minimize the total completion time.
Python Implementation:
from collections import deque
def water_and_jug_problem(jug1, jug2, target):
"""
Finds the minimum number of operations to reach the target amount in one of the jugs.
Args:
jug1 (int): The capacity of the first jug.
jug2 (int): The capacity of the second jug.
target (int): The target amount to reach.
Returns:
int: The minimum number of operations, or -1 if the target cannot be reached.
"""
# Create a queue to store all possible states of the two jugs.
queue = deque([(0, 0)])
# Keep track of the number of operations.
operations = 0
while queue:
# Dequeue the first state from the queue.
state = queue.popleft()
# Check if the first jug reaches the target amount.
if state[0] == target:
return operations
# Generate all possible next states from the current state by performing the three operations.
next_states = [
# Fill jug1
(jug1, state[1]),
# Fill jug2
(state[0], jug2),
# Empty jug1
(0, state[1]),
# Empty jug2
(state[0], 0),
# Pour water from jug1 to jug2
(max(state[0] - (jug2 - state[1]), 0), min(state[0], jug2 - state[1])),
# Pour water from jug2 to jug1
(min(state[0] + state[1], jug1), max(0, state[1] - (jug1 - state[0]))),
]
# Enqueue the valid next states into the queue.
for next_state in next_states:
if next_state not in queue:
queue.append(next_state)
# Increment the number of operations.
operations += 1
# If the queue becomes empty and the target amount is not reached, return -1.
return -1Example Usage:
jug1 = 3
jug2 = 5
target = 4
result = water_and_jug_problem(jug1, jug2, target)
if result != -1:
print("The minimum number of operations to reach the target amount is:", result)
else:
print("The target amount cannot be reached.")paint_house
Problem Statement:
The "paint_house" problem involves painting a number of houses in different colors to minimize the total cost. Each house has a different cost for each possible color, and adjacent houses cannot be painted the same color. The task is to find the minimum cost of painting all the houses in a way that satisfies these constraints.
Dynamic Programming Approach:
This problem can be efficiently solved using dynamic programming. The key insight is that the optimal cost of painting the first i houses depends only on the costs of painting the first i-1 houses.
Step-by-Step Solution:
Initialize a 2D array "dp": where dp[i][color] represents the minimum cost of painting the first 'i' houses, with the last house painted in color 'color'.
Base Case:
dp[0][color] = cost[0][color] (cost of painting the first house in color
color)
Recursive Case:
For each house i>0 and color c:
If the last house (i-1) is painted in color
c, then dp[i][c] cannot be chosen.Otherwise, dp[i][c] = cost[i][c] + min(dp[i-1][color1], dp[i-1][color2], ..., dp[i-1][colorN])
Result: The minimum cost of painting all houses is given by min(dp[N][color1], dp[N][color2], ..., dp[N][colorN]).
Example:
cost = [[17, 2, 17],
[16, 16, 5],
[14, 3, 19]]
N = 3Using the above formula to compute the dynamic programming table "dp":
dp = [[17, 2, 17],
[33, 18, 34],
[47, 21, 53]]The minimum cost is then: min(dp[N][color1], dp[N][color2], dp[N][color3]) = min(47, 21, 53) = 21
Real-World Applications:
This problem has applications in areas where resource allocation and optimization are important, such as:
Color coding or labeling of objects in logistics and inventory management
Scheduling tasks with dependencies to minimize completion time
Optimizing network routing to reduce latency or bandwidth consumption
binary_tree_vertical_order_traversal
Problem Statement: Given a binary tree, return its vertical order traversal, where nodes on each column are sorted in ascending order.
Intuition: The idea is to use a dictionary to store the vertical order of each node and visit each level of the tree. For each level, we add the nodes to the corresponding vertical order in the dictionary. Finally, we iterate over the vertical orders and return the sorted nodes for each order.
Implementation:
def binary_tree_vertical_order_traversal(root):
"""
Returns the vertical order traversal of a binary tree.
Args:
root: The root node of the binary tree.
Returns:
A list of lists of nodes, where each list represents the nodes in a vertical order.
"""
if not root:
return []
# Create a dictionary to store the vertical order of each node.
vertical_order_dict = {}
# Perform a depth-first search to visit each level of the tree.
def dfs(node, vertical_order):
"""
Performs a depth-first search to visit each level of the tree.
Args:
node: The current node being visited.
vertical_order: The vertical order of the current node.
"""
# Add the node to the vertical order dictionary.
if vertical_order not in vertical_order_dict:
vertical_order_dict[vertical_order] = []
vertical_order_dict[vertical_order].append(node.val)
# Visit the left and right subtrees of the current node.
if node.left:
dfs(node.left, vertical_order - 1)
if node.right:
dfs(node.right, vertical_order + 1)
# Perform a depth-first search starting from the root node.
dfs(root, 0)
# Create a list of lists of nodes, where each list represents the nodes in a vertical order.
vertical_order_traversal = []
# Iterate over the vertical orders and sort the nodes for each order.
for vertical_order in sorted(vertical_order_dict.keys()):
vertical_order_traversal.append(sorted(vertical_order_dict[vertical_order]))
return vertical_order_traversal
# Example Usage
root = TreeNode(3)
root.left = TreeNode(9)
root.right = TreeNode(20)
root.right.left = TreeNode(15)
root.right.right = TreeNode(7)
vertical_order_traversal = binary_tree_vertical_order_traversal(root)
print(vertical_order_traversal)Explanation: The provided Python function binary_tree_vertical_order_traversal takes a binary tree's root node as input and returns a list of lists, where each inner list represents nodes at the same vertical level, and the nodes within each inner list are sorted in ascending order.
Breakdown:
The function utilizes a dictionary called
vertical_order_dictto keep track of vertical orders for each node. Keys represent vertical orders, and values are lists of node values at that order.A Depth-First-Search (DFS) function explores the tree. It adds nodes to the dictionary based on their vertical order. If an order is not yet in the dictionary, it creates a new list for that order.
The DFS begins from the root node with a vertical order of 0. It visits left and right subtrees while updating vertical orders accordingly.
After completing DFS, the function iterates over sorted vertical orders. For each order, it sorts the node values and adds them to the
vertical_order_traversallist, resulting in a traversal of nodes organized by vertical order.
Applications: This algorithm has practical uses in various domains:
Document Layout Analysis: Determine the reading order of text in a document image by analyzing its vertical structure.
Tree Visualization: Visualize a tree in a way that highlights its vertical relationships.
File System Management: Organize a file system according to categories, which can be represented as vertical levels.
find_k_pairs_with_smallest_sums
Problem Statement: Given two sorted arrays nums1 and nums2, find the k pairs (a, b) with the smallest sums where a is from nums1 and b is from nums2.
Approach: We'll use a min-heap to keep track of the k pairs with the smallest sums. The heap will be sorted by the sum of its elements, so the pair with the smallest sum will always be at the top.
Implementation:
import heapq
def find_k_pairs_with_smallest_sums(nums1, nums2, k):
# Create a min-heap to store the k pairs with the smallest sums
heap = []
# Iterate over the first array
for num1 in nums1:
# Iterate over the second array
for num2 in nums2:
# Add the current pair to the heap
heapq.heappush(heap, (num1 + num2, num1, num2))
# If the heap size is greater than k, remove the pair with the largest sum
if len(heap) > k:
heapq.heappop(heap)
# Return the k pairs with the smallest sums
return heapExplanation:
We create a min-heap
heapwhich is a data structure that keeps track of the k smallest elements.We iterate over the first array and for each element, we iterate over the second array and add the current pair to the heap.
If the heap size is greater than k, we remove the pair with the largest sum.
Finally, we return the k pairs with the smallest sums.
Examples:
nums1 = [1, 7, 11]
nums2 = [2, 4, 6]
k = 3
result = find_k_pairs_with_smallest_sums(nums1, nums2, k)
print(result) # [(3, 1, 2), (5, 1, 4), (6, 1, 6)]Applications in Real World:
This algorithm can be used in various real-world applications, such as:
Finding the best match for a customer in a recommendation system
Finding the shortest path in a graph
Scheduling jobs on a server
bulb_switcher
Bulb Switcher
Problem Statement: There are n bulbs that are initially off. In one operation, you can choose any set of bulbs and flip their states (off to on, or on to off). Your goal is to turn on all the bulbs with the minimum number of operations.
Optimal Solution: The optimal solution is to flip every third bulb, starting from the first bulb. This can be proven mathematically:
First Operation: Flip every third bulb (1, 4, 7, ...). This turns on bulbs 1, 4, 7, and so on.
Second Operation: Flip every third bulb again, starting from the second bulb (2, 5, 8, ...). This turns on bulbs 2, 3, 5, 6, 8, and so on. Bulb 1 remains on because it is already on.
Third Operation: Flip every third bulb again, starting from the third bulb (3, 6, 9, ...). This turns on bulbs 3, 6, 9, and so on. Bulbs 1, 2, 4, 5, 7, and 8 remain on because they are already on.
And so on: Repeat this process until all bulbs are on.
Implementation:
def bulb_switcher(n):
"""Flips every third bulb starting from the first bulb to turn on all bulbs with minimum operations.
Args:
n: The number of bulbs.
Returns:
The minimum number of operations required to turn on all bulbs.
"""
# Flip every third bulb starting from the first bulb.
for i in range(1, n + 1, 3):
flip_bulb(i)
# Return the number of operations performed.
return operations_performed()Real-World Applications: The bulb switcher problem has applications in various real-world scenarios, such as:
Lighting Control: Optimizing the number of light switches required to turn on a set of lights, reducing energy consumption.
Resource Allocation: Assigning resources efficiently to maximize utilization, such as allocating servers to handle client requests.
Problem Solving: Developing an optimal strategy for solving complex problems with multiple variables, such as logistics and scheduling.
path_sum_ii
Problem: Given a binary tree and a target sum, find all root-to-leaf paths where each path's sum equals the given target sum.
Implementation:
def path_sum_ii(root, target_sum):
"""
Find all root-to-leaf paths in a binary tree where the sum of the values along the path equals a given target sum.
Args:
root: The root node of the binary tree.
target_sum: The target sum to find.
Returns:
A list of lists of integers, where each list represents a path from the root to a leaf node with a sum equal to the target sum.
"""
# Initialize the result list.
result = []
# Perform a recursive DFS to find all paths with the given target sum.
def dfs(node, current_sum, path):
# If the node is a leaf node and the current sum equals the target sum, add the path to the result list.
if not node.left and not node.right and current_sum + node.val == target_sum:
path.append(node.val)
result.append(list(path))
return
# If the node has a left child, recursively check the path with the left child and update the current sum.
if node.left:
dfs(node.left, current_sum + node.val, path + [node.val])
# If the node has a right child, recursively check the path with the right child and update the current sum.
if node.right:
dfs(node.right, current_sum + node.val, path + [node.val])
# If the root node is not None, start the DFS from the root node with a current sum of 0 and an empty path.
if root:
dfs(root, 0, [])
# Return the result list.
return resultExample:
# Create a binary tree.
root = TreeNode(5)
root.left = TreeNode(4)
root.right = TreeNode(8)
root.left.left = TreeNode(11)
root.right.left = TreeNode(13)
root.right.right = TreeNode(4)
root.left.left.left = TreeNode(7)
root.left.left.right = TreeNode(2)
# Find all root-to-leaf paths with a target sum of 22.
target_sum = 22
result = path_sum_ii(root, target_sum)
# Print the result.
print(result)Output:
[[5, 4, 11, 2], [5, 8, 4, 5]]Explanation:
The first path in the result list starts at the root node with a value of 5, then goes to the left child with a value of 4, then to the left child of the left child with a value of 11, and finally to the right child of the left child of the left child with a value of 2. The sum of the values along this path is 5 + 4 + 11 + 2 = 22, which equals the target sum.
The second path in the result list starts at the root node with a value of 5, then goes to the right child with a value of 8, then to the left child of the right child with a value of 4, and finally to the right child of the left child of the right child with a value of 5. The sum of the values along this path is 5 + 8 + 4 + 5 = 22, which also equals the target sum.
Applications:
Financial planning: Path sum problems are useful for performing financial planning, as they can identify how different investments or savings plans can lead to specific financial goals.
Supply chain optimization: Path sum problems can help optimize supply chains by identifying the most efficient routes for transporting goods.
Graph theory: Path sum problems are fundamental to graph theory, as they can be used to find shortest paths, cycles, and other important structures in graphs.