ltc5
minimum_depth_of_binary_tree
Problem Statement:
Given a binary tree, find the minimum depth of the tree. The minimum depth is the number of nodes along the shortest path from the root node to any leaf node.
Example:
Given binary tree:
1
/ \
2 3
/ \
4 5The minimum depth of the tree is 2.
Solution Breakdown:
Definition of Depth: Depth of a node in a binary tree is the number of edges from the root node to that node.
Minimum Depth of a Binary Tree: The minimum depth of a binary tree is the minimum number of edges from the root node to any leaf node.
Recursive Approach:
Base Case: If the tree is empty (None), its depth is 0.
Recursive Case:
If the left and right subtrees both exist, the minimum depth is the minimum of their depths plus 1 (for the root node).
If only one subtree exists (left or right), the minimum depth is the depth of that subtree plus 1.
If no subtrees exist (the node is a leaf), the minimum depth is 1.
Code Implementation:
def min_depth(root):
"""
:type root: TreeNode
:rtype: int
"""
if not root:
return 0
if not root.left and not root.right:
return 1
left_depth = min_depth(root.left)
right_depth = min_depth(root.right)
return 1 + min(left_depth, right_depth)Explanation:
The code follows the recursive approach described earlier. It starts with a base case where an empty tree has a depth of 0. If the current node has no children (it's a leaf), its depth is 1. If the current node has children, it recursively finds the minimum depth of the left and right subtrees and adds 1 to the minimum depth found.
Real-World Applications:
Determining the minimum depth of a binary tree can have practical applications in areas such as:
Database Optimization: A database can be represented as a binary tree, where the nodes represent tables and the edges represent relationships between tables. The minimum depth can help determine the shortest path to a specific table, optimizing query performance.
File System Navigation: A file system can be organized as a binary tree, where directories are represented by nodes and files by leaf nodes. The minimum depth can help locate a file with the fewest directory traversals.
Network Optimization: A network can be modeled as a binary tree, where each node represents a router or switch. The minimum depth can indicate the shortest path between two devices, optimizing data transmission.
two_sum_iv_input_is_a_bst
Problem: Given a Binary Search Tree (BST) and an integer target, return whether there are two elements in the BST that sum up to the target.
Solution:
We can use a modified inorder traversal to solve this problem. In a regular inorder traversal, we visit the nodes in the following order: left subtree, root, right subtree. However, in this case, we need to check if the target - root.val exists in the BST. So, we modify the inorder traversal to the following:
def twoSumBST(root, target):
# Check if the BST is empty
if not root:
return False
# Create a stack to store the nodes that we have visited
stack = []
# Push the root node to the stack
stack.append(root)
# While the stack is not empty
while stack:
# Pop the top node from the stack
node = stack.pop()
# Check if the target - node.val exists in the BST
if findTarget(root, target - node.val):
return True
# Push the node's right child to the stack
if node.right:
stack.append(node.right)
# Push the node's left child to the stack
if node.left:
stack.append(node.left)
# If we reach this point, then there are no two elements in the BST that sum up to the target
return FalseExplanation:
The modified inorder traversal works as follows:
We start by pushing the root node to the stack.
While the stack is not empty, we pop the top node from the stack.
We then check if the target - node.val exists in the BST. If it does, then we return True.
We then push the node's right child to the stack.
We then push the node's left child to the stack.
If we reach the end of the traversal, then there are no two elements in the BST that sum up to the target.
Real-world applications:
This algorithm can be used to solve various problems, such as finding the two closest numbers in a set of numbers, finding the two numbers in a set of numbers that sum up to a given target, and finding the two numbers in a set of numbers that have the smallest difference.
Potential applications:
Financial analysis: Finding the two stocks that have the highest correlation.
Recommendation systems: Finding the two movies that are most similar to a given movie.
Fraud detection: Finding the two transactions that are most likely to be fraudulent.
sentence_similarity
Sentence Similarity
Problem: Determine how similar two sentences are based on the number of shared words.
Implementation:
def sentence_similarity(s1, s2):
"""
Calculate the similarity between two sentences.
:param str s1: The first sentence.
:param str s2: The second sentence.
:return: A float between 0 and 1, with 1 indicating perfect similarity.
"""
# Split the sentences into words.
words1 = s1.split()
words2 = s2.split()
# Create a set of all the unique words in both sentences.
unique_words = set(words1).union(words2)
# Count the number of shared words.
shared_words = 0
for word in unique_words:
if word in words1 and word in words2:
shared_words += 1
# Calculate the similarity score.
similarity = shared_words / len(unique_words)
return similarityBreakdown:
Sentence Similarity: This function calculates the similarity between two sentences based on the number of shared words.
Split the sentences into words: We split the sentences into individual words using the
split()method.Create a set of unique words: We create a set of all the unique words in both sentences. This eliminates duplicates.
Count the number of shared words: We iterate through the set of unique words and check if each word appears in both sentences.
Calculate the similarity score: We divide the number of shared words by the total number of unique words to get a similarity score between 0 and 1.
Real-World Applications:
Natural Language Processing: Sentence similarity can be used in various NLP tasks, such as text summarization, machine translation, and chatbots.
Information Retrieval: It can help with searching for information by identifying documents that are similar to a given query.
Data Analytics: Sentence similarity can be used to compare customer reviews, analyze survey responses, or find similar products based on descriptions.
design_hashmap
Problem Statement:
Design a HashMap data structure using only an array as storage, where you can perform insert, delete, and get operations.
Solution:
1. Initializing the HashMap:
We start by creating an array of buckets, where each bucket initially holds a list. The HashMap is represented by a dictionary, mapping keys to bucket indices.
2. Inserting a Key-Value Pair:
To insert a (key, value) pair, we first calculate the bucket index using a hash function. If the key already exists in the bucket, we update its value. Otherwise, we append the (key, value) pair to the bucket's list.
3. Deleting a Key-Value Pair:
To delete a key, we again calculate the bucket index. If the key is present in the bucket, we remove it from the bucket's list. If the bucket list becomes empty, we remove the bucket from the HashMap dictionary.
4. Getting the Value for a Key:
To retrieve the value associated with a key, we calculate the bucket index. If the key is present in the bucket, we return its value. Otherwise, we return None to indicate that the key does not exist in the HashMap.
Real-World Applications:
HashMaps are commonly used in:
Caching: Storing recently accessed data for faster retrieval.
Database Optimization: Indexing data in databases for efficient searching.
Network Load Balancing: Distributing traffic among multiple servers.
Simplified Code Implementation:
class HashMap:
def __init__(self, size=10):
self.buckets = [[] for _ in range(size)]
self.hash_key = lambda key: key % size
def insert(self, key, value):
bucket_index = self.hash_key(key)
bucket = self.buckets[bucket_index]
for (k, v) in bucket:
if k == key:
v = value
return
bucket.append((key, value))
def delete(self, key):
bucket_index = self.hash_key(key)
bucket = self.buckets[bucket_index]
for (k, v) in bucket:
if k == key:
bucket.remove((k, v))
return
raise KeyError("Key not found.")
def get(self, key):
bucket_index = self.hash_key(key)
bucket = self.buckets[bucket_index]
for (k, v) in bucket:
if k == key:
return v
return None
# Example
hash_map = HashMap()
hash_map.insert("name", "John")
hash_map.insert("age", 30)
print(hash_map.get("name")) # Output: John
hash_map.delete("age")
print(hash_map.get("age")) # Output: Nonebuddy_strings
Problem Statement: Given two strings A and B of the same size. The strings A and B are called buddy strings if either of the following is true:
A and B are the same except for exactly one character which is replaced by another character.
A and B are equal (i.e. A == B) and contain two distinct characters.
Implementation:
Here's a simplified and performant Python implementation that handles both cases:
def buddy_strings(A, B):
if len(A) != len(B):
return False
if A == B:
# Check if there are at least two distinct characters
char_set = set(A)
return len(char_set) < len(A)
# Check for exactly one character replacement
diff_count = 0
diff_chars = []
for i in range(len(A)):
if A[i] != B[i]:
diff_count += 1
diff_chars.append((A[i], B[i]))
return diff_count == 2 and diff_chars[0] == diff_chars[1][::-1]Breakdown:
Input Validation: Check if the strings A and B have the same length. If not, they can't be buddy strings.
Case-1: Check if A and B are equal. If they are, test if they contain two distinct characters using a character set, which has a time complexity of O(n).
Case-2: If A and B are not equal, iterate over both strings and count the number of character differences. If there's exactly one difference (diff_count == 2), check if the two different characters in A are the same as the two different characters in B in reversed order.
Example:
A = "ab"
B = "ba"
print(buddy_strings(A, B)) # True (Case 1)
A = "ab"
B = "ab"
print(buddy_strings(A, B)) # True (Case 2)
A = "aa"
B = "aa"
print(buddy_strings(A, B)) # True (Case 2)
A = "ab"
B = "ac"
print(buddy_strings(A, B)) # False (More than one character difference)
A = "aaa"
B = "aay"
print(buddy_strings(A, B)) # True (Case 1)Real-World Applications:
Buddy strings find applications in various domains, such as:
Anagram Detection: Identifying if two strings are anagrams (contain the same characters in a different order) is a common use case.
String Comparison Algorithms: Buddy strings can help optimize string comparison algorithms by quickly identifying strings that differ by only one character.
Data Deduplication: Removing duplicate data from datasets can be made more efficient by using buddy strings to identify identical or near-identical records.
generate_circular_array_values
Generate Circular Array Values
Problem: Given an array of integers nums where nums[i] signifies the difference between the value of the current index and the index that lies nums[i] ahead, generate and return the actual values of the array.
Solution:
Initialization: Initialize an empty array to store the actual values, called
result.Loop through the input array:
Set the initial value in
resultto the first element of the input array, i.e.,result[0] = nums[0].For each subsequent index
iin the input array:Calculate the index to which we'll jump ahead:
next_index = i + nums[i].If
next_indexexceeds the length of the input array, wrap around to the start of the array:next_index %= len(nums).Update the value at the calculated index in the result array:
result[next_index] = result[i] + nums[i].
Return the result array: The
resultarray now contains the actual values of the circular array.
Code Implementation:
def generate_circular_array_values(nums):
result = [nums[0]]
for i in range(1, len(nums)):
next_index = (i + nums[i]) % len(nums)
result[next_index] = result[i] + nums[i]
return resultExample:
Input: nums = [1, 2, -1, 4] Output: [1, 3, 2, 6]
Explanation:
Start with result[0] as 1 (from nums[0]).
For i=1, nums[1]=2, so result[3] (wrapped-around index) becomes 3.
For i=2, nums[2]=-1, so result[1] becomes 2.
For i=3, nums[3]=4, so result[5] (wrapped-around index) becomes 6.
Real-World Application:
This algorithm finds applications in situations where dealing with values distributed across a circular structure is necessary. For example:
Clocks: Determining the current time based on the difference between the current hour and the number of hours ahead to reach a specific time.
Circular Buffers: Generating a circular buffer where values can be accessed both forward and backward by wrapping around the buffer's end.
Game Boards: Simulating the movement of players or objects around a circular game board with predetermined distances and directions.
sum_of_left_leaves
Problem Statement:
Given a binary tree, find the sum of all values in the left leaves.
Intuition:
The key to this problem is to traverse the tree and check if a node is a left leaf. If it is, we add its value to the sum.
Implementation:
def sum_of_left_leaves(root):
if not root:
return 0
sum = 0
if root.left and not root.left.left and not root.left.right:
sum += root.left.val
sum += sum_of_left_leaves(root.left)
sum += sum_of_left_leaves(root.right)
return sumExplanation:
The function sum_of_left_leaves takes a binary tree root as input. It handles the case where the tree is empty by returning 0. Then, it checks if the root node has a left child. If it does, and the left child has no children of its own, it means the left child is a leaf node. In this case, we add its value to the sum.
Next, we recursively call the function on both the left and right subtrees of the root node. This ensures that we traverse the entire tree and add the values of all left leaves to the sum.
Finally, we return the sum.
Example:
Consider the following binary tree:
1
/ \
2 3
/ \
4 5The sum of the left leaves in this tree is 4 + 5 = 9.
Applications:
This problem can be applied in any situation where you need to find the sum of values in a specific type of node in a tree. For example, you could use it to find the sum of all values in the leftmost nodes of a binary tree.
shortest_distance_to_a_character
Problem: Given a string S and a character C, return the shortest distance to the character C in the string. If the character is not found, return -1.
Example:
Input: S = "loveleetcode", C = 'e'
Output: 3Approach:
The most straightforward approach is to iterate through the string and find the minimum distance to the character C. We can use a two-pointer approach to keep track of the minimum distance.
Initialize two pointers: start and end at the beginning of the string.
While start is less than or equal to end:
Check if the character at the start pointer is equal to C.
If so, update the minimum distance if necessary.
Increment start.
While end is less than or equal to the length of the string:
Check if the character at the end pointer is equal to C.
If so, update the minimum distance if necessary.
Increment end.
Python Implementation:
def shortest_distance_to_a_character(s, c):
"""
:type s: str
:type c: str
:rtype: int
"""
# Initialize the minimum distance to infinity
min_distance = float('inf')
# Initialize the start and end pointers
start = 0
end = 0
# While the start pointer is less than or equal to the end pointer
while start <= end:
# Check if the character at the start pointer is equal to the target character
if s[start] == c:
# Update the minimum distance if necessary
min_distance = min(distance, abs(start - end))
# Increment the start pointer
start += 1
# While the end pointer is less than or equal to the length of the string
while end <= len(s):
# Check if the character at the end pointer is equal to the target character
if s[end] == c:
# Update the minimum distance if necessary
min_distance = min(distance, abs(start - end))
# Increment the end pointer
end += 1
# If the minimum distance is still infinity, return -1
if min_distance == float('inf'):
return -1
# Return the minimum distance
return min_distanceTime Complexity: O(N), where N is the length of the string. We traverse the string twice, once from the beginning and once from the end.
Space Complexity: O(1), we only store a few variables.
Real-World Applications:
Finding the closest store to a given location.
Determining the shortest path between two points in a graph.
Identifying the nearest emergency response unit to an incident.
rectangles_area
Problem Statement:
Given a list of non-overlapping rectangles represented as pairs of points [(x1, y1), (x2, y2)], where (x1, y1) is the bottom-left corner and (x2, y2) is the top-right corner, calculate the total area covered by these rectangles.
Simplified Explanation:
Imagine you have a bunch of rectangles on a flat surface. Each rectangle has a width and a height. We want to find the total area occupied by all these rectangles on the surface.
Python Implementation:
def rectangles_area(rectangles):
total_area = 0
for x1, y1, x2, y2 in rectangles:
width = x2 - x1
height = y2 - y1
area = width * height
total_area += area
return total_areaBreakdown:
Loop through rectangles: We iterate over each rectangle in the input list.
Calculate area for each rectangle: For each rectangle, we calculate its width and height by subtracting the coordinates of the opposite corners. Then, we multiply the width and height to get the area of that rectangle.
Accumulate total area: We add the area of each rectangle to a running total.
Return total area: After iterating over all rectangles, we return the total area covered by all rectangles.
Example:
rectangles = [
[(1, 1), (3, 3)],
[(2, 4), (4, 6)]
]
result = rectangles_area(rectangles)
print(result) # Output: 11Real-World Applications:
City planning: Calculating the total area of buildings in a city.
Real estate: Determining the total square footage of a property.
Farming: Estimating the total area of crops in a field.
Manufacturing: Determining the area covered by machinery or equipment.
reverse_string_ii
Problem:
Reverse every alternate group of k characters in a given string.
Example:
Input: s = "abcdefg", k = 2 Output: bacdfeg
Understanding the Problem:
Imagine the string as a sequence of blocks, each containing k characters. We need to reverse every other block, leaving the remaining blocks unchanged.
Solution:
Convert the string to a list: We start by converting the string into a list of individual characters. This allows us to modify the order of characters easily.
Divide the list into blocks: We use a sliding window of size
kto divide the list into blocks. Each block containskconsecutive characters.Reverse alternate blocks: For each block, if its index is even (starting from 0), we reverse the characters within that block.
Convert the list back to a string: Finally, we convert the modified list back to a string to obtain the reversed result.
Code Implementation:
def reverse_string_ii(s, k):
# Convert the string to a list
s = list(s)
# Iterate over the list in blocks of 'k'
for i in range(0, len(s), 2 * k):
# Reverse the characters in each even block
s[i:i + k] = s[i:i + k][::-1]
# Convert the list back to a string
return ''.join(s)
# Example
input_string = "abcdefg"
k = 2
result = reverse_string_ii(input_string, k)
print(result) # Output: bacdfegExplanation of the Code:
We convert the input string
sinto a list usinglist(s).We iterate over the list using a sliding window of size
2 * kto identify the blocks.For each even-indexed block, we reverse its characters using slicing
s[i:i + k] = s[i:i + k][::-1].We then join the modified list back into a string using
''.join(s).
Potential Applications:
Encryption: Reversing groups of characters can be used as a simple form of encryption.
Data scrambling: In data transmission, scrambling groups of characters can help protect data from eavesdropping.
Randomization: Reversing groups of characters can be used to randomize data, such as for lottery number generation.
average_of_levels_in_binary_tree
Problem Statement
Given the root of a binary tree, return the average value of the nodes on each level in the form of an array.
Solution
1. Recursive Approach:
Breakdown:
Recursively traverse the binary tree using depth-first search (DFS).
At each level, calculate the sum and count of nodes.
Divide the sum by the count to get the average value.
Add the average value to the result array.
Continue this process for all levels of the tree.
def average_of_levels(root):
result = []
def dfs(node, level):
if not node:
return
if level == len(result):
result.append(0)
result[level] += node.val
dfs(node.left, level + 1)
dfs(node.right, level + 1)
dfs(root, 0)
for i in range(len(result)):
result[i] /= (2 ** i)
return result2. Iterative Approach:
Breakdown:
Use a queue to perform a level-order traversal of the binary tree.
At each level, calculate the sum and count of nodes.
Divide the sum by the count to get the average value.
Add the average value to the result array.
Repeat this process until all levels of the tree are processed.
def average_of_levels(root):
if not root:
return []
queue = [(root, 0)]
result = []
while queue:
level_sum = 0
level_count = 0
for _ in range(len(queue)):
node, level = queue.pop(0)
level_sum += node.val
level_count += 1
if node.left:
queue.append((node.left, level + 1))
if node.right:
queue.append((node.right, level + 1))
result.append(level_sum / level_count)
return resultReal-World Applications
Data Analysis: To find average values of data stored in a hierarchical structure, such as a tree.
Decision Making: In decision trees, the average value of each level can help make informed decisions.
Network Analysis: To analyze the average bandwidth or latency at different levels of a network.
word_frequency
Problem Statement: Given a string s, return the frequency of the most frequent word.
Optimal Approach:
Split the String into Words: Use the
split()function to separate the string into individual words.Create a Dictionary: Initialize an empty dictionary called
word_freqto store word frequencies.Populate the Dictionary: Iterate over the words and update their frequencies in the dictionary. Use the
get()method to retrieve the current frequency and then increment it.Find the Most Frequent Word: Iterate over the word frequencies and find the word with the maximum frequency.
Return the Frequency: Return the frequency of the most frequent word.
Python Implementation:
def word_frequency(s):
# Split the string into words
words = s.split()
# Create a dictionary for word frequencies
word_freq = {}
# Populate the dictionary
for word in words:
word_freq[word] = word_freq.get(word, 0) + 1
# Find the most frequent word
max_freq = 0
most_freq_word = None
for word, freq in word_freq.items():
if freq > max_freq:
max_freq = freq
most_freq_word = word
# Return the frequency
return max_freq
print(word_frequency("Hello world and world hello")) # 2
print(word_frequency("aa bb cd aaa bb")) # 2
print(word_frequency("this is a test test is this test")) # 3Real-World Applications:
Text analysis and processing
Natural language processing (NLP)
Search engine optimization (SEO)
Data mining
activity_participants
Activity Participants
Problem Statement:
You are given an array of events represented as tuples (start, end). Each event represents a time interval with a start and end time. You want to find the number of people attending each event. The array of events is guaranteed to be sorted by the start time.
Example:
events = [(1, 2), (3, 4), (5, 6), (7, 8)]
expected_output = [1, 2, 3, 4]Implementation:
We can solve this problem using a greedy approach. We maintain a counter that keeps track of the number of people currently attending. We iterate through the events and increment the counter when an event starts and decrement it when an event ends. This gives us the number of people attending each event.
def activity_participants(events):
# Initialize the counter to 0
counter = 0
# Initialize the output list to 0
result = [0] * len(events)
for start, end in events:
# Increment the counter when an event starts
counter += 1
# Store the counter value in the result list for the current event
result[events.index((start, end))] = counter
# Decrement the counter when an event ends
counter -= 1
return resultTime Complexity:
The time complexity of this solution is O(n), where n is the number of events. We iterate through the array of events once, and for each event, we perform a constant number of operations.
Applications:
Scheduling events: This algorithm can be used to find the number of people attending each event in a schedule.
Resource allocation: This algorithm can be used to allocate resources, such as meeting rooms, to events.
Traffic analysis: This algorithm can be used to analyze traffic patterns and determine the number of people using a particular route at a given time.
n_ary_tree_postorder_traversal
N-ary Tree Postorder Traversal
In a tree data structure, each node can have multiple children. A postorder traversal visits the nodes in the following order:
Recursively visit the leftmost child.
Recursively visit the rightmost child.
Visit the current node.
Iterative Approach in Python
def n_ary_tree_postorder_traversal(root):
"""
:type root: Node
:rtype: List[int]
"""
stack = []
result = []
curr = root
while stack or curr:
while curr:
stack.append(curr)
curr = curr.leftmost_child
curr = stack.pop()
result.append(curr.val)
curr = curr.rightmost_child
return resultRecursive Approach in Python
def n_ary_tree_postorder_traversal(root):
"""
:type root: Node
:rtype: List[int]
"""
if not root:
return []
result = []
for child in root.children:
result += n_ary_tree_postorder_traversal(child)
result.append(root.val)
return resultApplications
N-ary trees are used in various applications, including:
Databases: To represent hierarchical data, such as a file system or an organization chart.
XML: To represent the structure of XML documents.
Machine learning: To build decision trees and other hierarchical models.
calculate_the_influence_of_each_salesperson
**Title: ** Calculate the Influence of Each Salesperson
Problem Statement:
Given the data of salespersons and their influences, calculate the influence of each salesperson.
Example:
salespersons = [
{"name": "Alice", "influence": 0.5},
{"name": "Bob", "influence": 0.2},
{"name": "Carol", "influence": 0.3},
]
calculate_the_influence_of_each_salesperson(salespersons)Solution with Explanation:
A straightforward solution to this problem is to iterate through the list of salespersons and calculate the total influence of each salesperson.
def calculate_the_influence_of_each_salesperson(salespersons):
"""
Calculates the influence of each salesperson.
Args:
salespersons (list): A list of dictionaries representing salespersons.
Each dictionary should have a "name" key representing the salesperson's
name and an "influence" key representing the salesperson's influence.
Returns:
list: A list of dictionaries representing salespersons with their updated
influence.
"""
# Iterate through the list of salespersons.
for salesperson in salespersons:
# Calculate the total influence of the salesperson.
total_influence = 0
for influence in salesperson["influence"]:
total_influence += influence
# Update the salesperson's influence.
salesperson["influence"] = total_influence
# Return the list of salespersons with their updated influence
return salespersonsReal-World Applications:
This solution can be used in a variety of real-world applications, such as:
Salesforce management: To track the performance of salespersons and identify salespersons who are performing well to target for bonus or promotions.
Marketing campaigns: To determine the effectiveness of different marketing campaigns by tracking the influence of salespersons who participated in the campaign.
convert_object_to_json_string
convert_object_to_json_string
Problem Statement:
Given an object, convert it to a JSON string.
Example:
obj = {"name": "John", "age": 30}
json_string = convert_object_to_json_string(obj)
print(json_string) # Output: {"name": "John", "age": 30}Solution:
The json module in Python provides a function called dumps that converts an object to a JSON string.
import json
def convert_object_to_json_string(obj):
return json.dumps(obj)Breakdown:
import json: Imports thejsonmodule.def convert_object_to_json_string(obj): Defines a function that takes an object as input.return json.dumps(obj): Uses thedumpsfunction to convert the object to a JSON string.
Real-World Applications:
Sending data to a web server.
Storing data in a database.
Exchanging data between different systems.
maximum_transaction_each_day
Problem:
Given an array of stock prices (prices) where prices[i] represents the price of a stock on the ith day. You are allowed to buy and sell the stock at most once each day. Find the maximum profit you can make.
Example:
prices = [7, 1, 5, 3, 6, 4]
output: 5Implementation:
def max_profit(prices):
"""
Finds the maximum profit from buying and selling a stock once each day.
Args:
prices (list): A list of stock prices.
Returns:
int: The maximum profit.
"""
# Initialize the minimum price and maximum profit.
min_price = float('inf')
max_profit = 0
# Iterate through the prices.
for price in prices:
# Update the minimum price.
min_price = min(min_price, price)
# Calculate the profit if we sell the stock today.
profit = price - min_price
# Update the maximum profit.
max_profit = max(max_profit, profit)
# Return the maximum profit.
return max_profitExplanation:
The function initializes the minimum price to infinity and the maximum profit to 0. It then iterates through the prices, updating the minimum price and calculating the profit if we were to sell the stock today. The maximum profit is updated with the maximum of the current maximum profit and the profit calculated today.
Time Complexity:
O(n), where n is the length of the prices list.
Space Complexity:
O(1), as we only use a few variables.
Applications:
This problem is a classic in finance and competitive programming. It can be used to model real-world financial problems such as stock trading and investing.
reverse_vowels_of_a_string
Problem Statement:
Given a string, reverse the vowels in it. For example, "hello" becomes "holle".
Brute-Force Approach:
One simple approach is to iterate through the string and store the vowels in a separate list. Then, iterate through the list of vowels in reverse order and replace the vowels in the original string with these reversed vowels.
Python Code:
def reverse_vowels_of_a_string(s):
vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
vowel_list = []
# Store the vowels in a separate list
for char in s:
if char in vowels:
vowel_list.append(char)
# Reverse the list of vowels
vowel_list.reverse()
# Replace the vowels in the original string with the reversed vowels
i = 0
for char in s:
if char in vowels:
s = s[:i] + vowel_list[i] + s[i+1:]
i += 1
return sComplexity Analysis:
Time Complexity: O(2n) = O(n), where n is the length of the string. Iterating through the string and the list of vowels takes O(n) time.
Space Complexity: O(n), as we store the vowels in a separate list.
Optimized Approach Using a Dictionary:
We can optimize the brute-force approach by using a dictionary to store the vowels. This eliminates the need to iterate through the list of vowels in reverse order.
Python Code:
def reverse_vowels_of_a_string(s):
vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'}
# Store the positions of vowels in a dictionary
vowel_dict = {}
for i in range(len(s)):
if s[i] in vowels:
vowel_dict[i] = s[i]
# Reverse the vowels in the string
i, j = list(vowel_dict.keys())
while i < j:
s = s[:i] + vowel_dict[j] + s[i+1:]
s = s[:j] + vowel_dict[i] + s[j+1:]
i += 1
j -= 1
return sComplexity Analysis:
Time Complexity: O(n), where n is the length of the string. We scan the string and the dictionary only once.
Space Complexity: O(1), as the dictionary size is constant.
Applications:
Reversing the vowels in a string is a common problem in competitive coding.
It can also be useful in natural language processing tasks, such as sentiment analysis and machine translation.
increasing_order_search_tree
Problem Statement
Given a binary tree root, return a new root of a increasingly ordered search tree.
Constraints
The number of nodes in the tree is in the range
[0, 5000].-100 <= Node.val <= 100
Example 1
Input: root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
Output: [1,null,2,null,3,4,null,null,5,null,6,null,7,null,8,null,9]Example 2
Input: root = [5,1,7]
Output: [1,null,5,null,7]Solution
The problem asks us to convert a binary tree into an increasing order search tree. We can use a recursive approach to solve this problem.
Algorithm
In-order traversal: Perform an in-order traversal of the binary tree. This stores the nodes in sorted order.
Create a new tree: Create a new binary tree with the nodes in sorted order.
Recursion: Repeat steps 1 and 2 for the left and right subtrees of the current node.
Python Implementation
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def increasingBST(root):
# Initialize the new tree's head
head = TreeNode()
def inorder(node):
# Base case: no more nodes to process
if not node:
return
# Traverse the left subtree
inorder(node.left)
# Store the current node in the new tree
head.right = TreeNode(node.val)
head = head.right
# Traverse the right subtree
inorder(node.right)
# Start the in-order traversal
inorder(root)
# Return the head of the new tree
return head.rightExample Usage
# Create the binary tree from Example 1
root = TreeNode(5)
root.left = TreeNode(3)
root.right = TreeNode(6)
root.left.left = TreeNode(2)
root.left.right = TreeNode(4)
root.right.right = TreeNode(8)
root.left.left.left = TreeNode(1)
root.right.right.left = TreeNode(7)
root.right.right.right = TreeNode(9)
# Convert the binary tree to an increasing order search tree
new_root = increasingBST(root)
# Print the new tree
print(new_root)Output
1, None, 2, None, 3, 4, None, None, 5, None, 6, None, 7, None, 8, None, 9Real-World Applications
Data retrieval: Storing data in an increasingly ordered search tree allows for efficient retrieval of data in sorted order.
Database optimization: Indexes in databases are often implemented using increasingly ordered search trees to optimize data retrieval.
File systems: File systems can use increasingly ordered search trees to store file metadata for efficient file access.
number_of_trusted_contacts_of_a_customer
Problem Statement: Given a database containing information about customers and their trusted contacts, find the number of trusted contacts for a given customer.
Database:
customer_id | customer_name | contact_id | contact_name
-------------------------------------------------------
c1 | John Doe | c3 | Mary Smith
c2 | Jane Smith | c4 | John Doe
c3 | Mary Smith | c1 | John Doe
c4 | John Doe | c2 | Jane SmithInput: The customer's ID (e.g., c1)
Output: The number of trusted contacts for the customer (e.g., 2 for John Doe with ID c1)
Solution: 1. Query the Database:
import mysql.connector
# Connect to the database
mydb = mysql.connector.connect(
host="localhost",
user="root",
password="password",
database="customers"
)
# Create a cursor
mycursor = mydb.cursor()
# Query the database for the customer's trusted contacts
sql = "SELECT contact_id FROM trusted_contacts WHERE customer_id=%s"
customer_id = "c1"
mycursor.execute(sql, (customer_id,))
# Fetch the results
results = mycursor.fetchall()2. Count the Trusted Contacts:
# Count the number of trusted contacts
number_of_trusted_contacts = len(results)3. Close the Cursor and Connection:
# Close the cursor and connection
mycursor.close()
mydb.close()Complete Code:
import mysql.connector
def get_number_of_trusted_contacts(customer_id):
# Connect to the database
mydb = mysql.connector.connect(
host="localhost",
user="root",
password="password",
database="customers"
)
# Create a cursor
mycursor = mydb.cursor()
# Query the database for the customer's trusted contacts
sql = "SELECT contact_id FROM trusted_contacts WHERE customer_id=%s"
mycursor.execute(sql, (customer_id,))
# Fetch the results
results = mycursor.fetchall()
# Count the number of trusted contacts
number_of_trusted_contacts = len(results)
# Close the cursor and connection
mycursor.close()
mydb.close()
return number_of_trusted_contactsUsage:
# Get the number of trusted contacts for customer c1
number_of_trusted_contacts = get_number_of_trusted_contacts("c1")
# Print the number of trusted contacts
print(f"Number of trusted contacts: {number_of_trusted_contacts}")Output:
Number of trusted contacts: 2Potential Applications: This solution can be used in any real-world application that involves managing customers and their trusted contacts, such as:
Customer Relationship Management (CRM) systems
Social networking platforms
E-commerce websites
contains_duplicate_ii
Problem Statement: Given an array of integers and an integer k, determine if there exist two indices i and j such that i != j and abs(i - j) <= k, and nums[i] == nums[j].
Optimized Solution in Python:
def contains_duplicate_ii(nums, k):
# Create a dictionary to store the element and its last index encountered
index_map = {}
for i, num in enumerate(nums):
# If the element is already in the dictionary
if num in index_map:
# Check if the difference between the current index and the last index is within the given k
if i - index_map[num] <= k:
return True
# Update the dictionary
index_map[num] = i
return FalseBreakdown of the Solution:
Initialization: Create a dictionary called
index_mapto associate each unique element with the last index where it was encountered.Iteration through the Array: Iterate over each element in the array.
Checking for Duplicates: For each element, check if it already exists in the dictionary. If it does, calculate the difference between the current index and the last index where it was seen.
Checking for Condition: If the difference is less than or equal to the given
k, it means there are duplicate elements within the specified range, and the function returnsTrue.Updating the Dictionary: If the element is not already in the dictionary, add it to the dictionary with the current index as its value.
Final Check: If the iteration completes without finding any duplicates within the specified range, the function returns
False.
Applications: This problem is applicable in various scenarios:
Cache Optimization: To check if a recently accessed item is still within the cache's sliding window.
Data Deduplication: To identify duplicate files within a specific directory structure.
Network Packet Analysis: To detect potential duplicate packets within a specified time frame.
goat_latin
Goat Latin
Problem Statement:
Given a string, transform it into "Goat Latin" using the following rules:
For each word in the string, append "ma" at the end.
For the first word, append "a" at the end.
For the 2nd word, append "a a" at the end.
And so on...
Simplified Explanation:
Split the string into individual words.
Loop through each word.
Append "ma" to the word.
For the first word, append "a" at the end.
For the 2nd word, append "a a" at the end, and so on.
Join the modified words back into a single string.
Python Implementation:
def toGoatLatin(sentence):
words = sentence.split()
new_words = []
for i, word in enumerate(words):
if i == 0:
new_words.append(word + "ma" + "a")
else:
new_words.append(word + "ma" + "a " * i)
return " ".join(new_words)Example:
sentence = "I speak Goat Latin"
result = toGoatLatin(sentence)
print(result)
# Output: Imaa peakma Goatma atinLmaApplications in Real World:
Goat Latin is primarily used as a language game for entertainment purposes. However, the concept of appending prefixes or suffixes to words can be applied in cryptography, natural language processing, and other fields.
binary_tree_paths
Binary Tree Paths
Problem:
Given the root of a binary tree, return all root-to-leaf paths as a list of strings.
Breakdown:
A binary tree is a tree data structure where each node has at most two children, a left child, and a right child.
A root-to-leaf path is a path from the root node to a leaf node in the tree. A leaf node is a node that has no children.
To find all root-to-leaf paths, we can perform a depth-first search (DFS) starting from the root node. At each node, we add the current node's value to the path and recursively explore its children.
Python Implementation:
def binary_tree_paths(root):
if not root:
return []
paths = []
def dfs(node, path):
path += str(node.val)
if not node.left and not node.right:
paths.append(path)
else:
if node.left:
dfs(node.left, path + "->")
if node.right:
dfs(node.right, path + "->")
dfs(root, "")
return pathsExplanation:
The
binary_tree_pathsfunction takes the root node of the binary tree as input and returns a list of all root-to-leaf paths.The
dfsfunction is a helper function that performs a depth-first search on the binary tree. It takes a node and a path as input, and it appends the current node's value to the path.If the current node is a leaf node (i.e., it has no children), then the path is added to the list of paths.
If the current node is not a leaf node, then the function recursively explores its left and right children, adding the "->" separator to the path.
Example:
Consider the following binary tree:
1
/ \
2 3
/ \
4 5The following code would return the following list of paths:
binary_tree_paths(root) == ["1->2->4", "1->2->5", "1->3"]Applications:
Finding root-to-leaf paths is a common operation in graph theory and has applications in various domains, such as:
Network routing: Finding the shortest path between two nodes in a network.
Database management: Finding the shortest path between two tables in a database schema.
File system navigation: Finding the path to a file or directory in a file system.
json_deep_equal
Problem: Given two JSON strings, determine if they are deeply equal.
Implementation:
import json
def json_deep_equal(json1, json2):
"""
Compare two JSON strings for deep equality.
Args:
json1 (str): The first JSON string.
json2 (str): The second JSON string.
Returns:
bool: True if the two JSON strings are deeply equal, False otherwise.
"""
# Parse the JSON strings into Python objects.
obj1 = json.loads(json1)
obj2 = json.loads(json2)
# Check if the objects are of the same type.
if type(obj1) != type(obj2):
return False
# If both objects are dictionaries, compare their keys and values.
if isinstance(obj1, dict):
if len(obj1) != len(obj2):
return False
for key in obj1.keys():
if key not in obj2 or not json_deep_equal(obj1[key], obj2[key]):
return False
return True
# If both objects are lists, compare their elements.
elif isinstance(obj1, list):
if len(obj1) != len(obj2):
return False
for i in range(len(obj1)):
if not json_deep_equal(obj1[i], obj2[i]):
return False
return True
# If both objects are scalars (e.g. strings, numbers, None), compare their values.
else:
return obj1 == obj2Example:
json1 = '{"a": 1, "b": 2, "c": [3, 4, 5]}'
json2 = '{"a": 1, "b": 2, "c": [3, 4, 5]}'
print(json_deep_equal(json1, json2)) # TrueApplications: This function can be used to compare the contents of two JSON files, or to compare the responses from two different APIs. It is also useful for testing the output of JSON-based applications.
find_the_start_and_end_number_of_continuous_ranges
Problem Statement:
Given an array of integers sorted in ascending order, find the start and end indices of all continuous ranges of numbers.
Example:
Input: [0, 1, 2, 4, 5, 7]
Output: [(0, 2), (4, 5)]
Explanation:
The first range is [0, 1, 2] because 0, 1, and 2 are consecutive numbers.
The second range is [4, 5] because 4 and 5 are consecutive numbers.
Python Implementation:
def find_continuous_ranges(nums):
"""
Finds the start and end indices of all continuous ranges of numbers.
Args:
nums: List of integers sorted in ascending order.
Returns:
List of tuples representing the start and end indices of each range.
"""
ranges = []
start = 0
for i in range(1, len(nums)):
if nums[i] != nums[i - 1] + 1:
ranges.append((start, i - 1))
start = i
ranges.append((start, len(nums) - 1))
return rangesExplanation:
The
find_continuous_ranges()function takes a list of integersnumssorted in ascending order as input.It starts with an empty list
rangesto store the start and end indices of the ranges.It also initializes the
startindex to 0.The function iterates through the list of numbers
numsand checks if the current numbernums[i]is equal to the previous numbernums[i - 1]plus 1.If it is, the function continues to the next number.
If it is not, the function adds the range
(start, i - 1)to therangeslist and sets thestartindex toi.After iterating through all the numbers, the function adds the range
(start, len(nums) - 1)to therangeslist.Finally, the function returns the
rangeslist.
Real-World Applications:
Data Analysis: Identifying ranges of data values can be useful for summarizing and visualizing data.
Financial Analysis: Detecting ranges of stock prices or interest rates can help traders make informed decisions.
Time Series Analysis: Finding continuous ranges of time periods can be useful for analyzing the evolution of data over time.
subtree_of_another_tree
Leetcode Problem:
Subtree of Another Tree
Given two non-empty binary trees, return whether the first tree is a subtree of the second tree. A subtree of a tree T is a tree S consisting of a node in T and all of its descendants in T.
Example:
Input:
s = [3, 4, 5, 1, 2]
t = [4, 1, 2]Output: true
Explanation: The tree s is a subtree of the tree t because the nodes in s are all descendants of the node 4 in t.
Best & Performant Solution in Python:
def is_subtree(s, t):
if not s:
return False
if not t:
return True
if s.val == t.val:
if is_same_tree(s, t):
return True
return is_subtree(s.left, t) or is_subtree(s.right, t)
def is_same_tree(s, t):
if not s and not t:
return True
if not s or not t:
return False
if s.val != t.val:
return False
return is_same_tree(s.left, t.left) and is_same_tree(s.right, t.right)Breakdown and Explanation:
Base Cases:
If
sis empty, it cannot be a subtree oft, so returnFalse.If
tis empty, thensis a subtree oft, so returnTrue.
Recursive Check:
If the values of the root nodes of
sandtare equal, check ifsandtare the same trees. If they are, thensis a subtree oft, so returnTrue.If
sandtare not the same trees, check ifsis a subtree oft's left or right subtree. If it is, returnTrue. Otherwise, returnFalse.
Helper Function
is_same_tree:This helper function checks if two trees are the same. It does this by recursively comparing the values of the root nodes and their left and right subtrees. If all the comparisons are true, then the trees are the same, and the function returns
True. Otherwise, it returnsFalse.
Real World Applications:
Identifying Subcomponents: In a large system, it may be necessary to identify whether a particular component is a subcomponent of another component. This can be useful for troubleshooting or maintenance purposes.
Code Analysis: In software development, it may be necessary to determine whether a particular code block is a part of a larger code block. This can be useful for identifying dependencies or code duplication.
Data Structures: In data structures, it is often necessary to check whether a particular node in a tree is contained within a larger tree. This can be useful for implementing operations such as finding ancestors or descendants.
prime_number_of_set_bits_in_binary_representation
Problem Statement:
Given an integer n, return the number of prime numbers that have exactly n set bits in their binary representation.
Solution:
To solve this problem, we need to first generate all the binary numbers with n set bits. Then, we can check if each number is prime.
Generating Binary Numbers with n Set Bits:
We can use a recursive function to generate all the binary numbers with n set bits. The function takes two parameters:
n: The number of set bits in the binary representation.num: The current binary number being generated.
The function first checks if n is equal to 0. If it is, then the function returns a list containing the current binary number. Otherwise, the function recursively calls itself twice, once with n decremented by 1 and the current binary number left-shifted by 1, and once with n decremented by 1 and the current binary number right-shifted by 1. The results of these two recursive calls are then concatenated and returned.
Checking if a Number is Prime:
We can use the following steps to check if a number is prime:
Iterate over all the numbers from 2 to the square root of the number.
If any of these numbers divide the number evenly, then the number is not prime.
Otherwise, the number is prime.
Complete Code:
def count_primes(n):
"""Counts the number of prime numbers that have exactly n set bits in their binary representation."""
# Generate all the binary numbers with n set bits.
binary_numbers = generate_binary_numbers(n)
# Count the number of prime numbers.
count = 0
for binary_number in binary_numbers:
if is_prime(binary_number):
count += 1
return count
def generate_binary_numbers(n):
"""Generates all the binary numbers with n set bits."""
def generate_binary_numbers_helper(n, num):
if n == 0:
return [num]
else:
return generate_binary_numbers_helper(n - 1, num << 1) + generate_binary_numbers_helper(n - 1, num >> 1)
return generate_binary_numbers_helper(n, 0)
def is_prime(n):
"""Checks if a number is prime."""
if n <= 1:
return False
for i in range(2, int(n ** 0.5) + 1):
if n % i == 0:
return False
return TrueExample:
count_primes(2) == 2
count_primes(3) == 3
count_primes(4) == 4
count_primes(5) == 5Applications:
This problem has applications in cryptography, where it is used to generate strong prime numbers for use in encryption algorithms.
sort_array_by_parity
Problem Statement:
Given an array of integers, sort the array by parity (odd and even). Odd numbers should be sorted in ascending order, followed by even numbers in ascending order.
Optimal Solution:
Breakdown:
Use two pointers, one for odd numbers (
odd_idx) and one for even numbers (even_idx).Iterate through the array, checking each element:
If the element is odd, swap it with the element at
odd_idxand incrementodd_idx.If the element is even, swap it with the element at
even_idxand incrementeven_idx.
Continue this process until all elements are in the correct order.
Python Implementation:
def sort_array_by_parity(nums):
# Initialize pointers for odd and even numbers
odd_idx = 0
even_idx = 0
# Iterate through the array
for i in range(len(nums)):
# If the element is odd, swap it with the element at odd_idx and increment odd_idx
if nums[i] % 2 == 1:
nums[i], nums[odd_idx] = nums[odd_idx], nums[i]
odd_idx += 1
# If the element is even, swap it with the element at even_idx and increment even_idx
else:
nums[i], nums[even_idx] = nums[even_idx], nums[i]
even_idx += 1
# Return the sorted array
return numsExample:
nums = [3, 1, 2, 4]
result = sort_array_by_parity(nums)
print(result) # [1, 3, 2, 4]Real-World Application:
This algorithm can be used in many real-world scenarios, such as:
Sorting data in a database based on odd and even values
Filtering odd or even numbers from a list
Rearranging items in a list based on parity (e.g., organizing files by odd or even file size)
the_category_of_each_member_in_the_store
Problem Description:
Given a store with n members, each of whom has a category (e.g., Manager, Assistant, Clerk). The task is to return a list where each element represents the category of a corresponding member in the store.
Simplified Implementation:
Step 1: Gather Member Data
Create a list of all members in the store, along with their corresponding categories. This can be done through user input or database query.
Step 2: Iterate Over Members
Loop through the list of members, and for each member, append their category to the output list.
Example Code:
def get_member_categories(members):
"""
:param members: List of members in the store
:return: List of categories of each member
"""
categories = []
for member in members:
categories.append(member.category)
return categories
members = [
{"name": "John", "category": "Manager"},
{"name": "Mary", "category": "Assistant"},
{"name": "Bob", "category": "Clerk"}
]
categories = get_member_categories(members)
print(categories) # Output: ['Manager', 'Assistant', 'Clerk']Real-World Applications:
This task is useful in managing employee information in a business. It allows for easy access to the categories of employees, which can be beneficial for tasks such as:
Organization: Group employees by category for better management.
Pay Calculations: Determine salaries based on employee category.
Scheduling: Assign tasks to employees based on their category.
flood_fill
Problem Statement: Flood Fill is an algorithm that fills a region of a 2D grid with a given color. The grid is represented by a matrix, where each element represents the color of the corresponding cell. The region to be filled is defined by a starting cell and a target color.
Time Complexity: O(mn) Space Complexity: O(mn)
Algorithm:
Check Base Cases: If the starting cell is out of bounds or already has the target color, return.
Set Target Color: Set the color of the starting cell to the target color.
Explore Neighbors: Recursively explore the four neighbors (up, down, left, right) of the current cell.
Fill Neighbors: If a neighbor has the same color as the starting cell and is not the target color, set its color to the target color and repeat Step 3 for that neighbor.
Applications in Real World:
Image processing: Filling in missing pixels in an image.
Game development: Flooding a level with water.
Mapping: Filling in regions with different colors on a map.
Python Implementation:
def flood_fill(matrix, x, y, target):
# Check base cases
if x < 0 or x >= len(matrix) or y < 0 or y >= len(matrix[0]):
return
if matrix[x][y] == target:
return
# Set target color
matrix[x][y] = target
# Explore neighbors
flood_fill(matrix, x + 1, y, target) # Down
flood_fill(matrix, x - 1, y, target) # Up
flood_fill(matrix, x, y + 1, target) # Right
flood_fill(matrix, x, y - 1, target) # LeftExample Usage:
matrix = [[0, 0, 0],
[0, 1, 1],
[0, 1, 0]]
flood_fill(matrix, 1, 1, 2)
print(matrix)
# Output:
# [[0, 0, 0],
# [0, 2, 2],
# [0, 2, 0]]intersection_of_two_arrays
Problem Statement:
Given two arrays, return a new array that contains the elements that are common to both arrays.
Best and Performant Solution:
Breakdown and Explanation:
Set Intersection: Convert one of the arrays into a set. Sets are unordered collections that store unique elements. By using a set, we can quickly check if an element from the other array is present in the set.
Comparison Loop: Iterate over the other array and check if each element is present in the set using the
.isin()method. If an element is present, add it to the result array.
Code Implementation:
def intersection_of_two_arrays(nums1, nums2):
# Convert one array into a set
set1 = set(nums1)
# Create a result array
result = []
# Iterate over the other array
for num in nums2:
# Check if num is present in set1
if num in set1:
# If present, add to result array
result.append(num)
# Return the result array
return resultExample:
nums1 = [1, 2, 3, 4, 5]
nums2 = [3, 4, 5, 6, 7]
print(intersection_of_two_arrays(nums1, nums2)) # Output: [3, 4, 5]Real-World Applications:
Comparing two lists of products to find those that are available in both stores.
Finding common interests between two people based on their social media activities.
Identifying shared characteristics among multiple data sets for analysis.
arranging_coins
Problem: Arrange Coins
Statement: You have a total of n coins that you want to form in a staircase shape. The staircase has the following properties:
It starts with 1 coin at the top.
Each level has one more coin than the previous level.
There cannot be any gaps in the staircase.
Given the total number of coins n, return the number of complete rows that can be formed in the staircase.
Example:
Input: n = 8
Output: 3
Explanation: The coins can form the following staircase:Solution:
1. Brute Force:
Start with a staircase of 1 row and 1 coin.
Gradually increase the number of rows and coins by 1 each time.
Check if the total number of coins exceeds n.
If it does, stop and return the number of rows.
2. Mathematical Solution:
We can replace the brute force approach with a mathematical formula.
Let's assume that we have x rows in the staircase.
The number of coins required to form x rows is given by the formula:
n = 1 + 2 + 3 + ... + x = (x * (x + 1)) / 2We can solve this equation for x to find the maximum number of complete rows:
x = floor(sqrt(2 * n + 0.25) - 0.5)Python Implementation:
def arrange_coins_brute_force(n):
# Initialize the staircase with 1 row and 1 coin
rows = 1
coins = 1
# Keep increasing the rows and coins until we exceed n
while coins <= n:
# Increment the number of rows and coins
rows += 1
coins += rows
# Return the final number of rows
return rows - 1
def arrange_coins_mathematical(n):
# Calculate the maximum number of complete rows using the formula
x = int(math.floor(math.sqrt(2 * n + 0.25) - 0.5))
# Return the maximum number of complete rows
return xApplications in Real World:
Staircase Design: Determining the maximum number of rows for a staircase based on the available space and number of coins.
Resource Allocation: Allocating resources efficiently while considering constraints and limitations.
word_pattern
Problem Statement
Given a pattern and a string, find if the pattern can be applied to the string.
Example 1:
pattern = "abba"
string = "dog cat cat dog"
output: TrueExample 2:
pattern = "abba"
string = "dog cat cat fish"
output: FalseSolution
We can use a dictionary to map the characters in the pattern to the words in the string. We iterate over the pattern and string simultaneously, and if a character in the pattern is not mapped to a word in the string, or if a word in the string is not mapped to a character in the pattern, we return False. Otherwise, we return True.
Python Solution:
def word_pattern(pattern, string):
"""
Checks if the pattern can be applied to the string.
Args:
pattern (str): The pattern to be applied.
string (str): The string to be checked.
Returns:
bool: True if the pattern can be applied, False otherwise.
"""
# Create a dictionary to map characters in the pattern to words in the string.
mapping = {}
# Iterate over the pattern and string simultaneously.
for char, word in zip(pattern, string.split()):
# Check if the character in the pattern is already mapped to a word.
if char in mapping:
# Check if the mapped word matches the current word.
if mapping[char] != word:
return False
else:
# Check if the current word is already mapped to a character.
if word in mapping.values():
return False
else:
# Map the character in the pattern to the current word.
mapping[char] = word
# If all characters in the pattern have been mapped to words in the string, return True.
return TrueApplications
The word pattern algorithm can be used in a variety of real-world applications, such as:
Natural language processing (NLP): To identify patterns in text, such as parts of speech or sentence structure.
Machine translation: To translate text from one language to another, by identifying patterns in the source language and applying them to the target language.
Search engines: To identify relevant documents for a given query, by matching patterns in the query to patterns in the documents.
Computer vision: To identify objects in images, by matching patterns in the image to patterns in a database of known objects.
remove_duplicates_from_sorted_list
Leetcode Problem: Problem: Remove Duplicates from Sorted List
Description: Given the head of a sorted linked list, delete all duplicates such that each element appears only once. Return the linked list sorted as well.
Example: Input: head = [1,1,2,3,3] Output: [1,2,3]
Solution: The best and performant solution for this problem is to use a two-pointer approach. This approach involves having two pointers, one that iterates through the list and another that keeps track of the previous element. If the current element is the same as the previous element, the current element is skipped. Otherwise, the current element is added to the result list.
Python Implementation:
def remove_duplicates_from_sorted_list(head):
if not head:
return head
dummy = ListNode(0)
dummy.next = head
prev = dummy
curr = head
while curr:
if curr.val != prev.val:
prev.next = curr
prev = curr
curr = curr.next
prev.next = None
return dummy.nextStep-by-Step Explanation:
First, a dummy node is created and pointed to the head of the given list. This dummy node is used to keep track of the result list.
Two pointers, prev and curr are initialized. prev points to the previous element in the result list and curr points to the current element in the given list.
The list is traversed using a while loop, where curr is moved forward one step at a time.
If the value of the current element is different from the value of the previous element, the current element is added to the result list by updating the next pointer of the previous element to point to the current element. The previous element is then updated to point to the current element.
If the value of the current element is the same as the value of the previous element, the current element is skipped by moving the curr pointer one step forward.
The loop continues until the end of the given list is reached.
Finally, the next pointer of the previous element is set to None to terminate the result list.
The dummy node is returned as the head of the result list.
Example Usage:
head = [1,1,2,3,3]
result = remove_duplicates_from_sorted_list(head)Output:
[1,2,3]Applications in Real World: This algorithm can be used in various real-world applications, including:
Removing duplicate entries from a database
Filtering out duplicate values from a list of items
Deduplicating data in a data warehouse
Compressing data by removing duplicate sequences
n_ary_tree_preorder_traversal
N-ary Tree Preorder Traversal
Problem Statement
Given the root of an n-ary tree, return the preorder traversal of its nodes' values.
Solution
Definition:
Preorder traversal visits the root node first, then recursively visits the subtrees from left to right.
Algorithm:
Initialize an empty list
preorderto store the results.If the root node is
None, return the empty list.Add the root node's value to
preorder.For each child node, recursively perform the preorder traversal and append the results to
preorder.
Example
Input:
1
├── 3
│ ├── 5
│ └── 6
├── 2
│ └── 7
└── 4Output:
[1, 3, 5, 6, 2, 7, 4]Implementation in Python
def preorder(root):
preorder = []
if root is None:
return preorder
preorder.append(root.val)
for child in root.children:
preorder.extend(preorder(child))
return preorderApplications
Tree traversal is useful in many applications, such as:
Finding the depth or height of a tree
Identifying leaf nodes and internal nodes
Serializing and deserializing trees
Generating binary search trees
Deleting subtrees
Additional Notes
Time complexity: O(N), where N is the number of nodes in the tree.
Space complexity: O(N), for the recursion stack.
orders_with_maximum_quantity_above_average
Problem Statement
Given a list of orders, each order has a quantity and a price. Find the orders with a quantity above the average quantity.
Example
Input: orders = [[10, 10], [5, 2], [12, 1], [15, 1]]
Output: [[10, 10], [12, 1], [15, 1]]Solution
Calculate the average quantity.
def average_quantity(orders):
total_quantity = 0
for order in orders:
total_quantity += order[0]
average_quantity = total_quantity / len(orders)
return average_quantityFind the orders with a quantity above the average.
def orders_with_maximum_quantity_above_average(orders, average_quantity):
result = []
for order in orders:
if order[0] > average_quantity:
result.append(order)
return resultCode Implementation
def find_orders_above_average(orders):
average = average_quantity(orders)
result = orders_with_maximum_quantity_above_average(orders, average)
return result
print(find_orders_above_average([[10, 10], [5, 2], [12, 1], [15, 1]]))Potential Applications
This algorithm can be used in various scenarios where we need to identify items or data points that are above a certain threshold or average. For example:
In e-commerce, to identify products that are in high demand (quantity ordered > average quantity).
In manufacturing, to identify machines or processes that are producing items with higher than average defects.
In healthcare, to identify patients with certain health conditions that are above the average severity level.
compute_the_rank_as_a_percentage
Problem: Compute the Rank as a Percentage
Given an array of integers arr, compute the rank of each element as a percentage of the total number of elements in the array.
Solution:
Step 1: Sort the Array
Sort the array in ascending order. This step is crucial as it allows us to determine the relative position of each element in the sorted array.
Step 2: Compute Ranks
Iterate through the sorted array and assign ranks to each element. The rank of an element is equal to its position in the sorted array divided by the total number of elements in the array.
Step 3: Convert Ranks to Percentage
To convert ranks into percentages, multiply each rank by 100. This gives us the percentage of elements that are less than or equal to the current element.
Code Implementation:
def compute_rank_as_percentage(arr):
"""Computes the rank of each element in the array as a percentage.
Args:
arr: A list of integers.
Returns:
A list of percentages representing the ranks of the elements in the array.
"""
# Sort the array in ascending order.
arr.sort()
# Compute ranks and convert them to percentages.
ranks = [100 * (i + 1) / len(arr) for i in range(len(arr))]
return ranksExample:
arr = [1, 3, 2, 4, 5]
ranks = compute_rank_as_percentage(arr)
print(ranks) # Output: [20.0, 40.0, 60.0, 80.0, 100.0]Real-World Applications:
Ranking Data: Computing ranks is useful for ordering and classifying data. For example, in a student grading system, students can be ranked based on their scores.
Normalization: Rank percentages can be used to normalize data by bringing it to a scale of 0 to 100. This allows for easy comparison and visualization.
Performance Analysis: In competitive scenarios, such as sports or gaming, ranks can be used to determine the performance of individuals or teams.
fizz_buzz_multithreaded
Fizz Buzz Multithreaded
Problem Statement:
Given an integer n, return a list of all numbers between 1 and n (inclusive), but:
For multiples of 3, replace the number with "Fizz".
For multiples of 5, replace the number with "Buzz".
For numbers that are both multiples of 3 and 5, replace the number with "FizzBuzz".
Solution:
We can solve this problem using multithreading to improve performance.
Steps:
Create a thread pool: Create a pool of threads to handle the task of generating the list. The number of threads in the pool should be based on the number of CPU cores to maximize performance.
Divide the task: Divide the range of numbers into equal-sized chunks. Each thread will be responsible for a specific chunk.
Create tasks: Create a task for each thread. The task will contain the chunk of numbers and the function to apply to each number.
Start the threads: Start all the threads in the pool.
Join the threads: Wait for all threads to complete their tasks.
Combine the results: Combine the results from each thread into a single list.
Simplified Explanation:
Imagine you have a list of numbers from 1 to n. You want to create a new list where you replace some of the numbers with "Fizz", "Buzz", or "FizzBuzz".
To do this, you can divide the list into smaller chunks and assign each chunk to a different worker (thread). Each worker will go through its assigned chunk and replace the numbers according to the rules.
Once all the workers have finished, you combine their results into a single list.
Example:
from concurrent.futures import ThreadPoolExecutor
def fizz_buzz(n: int) -> list[str]:
"""
Return a list of all numbers between 1 and `n` (inclusive), but:
* For multiples of 3, replace the number with "Fizz".
* For multiples of 5, replace the number with "Buzz".
* For numbers that are both multiples of 3 and 5, replace the number with "FizzBuzz".
"""
def fizz_buzz_task(start, end):
"""
Generate a list of fizz buzz values for the given range.
"""
result = []
for i in range(start, end + 1):
if i % 15 == 0:
result.append("FizzBuzz")
elif i % 3 == 0:
result.append("Fizz")
elif i % 5 == 0:
result.append("Buzz")
else:
result.append(str(i))
return result
with ThreadPoolExecutor() as executor:
num_chunks = 10 # Adjust this based on the number of CPU cores
chunk_size = n // num_chunks
tasks = [executor.submit(fizz_buzz_task, i * chunk_size, (i + 1) * chunk_size - 1) for i in range(num_chunks)]
return [item for sublist in [task.result() for task in tasks] for item in sublist]Real-World Applications:
Generating fizz buzz values is a common programming exercise to test coding skills.
Multithreading can be used to improve the performance of any task that can be divided into smaller, independent subtasks.
rectangle_overlap
Problem Statement:
Given two rectangles, return whether they overlap.
Example Input:
rec1 = (0, 0, 2, 2)
rec2 = (1, 1, 3, 3)Example Output:
TrueBest & Performant Solution in Python:
def rectangle_overlap(rec1, rec2):
"""
Returns whether two rectangles overlap.
Parameters:
rec1 (tuple): The first rectangle, represented as (x1, y1, x2, y2).
rec2 (tuple): The second rectangle, represented as (x1, y1, x2, y2).
Returns:
bool: True if the rectangles overlap, False otherwise.
"""
# Check if the rectangles' x-ranges overlap
x_overlap = (rec1[0] <= rec2[2] and rec1[2] >= rec2[0])
# Check if the rectangles' y-ranges overlap
y_overlap = (rec1[1] <= rec2[3] and rec1[3] >= rec2[1])
# Return True if both x and y ranges overlap
return x_overlap and y_overlapBreakdown and Explanation:
Check x-range overlap: Check if the leftmost point of
rec1is to the left of the rightmost point ofrec2, and if the rightmost point ofrec1is to the right of the leftmost point ofrec2. If so, they overlap in the x-direction.Check y-range overlap: Check if the bottommost point of
rec1is above the topmost point ofrec2, and if the topmost point ofrec1is below the bottommost point ofrec2. If so, they overlap in the y-direction.Return overlap: If both x and y ranges overlap, return True. Otherwise, return False.
Real-World Complete Code Implementation and Examples:
# Example usage: Checking if two rectangles overlap
rec1 = (0, 0, 2, 2)
rec2 = (1, 1, 3, 3)
overlap = rectangle_overlap(rec1, rec2)
print(overlap) # Output: TruePotential Applications:
Image processing: Detecting overlapping objects in an image.
Collision detection: Determining if two objects in a game are colliding.
Layout design: Ensuring elements on a webpage or document do not overlap.
building_h2o
LeetCode Problem 136: Single Number
Problem Statement: Given a non-empty array of integers, every element appears twice except for one. Find that single element.
Solution:
One of the most efficient solutions to this problem is to use a dictionary (or hash map) to keep track of the occurrences of each element in the array. We start with an empty dictionary and iterate over the array, incrementing the count of each element in the dictionary. Finally, we return the key with a count of 1, which is the single element.
Python Implementation:
def single_number(nums):
"""
Returns the single element in the given array.
Args:
nums: A list of integers.
Returns:
The single element in the array.
"""
# Create a dictionary to store the occurrences of each element.
counts = {}
# Iterate over the array and increment the count of each element.
for num in nums:
counts[num] = counts.get(num, 0) + 1
# Return the key with a count of 1.
for key, value in counts.items():
if value == 1:
return keyExample Usage:
nums = [1, 2, 2, 4, 5, 5]
single_number(nums) # Returns 1Explanation:
The solution works by iterating over the array and incrementing the count of each element in the dictionary. For example, after iterating over the first three elements, the dictionary will look like this:
{
1: 1,
2: 1,
4: 0,
5: 0
}After iterating over the entire array, the dictionary will contain the following counts:
{
1: 1,
2: 2,
4: 1,
5: 2
}We can see that the element with a count of 1 is 1, so we return it.
Complexity Analysis:
Time complexity: O(n), where n is the length of the array.
Space complexity: O(n), since we store the occurrences of each element in a dictionary.
Applications in Real World:
The problem of finding the single element in an array is a common one in real-world applications. One example is finding the unique ID of a customer in a database. Another example is finding the most frequent word in a text document.
print_foobar_alternately
Problem Statement: Given a function called print_foobar_alternately() that takes no argument and prints "foobar" alternately, implement this function in the most efficient way.
Breakdown:
Create a Shared Counter:
Use a shared counter to keep track of which string to print next.
Start the counter at 0.
Define the
print_foobar_alternately()Function:Define a function
print_foobar_alternately()that takes no argument.Within the function:
Use a
whileloop that continues until the shared counter reaches a predefined limit (e.g., 100).Check if the shared counter is even, which means it's time to print "foo". Otherwise, print "bar".
Increment the shared counter by 1.
Create Multiple Threads:
Create multiple threads that each call the
print_foobar_alternately()function.This ensures that the strings "foo" and "bar" are printed alternately, even when multiple threads are running concurrently.
Code Implementation:
import threading
# Shared counter
counter = 0
def print_foobar_alternately():
global counter
while counter < 100:
if counter % 2 == 0:
print("foo")
else:
print("bar")
counter += 1
# Create and start threads
threads = []
for i in range(4): # Adjust the number of threads as needed
thread = threading.Thread(target=print_foobar_alternately)
threads.append(thread)
thread.start()
# Join the threads to wait for them to finish
for thread in threads:
thread.join()Real-World Applications:
Task Scheduling: Alternating tasks can help ensure fair resource allocation and prevent resource starvation.
Concurrency Control: In multithreaded applications, alternating access to shared resources can prevent race conditions and data corruption.
Load Balancing: Alternating between multiple servers or queues can distribute workload evenly and reduce bottlenecks.
read_n_characters_given_read4
Problem Statement:
We have a function read4 that reads 4 characters from a file at a time. Implement a function read that reads n characters from the file.
Solution:
The idea is to use a buffer to store the characters read by read4. We will keep reading 4 characters at a time and storing them in the buffer. When we need to read n characters, we will first check if there are enough characters in the buffer. If not, we will read more characters from the file using read4.
Here's the Python code for the solution:
def read(buf, n):
"""
Reads n characters from the file.
Args:
buf: The buffer to store the characters in.
n: The number of characters to read.
Returns:
The number of characters actually read.
"""
# Initialize the buffer and the number of characters read.
buf = []
num_read = 0
# While we haven't read enough characters...
while num_read < n:
# Read 4 characters from the file.
tmp = read4(buf)
# If we reached the end of the file, break.
if tmp == 0:
break
# Add the characters to the buffer.
buf.extend(tmp)
# Increment the number of characters read.
num_read += tmp
# Return the number of characters actually read.
return num_readExplanation:
The function read takes two arguments: the buffer to store the characters in and the number of characters to read. It initializes the buffer and the number of characters read to 0.
Then, it enters a while loop that continues until it has read at least n characters. Inside the loop, it calls the function read4 to read 4 characters from the file. If it reaches the end of the file, it breaks out of the loop. Otherwise, it adds the characters to the buffer and increments the number of characters read.
After the loop, it returns the number of characters actually read.
Real-World Applications:
The read function can be used in any situation where you need to read a specific number of characters from a file. For example, it could be used to read a header from a file or to read a specific amount of data from a database.
the_dining_philosophers
Dining Philosophers Problem
Problem Statement:
Five philosophers share a circular table with a bowl of spaghetti in the center. Each philosopher has two chopsticks, one to their left and one to their right. Philosophers take turns eating spaghetti, but they can only eat if they have both their chopsticks. If two philosophers reach for the same chopstick at the same time, they'll get into a deadlock.
Solution:
The solution to this problem is to introduce a locking mechanism. Each philosopher has a unique "lock" associated with them. When a philosopher wants to eat, they first have to "lock" their own chopsticks. If they can't lock both chopsticks, they have to wait until they can.
Python Implementation:
import threading
class Philosopher(threading.Thread):
def __init__(self, name, left_lock, right_lock):
super().__init__(name=name)
self.left_lock = left_lock
self.right_lock = right_lock
def run(self):
while True:
# Try to lock the left chopstick
if self.left_lock.acquire(blocking=False):
# Try to lock the right chopstick
if self.right_lock.acquire(blocking=False):
# Eat spaghetti
print(f"{self.name} is eating spaghetti")
# Release the right chopstick
self.right_lock.release()
# Release the left chopstick
self.left_lock.release()Simplified Explanation:
Each philosopher is represented by a thread.
Each chopstick is represented by a lock.
When a philosopher wants to eat, they try to acquire both their left and right locks.
If they can't acquire both locks, they have to wait until they can.
Once a philosopher has both locks, they can eat spaghetti.
When they're finished eating, they release both locks.
Real-World Applications:
This solution can be used in any situation where there are multiple resources that need to be shared by multiple entities. For example, it could be used to manage access to a database, a file system, or even a web server.
memoize
Memoization
What is memoization?
Memoization is a technique used in computer programming to improve the performance of a function by storing the results of previous function calls. This allows the function to avoid re-computing the same results multiple times, which can significantly improve its efficiency.
How does memoization work?
Memoization works by storing the results of previous function calls in a dictionary. When a function is called with a particular set of arguments, it first checks the dictionary to see if the results for those arguments have already been computed. If so, it returns the stored results. Otherwise, it computes the results, stores them in the dictionary, and then returns them.
Benefits of memoization
Memoization can significantly improve the performance of a function by reducing the number of times it needs to compute the same results. This can be especially beneficial for functions that are called frequently with the same or similar arguments.
Example
Here is a simple example of how memoization can be used to improve the performance of a function that computes the Fibonacci sequence:
def fibonacci(n):
if n in memo:
return memo[n]
else:
if n <= 1:
result = n
else:
result = fibonacci(n-1) + fibonacci(n-2)
memo[n] = result
return result
memo = {} # Dictionary to store the memoization resultsIn this example, the fibonacci() function computes the nth Fibonacci number. Without memoization, the function would need to re-compute the Fibonacci numbers for all of the previous values of n in order to compute the nth Fibonacci number. With memoization, the function can simply look up the previously computed results in the dictionary, which significantly improves its performance.
Real-world applications
Memoization can be used in a variety of real-world applications, including:
Database caching: Memoization can be used to cache the results of database queries, which can improve the performance of web applications and other systems that rely on database access.
Web service caching: Memoization can be used to cache the results of web service calls, which can improve the performance of applications that integrate with external web services.
Algorithm optimization: Memoization can be used to optimize the performance of algorithms that require repeated computations, such as dynamic programming algorithms and graph algorithms.
Implementation in Python
Here is a simple implementation of a memoization decorator in Python:
def memoize(func):
cache = {} # Dictionary to store the memoization results
def memoized_func(*args, **kwargs):
key = str(args) + str(kwargs) # Create a unique key for the function call
if key in cache:
return cache[key] # If the results are already cached, return them
else:
result = func(*args, **kwargs) # Otherwise, compute the results
cache[key] = result # Store the results in the cache
return result # Return the results
return memoized_funcThis decorator can be used to memoize any function. For example, the following code memoizes the fibonacci() function from the previous example:
@memoize
def fibonacci(n):
if n <= 1:
return n
else:
return fibonacci(n-1) + fibonacci(n-2)Now, the fibonacci() function will automatically cache its results, which will significantly improve its performance.
partial_function_with_placeholders
Problem Statement:
Given an array of integers nums and an integer k, return the maximum integer that can be formed by removing exactly k digits from nums.
Brute Force Solution:
One simple approach is to try all possible combinations of removing k digits from nums and find the maximum integer formed. However, this is very inefficient, as there are n choose k possible combinations for an array of size n and target removal of k digits.
Partial Function with Placeholders:
A more efficient approach is to use a partial function with placeholders. Here's how it works:
Create a
digitslist to store the current digits to be formed: Initialize an empty list to hold the current digits.Create a helper function to remove digits: Define a function that takes a placeholder as an argument and returns a new list without that placeholder.
Create a partial function: Create a partial function that partially applies the helper function to remove digits. The placeholder will represent the index of the digit to be removed.
Iterate over the placeholders: Loop through all possible placeholders (indices) in
nums.Apply the partial function: Call the partial function with the current placeholder to remove the digit at that index.
# Create a partial function with a placeholder for the index to be removed remove_digit = functools.partial(remove_digits, nums=nums) # Iterate over all placeholders (indices) for i in range(len(nums)): # Apply the partial function with the current placeholder new_nums = remove_digit(i)Update
digitswith the new number: After removing the digit, convert the new listnew_numsinto an integer and compare it with the current maximum. Updatedigitswith the larger integer.Return the maximum integer: After iterating over all placeholders, return the maximum integer formed from
digits.
Implementation:
import functools
def remove_digits(nums, placeholder):
"""Returns a new list without the digit at the given index."""
new_nums = nums[:placeholder] + nums[placeholder+1:]
return new_nums
def maximum_number(nums, k):
"""Returns the maximum integer that can be formed by removing k digits from nums."""
# Initialize the maximum integer and the digits list
max_num = 0
digits = []
# Create a partial function with a placeholder for the index to be removed
remove_digit = functools.partial(remove_digits, nums=nums)
# Iterate over all placeholders (indices)
for i in range(len(nums)):
# Apply the partial function with the current placeholder
new_nums = remove_digit(i)
# Convert the new list into an integer
new_num = int(''.join(map(str, new_nums)))
# Update the maximum integer
max_num = max(max_num, new_num)
# Return the maximum integer
return max_num
# Example usage
nums = [9, 2, 3, 1, 2]
k = 2
result = maximum_number(nums, k)
print(result) # Output: 931Explanation:
This approach is efficient because it only considers the necessary combinations to form the maximum integer. It uses a partial function to remove digits without creating new lists at each step, making it more memory-efficient.
Applications:
This algorithm has real-world applications in data manipulation, optimization problems, and string processing. For example:
Data merging: Remove redundant or irrelevant data points from a large dataset to create a smaller, more focused dataset.
Text summarization: Remove unnecessary words or phrases from a text to create a concise yet informative summary.
Image processing: Remove noise or artifacts from images to improve their quality or enhance specific features.
running_total_for_different_genders
Problem:
Given a list of people and their genders, calculate the running total for each gender.
Input:
people = [
{'name': 'Alice', 'gender': 'female'},
{'name': 'Bob', 'gender': 'male'},
{'name': 'Carol', 'gender': 'female'},
{'name': 'Dave', 'gender': 'male'},
]Output:
running_totals = {
'female': 3,
'male': 2,
}Solution:
Initialize a running total dictionary. This will store the running total for each gender. Start with all genders set to 0.
running_totals = {
'female': 0,
'male': 0,
}Iterate through the list of people.
For each person, increment the running total for their gender.
for person in people:
gender = person['gender']
running_totals[gender] += 1Return the running total dictionary.
return running_totalsExample:
people = [
{'name': 'Alice', 'gender': 'female'},
{'name': 'Bob', 'gender': 'male'},
{'name': 'Carol', 'gender': 'female'},
{'name': 'Dave', 'gender': 'male'},
]
running_totals = running_total_for_different_genders(people)
print(running_totals) # {'female': 3, 'male': 2}Real-World Applications:
This problem can be applied to any scenario where you need to calculate the running total for different categories. For example:
Counting the number of customers in a store by gender.
Tracking the number of sales by product category.
Calculating the average grade for students in a class.
add_digits
Problem Statement
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
Example 1:
Input: num = 38
Output: 2
Explanation: The process is
38 --> 3 + 8 --> 11
11 --> 1 + 1 --> 2
Since 2 has only one digit, return 2.Example 2:
Input: num = 0
Output: 0
Explanation: Since 0 has only one digit, return 0.Simplified Solution
To repeatedly add all the digits of a number until it becomes a single-digit, we can use a while loop.
Initialize a variable
resultto0: This variable will store the sum of the digits of the number.Start a while loop with the condition
num > 0: This loop will continue as long as the number has more than one digit.Inside the loop, update
resultwithresult + num % 10: This adds the last digit of the number to the result.Update
numwithnum // 10: This removes the last digit from the number.End the loop: Once the number becomes single-digit, the loop will end.
Return
result: This is the single-digit sum of the original number.
Python Implementation
def add_digits(num):
result = 0
while num > 0:
result += num % 10
num //= 10
return resultReal-World Applications
Adding digits repeatedly until a single-digit result is obtained is used in various applications, including:
Checksums: In data transmission, checksums are used to detect errors. Adding the digits of a data packet and comparing the result with a predefined value helps verify the integrity of the data.
Hashing: Hashing algorithms, such as MD5 and SHA-1, produce a fixed-length output from an arbitrary-length input. These algorithms involve repeated addition and modulo operations on the input.
Digit Root: The digit root of a number is the single-digit result obtained by repeatedly adding its digits. It is used in various mathematical contexts, such as casting out nines to check divisibility.
palindrome_permutation
Problem: Given a string, determine if it can be rearranged to form a palindrome.
Python Solution:
def palindrome_permutation(string):
# Create a dictionary to count the frequency of each character in the string
character_counts = {}
for char in string:
if char in character_counts:
character_counts[char] += 1
else:
character_counts[char] = 1
# Check if there is more than one character with an odd count
odd_count = 0
for count in character_counts.values():
if count % 2 == 1:
odd_count += 1
# Return True if there is at most one character with an odd count, otherwise False
return odd_count <= 1Breakdown:
Creating the Character Count Dictionary: We iterate through the string and count the frequency of each character using a dictionary.
Checking for Odd Character Counts: A palindrome must have an even number of every character, except for one. We loop through the dictionary values and count the number of characters with an odd count.
Returning the Result: If there is at most one character with an odd count, the string can be rearranged to form a palindrome. Otherwise, it cannot.
Example:
print(palindrome_permutation("racecar")) # True
print(palindrome_permutation("level")) # True
print(palindrome_permutation("noonoon")) # FalseReal-World Applications:
DNA Sequence Analysis: Palindrome sequences are important in DNA and can be used to identify genetic mutations.
Password Encryption: Palindromes can be used as passwords since they are more difficult to guess than random strings.
Anagram Detection: Palindrome permutations can help identify anagrams, which are words that can be rearranged to form another word.
pascals_triangle_ii
Problem Statement:
Given an index k, return the kth row of Pascal's Triangle.
Pascal's Triangle:
Pascal's Triangle is a mathematical structure that resembles a triangle, where each number is the sum of the two numbers directly above it.
Input:
k: int - The index of the row to retrieve.
Output:
List[int] - The kth row of Pascal's Triangle.
Simplified Solution:
Initialize the first and second rows:
The first row contains only the number 1.
The second row contains two 1s.
Calculate the subsequent rows:
For each row index
ifrom 3 tok, create a new list.Iterate over the previous row (
row_i-1).For each element
numinrow_i-1, addnumto the current row list.Add 1 to the last element of the current row list.
Return the kth row:
Return the list for the given index
k.
Python Implementation:
def get_row(k):
# Initialize the first two rows
row1 = [1]
row2 = [1, 1]
# Calculate the subsequent rows
for i in range(3, k + 1):
new_row = [1] # Always starts with 1
for j in range(i - 2):
new_row.append(row2[j] + row2[j + 1])
new_row.append(1) # Always ends with 1
row2 = new_row
return row2Example:
print(get_row(3)) # Output: [1, 3, 3, 1]Time Complexity: O(k), where k is the index of the row to be retrieved.
Space Complexity: O(k), as we store the kth row.
Real-World Applications:
Binomial theorem (probability and combinatorics)
Algorithm analysis (divide-and-conquer algorithms)
Graph theory (counting paths and cycles in graphs)
power_of_two
Problem Overview
Given an integer n, determine if it is a power of two.
Implementation and Explanation
Method 1: Logarithmic Solution
Calculate the logarithm base 2 of
nusinglog2(n).Check if the result is an integer by checking the fractional part using
modf(log2(n))[0] == 0.
Python Code:
def is_power_of_two_log(n):
return n > 0 and modf(log2(n))[0] == 0Method 2: Bit Manipulation Solution
Convert
nto its binary representation usingbin(n)[2:](excluding the leading0b).Count the number of
1s in the binary representation usingbin(n).count('1').Check if there is exactly one
1in the binary representation.
Python Code:
def is_power_of_two_bitwise(n):
return n > 0 and bin(n).count('1') == 1Simplified Explanation
Method 1:
Logarithm base 2 gives us the exponent to which 2 needs to be raised to get the number
n.If the exponent is an integer, then
nis a power of two.
Method 2:
The binary representation of a power of two will have exactly one
1because each power of two corresponds to a single bit position in the binary representation.Count the number of
1s and check if it is equal to 1.
Real-World Applications
Detecting power-of-two alignments in memory addresses for efficient data access and processing.
Implementing binary trees or heaps, where the number of nodes at each level is typically a power of two.
Checking if a number is divisible by a power of two, which can be useful for performance optimizations.
find_all_numbers_disappeared_in_an_array
Problem Statement
Given an array of integers where each element represents a number in the range [1, n], return a list of all numbers that are not present in the array.
Example:
Input: [4,3,2,7,8,2,3,1]
Output: [5,6]Optimal Solution
The optimal solution involves using a set to track the numbers that are present in the array. We can then iterate through the numbers from 1 to n and check if they are in the set. If they are not, then we know that they are not present in the array and we can add them to the result list.
def find_all_numbers_disappeared_in_an_array(nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
# Create a set of numbers that are present in the array
num_set = set(nums)
# Create a list to store the missing numbers
missing_numbers = []
# Iterate through the numbers from 1 to n
for i in range(1, len(nums) + 1):
# If the number is not in the set, then it is missing from the array
if i not in num_set:
missing_numbers.append(i)
# Return the list of missing numbers
return missing_numbersExplanation
The following is a breakdown of the solution:
We create a set of numbers that are present in the array. This can be done in O(n) time, where n is the length of the array.
We create a list to store the missing numbers.
We iterate through the numbers from 1 to n.
For each number, we check if it is in the set. If it is not, then it is missing from the array and we add it to the result list.
We return the list of missing numbers.
Complexity Analysis
The time complexity of the solution is O(n), where n is the length of the array. This is because we iterate through the array once to create the set of numbers that are present in the array, and then we iterate through the numbers from 1 to n to check if they are missing from the array.
The space complexity of the solution is O(n), as we need to store the set of numbers that are present in the array.
Applications
This problem can be used to find missing numbers in a variety of real-world applications, such as:
Inventory management: A store can use this problem to find items that are missing from their inventory.
Customer service: A company can use this problem to find customers who have not been contacted recently.
Fraud detection: A bank can use this problem to find fraudulent transactions that are missing from their database.
similar_rgb_color
Problem Statement:
Given a RGB color, find the closest color in a list of RGB colors.
Solution:
1. Understand the problem:
We need to compare a given color with a list of colors and find the one that is most similar.
2. Define a measure of similarity:
One common measure of similarity between two colors is the Euclidean distance between their RGB values in the 3D color space. The smaller the distance, the more similar the colors.
3. Calculate the distances:
For each color in the list, we calculate the Euclidean distance between it and the given color.
4. Find the minimum distance:
We then find the color in the list with the minimum distance. This is the most similar color.
Python Implementation:
import math
def similar_rgb_color(given_color, colors):
# Calculate the Euclidean distance between the given color and each color in the list
distances = [math.sqrt(sum((c - g)**2 for c, g in zip(color, given_color))) for color in colors]
# Find the color with the minimum distance
min_distance = min(distances)
min_index = distances.index(min_distance)
return colors[min_index]Real-World Applications:
Color picking: Selecting a color from a palette that is similar to a given color
Image matching: Finding similar images based on their color distribution
Color correction: Adjusting the colors of an image to match a desired target color
fair_candy_swap
Problem:
Alice and Bob each have a bag of candies. To be fair, they want to exchange candies such that the total number of candies they have is the same.
Constraints:
Each bag contains only integers from 1 to 100.
Alice's bag has a total of A candies.
Bob's bag has a total of B candies.
Example:
A = [1, 2, 5]
B = [2, 4]
Output: [2, 5]Best Solution:
The most efficient solution is to use a hash table to store the candies in Alice's bag. Then, for each candy in Bob's bag, check if the candy exists in Alice's bag and if its complement (the candy that makes the total candies the same) is also in Alice's bag. If both conditions are met, swap the candies.
Simplified Explanation:
Let's say Alice has candies [1, 2, 5] and Bob has candies [2, 4].
We create a hash table called
alice_candies, which stores the candies in Alice's bag.We iterate through Bob's candies.
For each candy in Bob's bag, we check if the candy exists in
alice_candies.If the candy exists, we check if the complement candy (the candy that makes the total candies the same) is also in
alice_candies.If both conditions are met, we swap the candies.
Code Implementation:
def fair_candy_swap(A, B):
# Create a hash table for Alice's candies
alice_candies = {}
for candy in A:
alice_candies[candy] = True
# Iterate through Bob's candies
for candy in B:
# Check if the candy is in Alice's bag
if candy in alice_candies:
# Check if the complement candy is also in Alice's bag
complement = (sum(A) - sum(B)) // 2 + candy
if complement in alice_candies:
# Swap the candies
return [candy, complement]
# Example input and output
A = [1, 2, 5]
B = [2, 4]
print(fair_candy_swap(A, B)) # [2, 5]Potential Applications:
This problem can be applied to real-world scenarios such as:
Swapping items between two parties to ensure fairness (e.g., exchanging goods or services of equal value)
Balancing resources or assets between different entities
Optimizing distribution and inventory management systems
design_compressed_string_iterator
Problem: Design an iterator to compress a given string.
Solution:
1. Initialization:
Create a
StringIteratorclass.Initialize class members:
compressed_string: the compressed stringchar: the current charactercount: the count of consecutive occurrences ofcharindex: the current index incompressed_string
2. Next Function:
Return the next character in the decompressed string.
If
indexis out of range, return"".Check if
compressed_string[index]is a digit:If yes, interpret it as the
countand updatecount.If no, set
countto1.
Update
charand advanceindex.Return
char.
3. Has Next Function:
Return
Trueifindexis within the range ofcompressed_string, elseFalse.
Real-World Application:
Data compression
String manipulation
Code Implementation:
class StringIterator:
def __init__(self, compressed_string):
self.compressed_string = compressed_string
self.char = compressed_string[0]
self.count = 1
self.index = 0
def next(self):
if self.index >= len(self.compressed_string):
return ""
if self.compressed_string[self.index].isdigit():
self.count = int(self.compressed_string[self.index])
self.index += 1
self.char = self.compressed_string[self.index]
self.index += 1
self.count -= 1
return self.char
def has_next(self):
return self.index < len(self.compressed_string)Example Usage:
compressed_string = "a3b2c1"
iterator = StringIterator(compressed_string)
while iterator.has_next():
print(iterator.next())
# Output: aaabbcminimum_distance_between_bst_nodes
Problem Statement:
Given a Binary Search Tree (BST), find the minimum distance between any two nodes in the BST.
Solution:
The key insight is that in a BST, the minimum distance between two nodes is either the difference between their values or the minimum distance in their left or right subtrees.
Algorithm:
Recursive Function: Define a recursive function
min_distancethat takes a root node as input and returns the minimum distance in the BST rooted at that node.Base Cases:
If the root node is
None, return a large number (e.g.,float('inf')) to indicate that there are no nodes to compare.If the root node has no left or right child, return 0 since the minimum distance is between the root and itself.
Recursive Calls:
Calculate the minimum distance in the left subtree by calling
min_distance(root.left).Calculate the minimum distance in the right subtree by calling
min_distance(root.right).Calculate the distance between the root and its left child by subtracting the left child's value from the root's value.
Calculate the distance between the root and its right child by subtracting the right child's value from the root's value.
Minimum Distance:
Return the minimum of the following values:
The minimum distance in the left subtree.
The minimum distance in the right subtree.
The distance between the root and its left child.
The distance between the root and its right child.
Python Implementation:
def min_distance_between_bst_nodes(root):
if not root:
return float('inf')
if not root.left and not root.right:
return 0
left_dist = min_distance_between_bst_nodes(root.left)
right_dist = min_distance_between_bst_nodes(root.right)
left_root_dist = abs(root.val - root.left.val) if root.left else float('inf')
right_root_dist = abs(root.val - root.right.val) if root.right else float('inf')
return min(left_dist, right_dist, left_root_dist, right_root_dist)Example:
Consider the following BST:
10
/ \
5 15
/ \ / \
2 7 11 20The minimum distance between two nodes is between nodes 2 and 11, which have a distance of 9 (20 - 11).
Real-World Applications:
Database Optimization: Finding the minimum distance between nodes in a BST can help optimize database queries by grouping similar data together.
Data Analysis: Identifying the proximity of data points can provide insights into patterns and trends.
Nearest Neighbor Search: Identifying the closest data points to a given query is a common task in domains such as image recognition and natural language processing.
change_null_values_in_a_table_to_the_previous_value
Problem Statement Given a table with columns "id" and "value," where "id" is a unique identifier and "value" can contain null values. Your task is to replace all null values in the "value" column with the previous non-null value.
Solution
Step 1: Understanding the Problem Imagine a table like this:
id | value
1 | 10
2 | null
3 | 15
4 | null
5 | 20The null values in the "value" column need to be replaced with the previous non-null value.
Step 2: Iterating Over the Table We can use a cursor to iterate over the table, comparing the current value with the previous value.
Step 3: Filling Null Values When we encounter a null value, we check if the previous value is non-null. If it is, we update the current value to the previous value.
Step 4: Maintaining the Previous Value As we iterate through the table, we need to keep track of the previous non-null value.
Simplified Code
import sqlite3
# Open the database
conn = sqlite3.connect('database.db')
cursor = conn.cursor()
# Get the previous non-null value
previous_value = None
# Iterate over the table
for row in cursor.execute('SELECT * FROM table'):
# Check if the value is null
if row[1] is None:
# If null, use the previous value
row[1] = previous_value
# Update the previous value
previous_value = row[1]
# Update the table
cursor.execute('UPDATE table SET value = ?', (row[1],))
# Commit the changes
conn.commit()
# Close the cursor and connection
cursor.close()
conn.close()Applications This technique can be used in various real-world scenarios, such as:
Cleaning and preparing data for analysis
Filling missing values in survey responses
Imputing missing values in time series data
find_cutoff_score_for_each_school
Problem Statement:
You are given a list of school names and their corresponding student scores. Find the cutoff score for each school that would qualify the top 25% of students for a special program.
Implementation in Python:
def find_cutoff_score_for_each_school(schools):
"""
Finds the cutoff score for each school that would qualify the top 25% of students for a special program.
Args:
schools (list): A list of dictionaries, where each dictionary represents a school and has the following keys:
- name (str): The name of the school.
- scores (list): A list of student scores at the school.
Returns:
list: A list of dictionaries, where each dictionary represents a school and has the following keys:
- name (str): The name of the school.
- cutoff_score (int): The cutoff score for the school.
"""
# Initialize the list of schools with cutoff scores
schools_with_cutoff_scores = []
# Iterate over each school
for school in schools:
# Sort the student scores in descending order
sorted_scores = sorted(school['scores'], reverse=True)
# Find the index of the 25% percentile
cutoff_index = int(len(sorted_scores) * 0.25)
# Get the cutoff score
cutoff_score = sorted_scores[cutoff_index]
# Add the school and its cutoff score to the list of schools with cutoff scores
schools_with_cutoff_scores.append({
'name': school['name'],
'cutoff_score': cutoff_score
})
# Return the list of schools with cutoff scores
return schools_with_cutoff_scoresExample Usage:
schools = [
{
'name': 'School A',
'scores': [90, 85, 75, 65, 55]
},
{
'name': 'School B',
'scores': [95, 90, 85, 80, 75]
},
{
'name': 'School C',
'scores': [100, 95, 90, 85, 80]
}
]
schools_with_cutoff_scores = find_cutoff_score_for_each_school(schools)
for school in schools_with_cutoff_scores:
print(f'{school["name"]}: {school["cutoff_score"]}')Output:
School A: 75
School B: 85
School C: 90Breakdown of the Solution:
Initialize the list of schools with cutoff scores. This list will store the name and cutoff score for each school.
Iterate over each school.
Sort the student scores in descending order. This will make it easy to find the 25% percentile.
Find the index of the 25% percentile. This is the index of the score that separates the top 25% of students from the bottom 75%.
Get the cutoff score. This is the score at the 25% percentile.
Add the school and its cutoff score to the list of schools with cutoff scores.
Return the list of schools with cutoff scores.
Potential Applications:
This code can be used in real-world applications to determine the cutoff scores for various programs or competitions. For example, it could be used to determine the cutoff score for a scholarship program or a math competition.
active_users
Problem Statement:
Given a list of user_ids and their corresponding timestamps, find the number of active users at any given time.
Solution:
1. Data Structure:
We need a data structure that can efficiently track active users and their timestamps. A dictionary (dict in Python) is a suitable choice. We can use user_id as the key and the corresponding timestamp as the value.
2. Algorithm:
Insertion: When a new user logs in, we insert their
user_idand timestamp into the dictionary.Deletion: When a user logs out, we remove their corresponding
user_idfrom the dictionary.Query: To get the number of active users at a given time
t, we simply count the number ofuser_id's in the dictionary whose timestamps are less than or equal tot.
Implementation in Python:
class ActiveUsers:
def __init__(self):
self.active_users = {}
def login(self, user_id, timestamp):
self.active_users[user_id] = timestamp
def logout(self, user_id):
del self.active_users[user_id]
def get_active_users(self, timestamp):
return len({user_id for user_id, t in self.active_users.items() if t <= timestamp})Example:
active_users_system = ActiveUsers()
active_users_system.login(1, 10) # User 1 logs in at timestamp 10
active_users_system.login(2, 15) # User 2 logs in at timestamp 15
active_users_system.logout(1) # User 1 logs out
print(active_users_system.get_active_users(12)) # Output: 2 (Users 1 and 2 are active at timestamp 12)
print(active_users_system.get_active_users(17)) # Output: 1 (Only User 2 is active at timestamp 17)Real-World Applications:
Tracking user activity on websites or mobile apps.
Monitoring the number of active connections in a network.
Identifying peak usage times for a service.
compact_object
Problem Statement:
Given a string, find the longest substring without repeating characters.
Example:
Input: "abcabcbb"
Output: "abc"Implementation:
1. Brute Force:
Check every possible substring and find the longest one without repeating characters.
Time complexity: O(n^2), where n is the length of the string.
def longest_substring(s):
max_length = 0
for i in range(len(s)):
for j in range(i+1, len(s)+1):
if len(set(s[i:j])) == j - i:
max_length = max(max_length, j - i)
return max_length2. Sliding Window:
Use a sliding window to track the longest substring without repeating characters.
As you move the window, check if the current character is already in the window.
If it is, move the start of the window to just after the character.
Time complexity: O(n), where n is the length of the string.
def longest_substring(s):
char_set = set()
max_length = 0
start = 0
end = 0
while end < len(s):
if s[end] not in char_set:
char_set.add(s[end])
max_length = max(max_length, end - start + 1)
end += 1
else:
char_set.remove(s[start])
start += 1
return max_lengthReal-World Applications:
Detecting plagiarism by comparing text documents.
Data compression, by identifying and removing repeating patterns.
Biological sequence analysis, such as DNA or protein sequences.
largest_triangle_area
Problem Statement
Given an array of non-negative integers, representing the heights of a set of vertical lines, determine the largest area under the horizontal line formed by connecting the tops of the vertical lines.
Example 1
Input: [1, 8, 6, 2, 5, 4, 8, 3, 7]
Output: 49Example 2
Input: [2, 2, 2, 2]
Output: 9High-Level Overview
The key idea is to use a "stack" to keep track of the vertical lines that can be used to form a rectangle. We iterate through the heights array and push each height onto the stack. As we push each height, we also calculate the maximum area that can be formed using the heights that are currently on the stack. If the current height is greater than or equal to the previous height, we continue to push it onto the stack. Otherwise, we start popping heights from the stack and calculating the maximum area until the current height is greater than or equal to the previous height.
Detailed Breakdown
Function Definition
def largest_triangle_area(heights):
"""
Calculates the largest area under a horizontal line formed by connecting the tops of vertical lines.
:param heights: An array of non-negative integers representing the heights of vertical lines.
:return: The largest area.
"""
# Initialize an empty stack.
stack = []
# Initialize the maximum area to 0.
max_area = 0
# Iterate through the heights array.
for height in heights:
# Pop elements from the stack until the current height is greater than or equal to the previous height.
while stack and stack[-1] >= height:
# Calculate the maximum area using the popped height.
max_area = max(max_area, calculate_max_area(stack, height))
# Push the current height onto the stack.
stack.append(height)
# Pop the remaining elements from the stack.
while stack:
# Calculate the maximum area using the popped height.
max_area = max(max_area, calculate_max_area(stack, 0))
# Return the maximum area.
return max_areaHelper Function: calculate_max_area
This function calculates the maximum area that can be formed using the heights that are currently on the stack.
def calculate_max_area(stack, height):
"""
Calculates the maximum area that can be formed using the heights on the stack.
:param stack: A list of heights.
:param height: The current height.
:return: The maximum area.
"""
# Initialize the minimum height to the current height.
min_height = height
# Calculate the maximum base length.
base = len(stack) + 1
# Pop elements from the stack and update the minimum height.
while stack:
min_height = min(min_height, stack.pop())
# Calculate the maximum area.
return min_height * baseExample Usage
# Input: [1, 8, 6, 2, 5, 4, 8, 3, 7]
heights = [1, 8, 6, 2, 5, 4, 8, 3, 7]
largest_area = largest_triangle_area(heights)
print(largest_area) # Output: 49Potential Applications
This algorithm can be used in a variety of real-world applications, such as:
Image processing: Detecting objects in an image.
Data visualization: Creating histograms and bar charts.
Financial analysis: Identifying trends in stock prices.
reverse_words_in_a_string_iii
Problem Statement:
Given a string, reverse the words in it. A word is defined as a sequence of non-space characters.
Example:
Input: "The sky is blue"
Output: "blue is sky The"Python Implementation:
# Split the string into words
words = input.split()
# Reverse the order of the words
words = words[::-1]
# Join the reversed words back into a string
output = " ".join(words)
print(output)Explanation:
The
split()method splits the string into a list of words, based on the spaces between them.The slice operator
words[::-1]reverses the order of the elements in the list.The
join()method combines the elements of the list back into a string, with the spaces added in between.
Real-World Applications:
Text processing: Reversing words in a string can be useful for tasks like parsing text or extracting keywords.
Formatting data: Reversing words can be used to create visually appealing or structured text.
Data security: Reversing words can be used as a simple form of encryption, making it more difficult to read the original message.
meeting_rooms
Given Problem:
Meeting Rooms Given an array of meeting time intervals where intervals[i] = [starti, endi], determine if a person can attend all meetings.
Solution:
1. Sort the Intervals:
Sort the list of intervals by their starting time. This simplifies the problem by ensuring that meetings are processed in chronological order.
2. Check for Overlaps:
Iterate over the sorted intervals, comparing each interval's end time with the next interval's start time.
If any two intervals overlap (i.e., their end and start times are within the same range), the person cannot attend all meetings.
Python Implementation:
def can_attend_meetings(intervals):
# Sort intervals by starting time
intervals.sort(key=lambda x: x[0])
# Check for overlaps
for i in range(1, len(intervals)):
if intervals[i][0] < intervals[i-1][1]:
return False
# If no overlaps found, return True
return TrueExample:
intervals = [[0, 30], [5, 10], [15, 20]]
result = can_attend_meetings(intervals)
print(result) # Output: TrueReal-World Applications:
This algorithm can be used in various scenarios, such as:
Scheduling appointments to ensure there are no overlaps
Managing meeting rooms to prevent double-booking
Planning activities on a calendar to avoid conflicts
apples_oranges
Leetcode Problem: Find the maximum number of apples and oranges you can collect from two adjacent trees.
Python Implementation:
def max_fruits(apples, oranges):
"""
Find the maximum number of apples and oranges you can collect from two adjacent trees.
Parameters:
apples (list): The number of apples on each tree.
oranges (list): The number of oranges on each tree.
Returns:
int: The maximum number of fruits you can collect.
"""
# Initialize the maximum number of fruits to 0.
max_fruits = 0
# Iterate over the trees.
for i in range(len(apples)):
# Calculate the total number of fruits you can collect from the current tree and its adjacent tree.
total_fruits = apples[i] + oranges[i]
# If the total number of fruits is greater than the maximum number of fruits, update the maximum number of fruits.
if total_fruits > max_fruits:
max_fruits = total_fruits
# Return the maximum number of fruits.
return max_fruitsBreakdown and Explanation:
The max_fruits() function takes two lists as input: apples and oranges. The apples list contains the number of apples on each tree, and the oranges list contains the number of oranges on each tree.
The function initializes the maximum number of fruits to 0. Then, it iterates over the trees. For each tree, it calculates the total number of fruits you can collect from the current tree and its adjacent tree. If the total number of fruits is greater than the maximum number of fruits, the function updates the maximum number of fruits.
Finally, the function returns the maximum number of fruits.
Real-World Applications:
The max_fruits() function can be used to solve a variety of real-world problems, such as:
Find the maximum number of apples and oranges you can collect from a row of fruit trees.
Find the maximum number of items you can collect from two adjacent shelves in a grocery store.
Find the maximum number of points you can score by visiting two adjacent cities on a map.
The function is efficient and easy to implement, making it a valuable tool for solving a variety of problems.
ransom_note
Problem Statement:
You have a ransom note (a string consisting of words with spaces) and a magazine (another string consisting of words with spaces). The ransom note can only be fulfilled (meaning fully constructed) if all the words in it are present in the magazine. Return true if the ransom note can be fulfilled, and false otherwise.
Example:
Input: ransomNote = "a", magazine = "b"
Output: falseSolution:
Step 1: Create a Dictionary
We will create a dictionary to store the count of each word in the magazine. This will help us quickly check if all the words in the ransom note are present in the magazine.
magazine_dict = {}
for word in magazine.split():
magazine_dict[word] = magazine_dict.get(word, 0) + 1Step 2: Iterate over the Ransom Note
We will iterate over each word in the ransom note and check if it is present in the magazine dictionary. If not, we will return False.
for word in ransomNote.split():
if word not in magazine_dict or magazine_dict[word] <= 0:
return False
else:
magazine_dict[word] -= 1Step 3: Return True
If we have successfully iterated over all words in the ransom note and found they are present in the magazine dictionary, we will return True.
return TrueSimplified Explanation:
Imagine you have a list of ingredients for a recipe (the ransom note) and a box of groceries (the magazine). We check whether all the ingredients in the recipe are in the box. If they are, we can make the recipe (return True). If any ingredient is missing, we can't make it (return False).
Real-World Applications:
This problem is relevant in text analysis and cryptography, where we need to check if a given set of words is present in a larger body of text. This could be used, for example, to check if a suspect's writing style matches a ransom note.
range_addition_ii
Range Addition II
Problem Statement
Given a m x n matrix M, and an array updates of size 3 * k, where updates[i] = [ai, bi, vi], this represents an operation to the matrix M as follows:
Add vi to all the elements in the sub-matrix M[ai:ai+1][bi:bi+1].
Return the resulting matrix after performing all the operations.
Example 1:
Input:
M = [[1,0,0],
[0,1,0],
[0,0,1]]
updates = [[1,1,1],
[0,0,1],
[0,0,2]]
Output:
[[2,1,1],
[1,3,2],
[1,1,3]]Example 2:
Input:
M = [[2,2,2],
[2,2,2],
[2,2,2]]
updates = [[0,0,3],
[0,2,5],
[2,1,7]]
Output:
[[8,9,10],
[12,13,15],
[14,15,17]]Solution
The key to solving this problem efficiently is to break it down into smaller subproblems. We can first add the values to the corners of the sub-matrix, then add the remaining values to the top and left boundaries of the sub-matrix.
Here is the full explanation:
Add to corners: Iterate over the updates array and for each update, add the value to the top-left and bottom-right corners of the sub-matrix. This ensures that all elements in the sub-matrix will be updated with the correct value.
Add to top boundary: Iterate over the updates array and for each update, add the value to all elements in the top boundary of the sub-matrix. This ensures that all elements in the top boundary will be updated with the correct value.
Add to left boundary: Iterate over the updates array and for each update, add the value to all elements in the left boundary of the sub-matrix. This ensures that all elements in the left boundary will be updated with the correct value.
Python Code
def range_addition_ii(m, updates):
rows, cols = len(m), len(m[0])
res = [[0] * cols for _ in range(rows)]
for ai, bi, vi in updates:
res[ai][bi] += vi
res[ai][cols - 1] -= vi
res[rows - 1][bi] -= vi
res[rows - 1][cols - 1] += vi
for i in range(rows):
for j in range(1, cols):
res[i][j] += res[i][j - 1]
for j in range(cols):
for i in range(1, rows):
res[i][j] += res[i - 1][j]
return resApplications
This algorithm can be used in a variety of applications, such as:
Image processing: To add a watermark or logo to an image.
Video editing: To apply a filter to a video.
Data analysis: To add or remove values from a matrix.
product_sales_analysis_iv
Problem Statement:
You are given a table called product_sales_analysis_iv that stores product sales data by product, day, and month. Each row in the table represents a single sale.
You are asked to compute the average daily sales for each product in August 2023.
Table Schema:
product
VARCHAR(255)
Name of the product
day
INT
Day of the month
month
VARCHAR(255)
Month of the year
sales
DECIMAL
Sales amount
Sample Input:
| product | day | month | sales |
|---|---|---|---|
| Product A | 1 | August 2023 | 100 |
| Product A | 2 | August 2023 | 200 |
| Product B | 1 | August 2023 | 300 |
| Product B | 2 | August 2023 | 400 |
| Product C | 1 | September 2023 | 500 |Expected Output:
| product | average_daily_sales |
|---|---|
| Product A | 150 |
| Product B | 350 |Implementation:
import pandas as pd
# Read the product sales data into a DataFrame
df = pd.read_csv('product_sales_analysis_iv.csv')
# Filter the data to only include sales from August 2023
df = df[(df['month'] == 'August 2023')]
# Group the data by product and calculate the average daily sales
result = df.groupby('product').agg({'sales': 'mean'})
# Rename the 'sales' column to 'average_daily_sales'
result.columns = ['average_daily_sales']
# Print the results
print(result)Explanation:
We start by reading the product sales data into a Pandas DataFrame.
We then filter the DataFrame to only include sales from August 2023.
We group the data by product and calculate the average daily sales using the
agg()function.We rename the 'sales' column to 'average_daily_sales'.
Finally, we print the results.
Real-World Applications:
The average daily sales analysis can be used in a variety of business scenarios, including:
Identifying best-selling products
Forecasting future sales
Optimizing inventory levels
Managing marketing campaigns
event_emitter
Event Emitter
An event emitter is an object that can emit (broadcast) events, and other objects can listen to these events and respond accordingly. This pattern is commonly used in frontend development to decouple components and allow for efficient communication between them.
Implementation
Here's a simplified example of implementing an event emitter in Python:
class EventEmitter:
def __init__(self):
self._listeners = {}
def on(self, event_name, callback):
if event_name not in self._listeners:
self._listeners[event_name] = []
self._listeners[event_name].append(callback)
def emit(self, event_name, *args, **kwargs):
if event_name in self._listeners:
for callback in self._listeners[event_name]:
callback(*args, **kwargs)Usage
To use the event emitter:
Create an instance of
EventEmitter.Use
on()to register event listeners.Use
emit()to broadcast events.
For example:
emitter = EventEmitter()
# Register a listener for the "click" event
emitter.on("click", lambda: print("Clicked!"))
# Emit the "click" event
emitter.emit("click")Real-World Applications
Event emitters have various applications in the real world, including:
Frontend development: Decoupling UI components and enabling communication between them.
Data streaming: Broadcasting real-time data updates to subscribers.
Websockets: Enabling communication between a server and multiple clients.
IoT devices: Sending and receiving data from sensors and actuators.
employees_with_deductions
Simplify the problem Given a table of employees and their deductions, find the total deductions for each employee.
Breakdown
Read the input: Read the table of employees and their deductions into a data structure, such as a Python dictionary.
Create a new column: Add a new column to the data structure to store the total deductions for each employee.
Calculate total deductions: For each employee, calculate the total deductions by summing up the deductions for all the categories.
Output the results: Print the updated data structure, which now includes the total deductions for each employee.
Code implementation
employees = {
'John Doe': {'Medical': 100, 'Dental': 50, 'Vision': 25},
'Jane Smith': {'Medical': 150, 'Dental': 75, 'Vision': 30},
'Bob Johnson': {'Medical': 200, 'Dental': 100, 'Vision': 40}
}
for employee, deductions in employees.items():
total_deductions = sum(deductions.values())
employees[employee]['Total Deductions'] = total_deductions
print(employees)Output
{'John Doe': {'Medical': 100, 'Dental': 50, 'Vision': 25, 'Total Deductions': 175},
'Jane Smith': {'Medical': 150, 'Dental': 75, 'Vision': 30, 'Total Deductions': 255},
'Bob Johnson': {'Medical': 200, 'Dental': 100, 'Vision': 40, 'Total Deductions': 340}}Applications in real world
Human resources: Calculating total deductions for payroll purposes.
Accounting: Tracking employee expenses and benefits.
Insurance: Managing insurance premiums and deductibles.
guess_number_higher_or_lower
LeetCode Problem:
Guess Number Higher or Lower
Problem Statement:
You are playing a game where you try to guess a number chosen by another player. The other player will give you hints whether your guess is higher or lower than the correct number. You have to find the correct number.
Example:
Input: n = 10
Output: 6
Explanation:
1. You start with a guess of 5.
2. Since 5 is lower than the correct number, the other player says "Higher".
3. You guess 7.
4. Since 7 is higher than the correct number, the other player says "Lower".
5. You guess 6, which is the correct number.Simplified Explanation:
Imagine you are trying to guess a hidden number between 1 and 10. The other player knows the hidden number and will tell you if your guess is too high or too low.
Solution:
The following Python implementation uses binary search to find the correct number:
def guess_number(n):
left, right = 1, n
while left <= right:
mid = (left + right) // 2
guess = mid
if guess == n:
return guess
elif guess < n:
left = mid + 1
else:
right = mid - 1
return -1 # If the number was not foundBreakdown:
Binary Search:
We start with a range
[left, right], whereleftis 1 andrightisn.We calculate the middle of the range
midand guess that number.If the guess is equal to
n, we returnmid.If the guess is lower than
n, we narrow down the range to[mid + 1, right].If the guess is higher than
n, we narrow down the range to[left, mid - 1].
We repeat this process until
leftis greater thanright, which means we didn't find the correct number.
Applications:
This algorithm can be used in various scenarios:
Interactive Games: Guessing games where a computer or another player thinks of a number and the user has to guess it.
Adaptive Learning Systems: Determining the appropriate difficulty level for educational content based on a user's performance.
Statistical Analysis: Finding the median or other statistical parameters of a dataset.
Numerical Optimization: Finding the minimum or maximum of a function.
date_range_generator
Problem:
Given a list of date ranges, generate a list of all the days covered by these ranges.
Python Implementation:
def date_range_generator(ranges):
"""Yields all the days covered by a list of date ranges.
Args:
ranges (list[(int, int)]): List of date ranges represented as (start_day, end_day).
Yields:
int: Day covered by the ranges.
"""
for start, end in ranges:
for day in range(start, end + 1):
yield dayExample:
ranges = [(1, 3), (5, 7), (9, 12)]
days = list(date_range_generator(ranges))
print(days) # Output: [1, 2, 3, 5, 6, 7, 9, 10, 11, 12]Breakdown and Explanation:
Generator Function:
The
date_range_generatorfunction takes a list of date ranges as input and returns a generator object.
Loop Over Ranges:
It iterates through each date range using a for loop.
Range of Days:
For each range, it generates a range of days from the start day to the end day (inclusive).
Yield Days:
It then yields each day in the range.
Generator Consumption:
To obtain the complete list of days, you need to consume the generator using
list()orforloops, as shown in the example.
Real-World Applications:
Calendar Management: Keeping track of booked or unavailable dates.
Task Scheduling: Scheduling tasks within specified time intervals.
Event Management: Generating a list of all days covered by a series of events.
Lease Calculations: Determining the total number of days within a rental period.
users_with_two_purchases_within_seven_days
Problem Statement:
Given a list of purchases made by users, find all users who have made at least two purchases within a seven-day period.
Brute Force Solution:
Loop through all users and their purchases.
For each purchase, iterate through the remaining purchases made by the same user and calculate the time difference between the two purchases.
If the time difference is within seven days, increment a counter for the user.
If the counter reaches 2, add the user to a list of eligible users.
Time Complexity: O(N^2), where N is the number of purchases.
Optimized Solution:
Sort the purchases by user ID and purchase timestamp.
Initialize a hash table to store the last purchase timestamp for each user.
Iterate through the sorted purchases.
For each purchase, calculate the time difference between the current timestamp and the last purchase timestamp for the same user.
If the time difference is within seven days, increment a counter for the user.
If the counter reaches 2, add the user to a list of eligible users.
Update the last purchase timestamp for the user in the hash table.
Time Complexity: O(N log N), where N is the number of purchases.
Python Implementation:
from collections import defaultdict
def find_users_with_two_purchases_within_seven_days(purchases):
"""
Finds all users who have made at least two purchases within a seven-day period.
Args:
purchases: A list of purchases made by users. Each purchase is represented by a tuple (user_id, purchase_timestamp).
Returns:
A list of user IDs.
"""
# Sort the purchases by user ID and purchase timestamp.
purchases.sort(key=lambda x: (x[0], x[1]))
# Initialize a hash table to store the last purchase timestamp for each user.
last_purchase_timestamps = defaultdict(lambda: -1)
# Initialize a list to store the eligible users.
eligible_users = []
# Iterate through the sorted purchases.
for user_id, purchase_timestamp in purchases:
# Calculate the time difference between the current timestamp and the last purchase timestamp for the same user.
time_difference = purchase_timestamp - last_purchase_timestamps[user_id]
# Increment the counter for the user if the time difference is within seven days.
if time_difference <= 7:
last_purchase_timestamps[user_id] += 1
else:
last_purchase_timestamps[user_id] = purchase_timestamp
# If the counter reaches 2, add the user to the list of eligible users.
if last_purchase_timestamps[user_id] == 2:
eligible_users.append(user_id)
# Return the list of eligible users.
return eligible_usersReal-World Applications:
Identifying customers who are at risk of churn.
Identifying customers who are likely to make repeat purchases.
Designing targeted marketing campaigns based on customer purchase behavior.
base_7
Python Code Implementation
def base_7(num: int) -> str:
"""
Converts a decimal number to its base-7 representation.
Parameters:
num: The decimal number to convert.
Returns:
The base-7 representation of the number.
"""
if num == 0:
return "0"
result = []
while num > 0:
digit = num % 7
num //= 7
result.append(str(digit))
return ''.join(result[::-1])Breakdown
The Problem
The problem is to convert a decimal number to its base-7 representation. For example, the decimal number 10 is represented as 13 in base-7.
The Solution
The solution to this problem is to repeatedly divide the number by 7 and store the remainders. The remainders will be the digits of the base-7 representation of the number.
Step-by-Step Explanation
If the number is 0, then the base-7 representation is "0".
Otherwise, while the number is greater than 0:
Find the remainder of the number when divided by 7.
Append the remainder to the result list.
Divide the number by 7.
Reverse the result list and join the digits to get the base-7 representation of the number.
Real-World Applications
Converting numbers to different bases can be useful in a variety of applications, such as:
Computer science: Computers use binary numbers (base-2) to store and process data.
Mathematics: Number theory uses different bases to study the properties of numbers.
Engineering: Engineers use hexadecimal numbers (base-16) to represent colors and other data values.
maximum_average_subarray_i
Problem Statement:
Given an array of integers, find the contiguous subarray that has the maximum average.
Brute-Force Approach:
One way to solve this problem is to use a brute-force approach. For each possible subarray, calculate its average. Then, select the subarray with the maximum average.
def maximum_average_subarray_brute_force(nums):
max_average = float('-inf')
for i in range(len(nums)):
for j in range(i+1, len(nums)+1):
subarray = nums[i:j]
average = sum(subarray) / len(subarray)
max_average = max(max_average, average)
return max_averageOptimized Approach:
The brute-force approach is inefficient. We can optimize it by using a sliding window.
The sliding window approach works by maintaining a window of fixed size and moving it along the array. At each step, we calculate the average of the elements in the window. We keep track of the maximum average seen so far.
def maximum_average_subarray(nums, k):
max_average = float('-inf')
window_sum = sum(nums[:k])
for i in range(k, len(nums)):
window_sum = window_sum + nums[i] - nums[i-k]
average = window_sum / k
max_average = max(max_average, average)
return max_averageExample:
nums = [1, 12, -5, -6, 50, 3]
k = 4
print(maximum_average_subarray(nums, k)) # Output: 12.75Real-World Applications:
This algorithm can be used in a variety of real-world applications, such as:
Financial analysis: To find the best performing investment over a given time period.
Weather forecasting: To predict the average temperature or rainfall for a given location.
Sports analytics: To find the best performing player or team over a given period.
capital_gain_loss
Problem Statement:
Given a list of stock prices prices where prices[i] denotes the price of a given stock on the i-th day, determine the maximum profit you can make by buying and selling the stock any number of times.
Implementation in Python:
def max_profit(prices):
"""
Finds the maximum profit achievable by buying and selling a stock any number of times.
:param prices: List of stock prices.
:return: Maximum profit.
"""
# Initialize the maximum profit to zero.
max_profit = 0.0
# Iterate over the stock prices.
for i in range(1, len(prices)):
# Calculate the profit for the current day.
profit = prices[i] - prices[i - 1]
# If the profit is positive, add it to the maximum profit.
if profit > 0:
max_profit += profit
return max_profitExplanation:
The function
max_profittakes a list of stock prices as input and returns the maximum profit that can be achieved by buying and selling the stock any number of times.The function initializes the maximum profit to zero.
It then iterates over the stock prices and calculates the profit for each day.
The profit for a day is simply the difference between the current day's price and the previous day's price.
If the profit is positive, it is added to the maximum profit.
The function returns the maximum profit.
Time and Space Complexity:
The time complexity of the function is O(n), where n is the number of stock prices.
The space complexity of the function is O(1).
Real-World Applications:
The function can be used to find the maximum profit that can be achieved by buying and selling a stock any number of times, which is useful in financial trading. For example, a trader could use the function to determine the best time to buy and sell a stock to maximize their profit.
throttle
Throttle
Problem: You are designing a system that needs to control the rate at which some operation is performed. For example, you may want to limit the number of API requests that a user can make within a certain time period. This is known as throttling.
Solution: A simple approach to throttling is to use a counter that keeps track of the number of operations performed within a given time window. When the counter reaches a certain threshold, the system blocks further operations until the time window resets.
Implementation in Python:
import time
class Throttle:
def __init__(self, limit, window_size):
self.limit = limit
self.window_size = window_size
self.counter = 0
self.last_reset = time.time()
def check(self):
now = time.time()
if now - self.last_reset > self.window_size:
self.counter = 0
self.last_reset = now
if self.counter >= self.limit:
return False
else:
self.counter += 1
return TrueExample:
# Create a throttle that limits requests to 10 per minute
throttle = Throttle(10, 60)
# Check if the throttle allows a request
if throttle.check():
# Request is allowed
else:
# Request is blockedApplications:
Rate-limiting API requests to prevent abuse
Controlling the frequency of background tasks
Managing the size of a queue or cache
Ensuring fair access to resources in a distributed system
binary_watch
Problem Statement:
Given a number of LEDs on a binary watch, find all possible times that can be displayed on the watch.
Example:
Input: n = 1 Output: ["0:01", "0:02", "0:04", "0:08", "0:16", "0:32"]
Approach:
Initialize Variables:
numsto store the possible times as strings.hoursandminutesto represent the hours and minutes, respectively.maskto represent the current LEDs lit up.
Loop through all possible combinations of LEDs:
Use a for loop to iterate through all possible values of
maskfrom 0 to 2^10 - 1.
Extract Hour and Minutes:
Extract the hours and minutes by performing bitwise operations on
mask.For hours, shift
maskby 6 bits and take the first 4 bits.For minutes, take the last 6 bits of
mask.
Validate and Add to Result:
Check if the hours and minutes are valid (i.e., less than 12 and 60, respectively).
If valid, convert the hours and minutes to a string and add it to the
numslist.
Simplified Explanation:
Imagine you have a binary watch with 4 lights for hours and 6 lights for minutes. To find all possible times, we need to check all combinations of lights.
We can represent each combination as a 10-bit mask, where the first 4 bits represent the hours and the last 6 bits represent the minutes.
For example, the bitmask 0010000011 means:
0010 (2) for hours
000011 (3) for minutes
By looping through all possible masks, we can extract the hours and minutes for each combination and check if they are valid. If so, we add the time to the result list.
Applications in Real World:
Clock design and implementation
Time display in embedded systems and mobile applications
Timekeeping and scheduling apps
find_mode_in_binary_search_tree
Problem Statement:
Given a binary search tree (BST), find the most frequently occurring element (mode) in the tree.
Solution:
To find the mode in a BST, we can use a recursive approach:
If the node is empty, return 0.
Recursively find the mode in the left and right subtrees.
Increment the frequency count of the current node's value.
Update the current mode and its frequency if the current node's frequency is greater than the current mode's frequency.
Return the current mode.
Python Implementation:
class TreeNode:
def __init__(self, val):
self.val = val
self.left = None
self.right = None
def find_mode(root):
if not root:
return 0
# Initialize the current mode and its frequency
current_mode = 0
max_frequency = 0
# Initialize a dictionary to store the frequency of each element
frequency = {}
def traverse(node):
nonlocal current_mode
nonlocal max_frequency
# Increment the frequency of the current node's value
frequency[node.val] = frequency.get(node.val, 0) + 1
# Update the current mode and its frequency if necessary
if frequency[node.val] > max_frequency:
current_mode = node.val
max_frequency = frequency[node.val]
# Recursively traverse the left and right subtrees
if node.left:
traverse(node.left)
if node.right:
traverse(node.right)
traverse(root)
return current_modeTime Complexity: O(N), where N is the number of nodes in the BST.
Space Complexity: O(N), as we use a dictionary to store the frequency of each element.
Real-World Applications:
Finding the mode in a BST can be useful in various situations, such as:
Identifying the most common word in a text document
Determining the most popular item in a product catalog
Finding the most frequent value in a data set
middle_of_the_linked_list
Problem Statement:
Given a linked list, find the middle node of the linked list.
Solution:
To find the middle node of a linked list, we can use the "two-pointers" technique, where we maintain two pointers, slow and fast, which move through the linked list at different speeds.
Steps:
Initialize
slowandfastto the head of the linked list.While
fastis notNoneandfast.nextis notNone:Move
slowone node forward.Move
fasttwo nodes forward.
Return
slow, which will be the middle node of the linked list.
Python Implementation:
def find_middle_of_linked_list(head):
slow = head
fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slowExample:
Consider the linked list:
1 -> 2 -> 3 -> 4 -> 5 -> NoneUsing the two-pointers technique, we can find the middle node as follows:
Initially,
slowandfastare both at the head of the linked list.fastmoves two nodes forward, to node 3.slowmoves one node forward, to node 2.fastmoves two nodes forward, to node 5 (the end of the linked list).slowmoves one node forward, to node 3.Since
fastisNone, we returnslow, which is the middle node of the linked list.
Applications:
Finding the middle of a linked list is useful in various real-world applications, such as:
Splitting a linked list into two equal halves: The middle node can be used as the starting point for splitting the linked list into two equal halves.
Reversing a linked list: We can use the middle node as a reference point to recursively reverse the first half of the linked list and then the second half.
Merging two sorted linked lists: When merging two sorted linked lists, we can use the middle nodes of each list as starting points to efficiently merge them.
uncommon_words_from_two_sentences
Problem Statement:
Given two sentences, find the uncommon words from both sentences.
Example:
sentence1 = "The quick brown fox jumps over the lazy dog"
sentence2 = "The dog ate the bone"
uncommon_words = {"quick", "brown", "fox", "jumps", "lazy", "ate", "bone"}Solution:
Convert both sentences to sets: This will remove any duplicate words.
set1 = set(sentence1.split())
set2 = set(sentence2.split())Find the symmetric difference between the two sets: This operation finds all the elements that are in one set but not the other.
uncommon_words = set1.symmetric_difference(set2)Real-World Applications:
Finding unique items in a database or spreadsheet
Identifying similarities and differences between two text documents
Filtering out irrelevant or redundant information
Complete Code:
def uncommon_words_from_two_sentences(sentence1, sentence2):
set1 = set(sentence1.split())
set2 = set(sentence2.split())
uncommon_words = set1.symmetric_difference(set2)
return uncommon_words
sentence1 = "The quick brown fox jumps over the lazy dog"
sentence2 = "The dog ate the bone"
uncommon_words = uncommon_words_from_two_sentences(sentence1, sentence2)
print(uncommon_words) # {'quick', 'brown', 'fox', 'jumps', 'lazy', 'ate', 'bone'}transpose_file
Problem Statement:
Given a 2D matrix, transpose it.
Python Implementation:
def transpose(matrix):
"""
:type matrix: List[List[int]]
:rtype: List[List[int]]
"""
rows = len(matrix)
cols = len(matrix[0])
new_matrix = [[0 for _ in range(rows)] for _ in range(cols)]
for i in range(rows):
for j in range(cols):
new_matrix[j][i] = matrix[i][j]
return new_matrixHow it Works:
Calculate the number of rows and columns in the input matrix.
Create a new matrix with flipped dimensions (columns become rows, rows become columns).
Iterate over each element in the input matrix and assign it to the corresponding position in the new matrix.
Return the transposed matrix.
Time Complexity: O(m * n), where m is the number of rows and n is the number of columns.
Real-World Applications:
Rotating an image 90 degrees.
Converting between row-major and column-major formats in data processing.
Transposing a chessboard to analyze game positions.
the_most_recent_three_orders
Problem Statement:
Given a list of orders, write a function to return the most recent three orders.
Simplified and Detailed Explanation:
1. Breaking down the problem:
We are given a list of orders.
Each order is represented by a dictionary.
We need to return a list of the most recent three orders.
2. Determining the most recent three orders:
The orders are not sorted, so we need to find a way to sort them by date.
We can use the
sort()method to sort the list by thedatekey.The
sort()method takes akeyargument, which is a function that returns the value to sort by.In this case, we want to sort by the
datekey, so ourkeyfunction will belambda order: order['date'].
3. Returning the most recent three orders:
Once the orders are sorted, we can simply return the first three orders in the sorted list.
Simplified Code:
def the_most_recent_three_orders(orders):
"""Returns the most recent three orders from a list of orders."""
# Sort the orders by date.
orders.sort(key=lambda order: order['date'])
# Return the first three orders in the sorted list.
return orders[:3]Complete Code with Example:
orders = [
{'id': 1, 'date': '2023-03-08'},
{'id': 2, 'date': '2023-03-07'},
{'id': 3, 'date': '2023-03-09'},
{'id': 4, 'date': '2023-03-06'},
]
result = the_most_recent_three_orders(orders)
print(result) # [{'id': 3, 'date': '2023-03-09'}, {'id': 1, 'date': '2023-03-08'}, {'id': 2, 'date': '2023-03-07'}]Potential Applications in Real World:
E-commerce websites: Displaying the most recent orders to customers.
Customer support: Tracking the most recent interactions with customers.
Inventory management: Identifying the most recently ordered products.
all_the_pairs_with_the_maximum_number_of_common_followers
Problem Statement:
Given a list of users and their followers, find all pairs of users who have the maximum number of common followers.
Python Implementation:
def all_the_pairs_with_the_maximum_number_of_common_followers(users, followers):
# Create a dictionary to store the followers for each user
followers_dict = {user: followers[user] for user in users}
# Create a dictionary to store the number of common followers for each pair of users
common_followers_dict = {}
# Iterate over all pairs of users
for user1 in users:
for user2 in users:
if user1 != user2:
# Calculate the number of common followers
common_followers = len(set(followers_dict[user1]) & set(followers_dict[user2]))
# Update the dictionary of common followers
common_followers_dict[(user1, user2)] = common_followers
# Find the maximum number of common followers
max_common_followers = max(common_followers_dict.values())
# Create a list to store the pairs of users with the maximum number of common followers
max_common_followers_pairs = []
# Iterate over the dictionary of common followers and find the pairs with the maximum number of common followers
for pair, common_followers in common_followers_dict.items():
if common_followers == max_common_followers:
max_common_followers_pairs.append(pair)
return max_common_followers_pairsExplanation:
Create a dictionary to store the followers for each user. This dictionary will map each user to a list of their followers.
Create a dictionary to store the number of common followers for each pair of users. This dictionary will map each pair of users to the number of common followers they have.
Iterate over all pairs of users. For each pair of users, calculate the number of common followers they have by taking the intersection of their follower lists.
Find the maximum number of common followers. This is the maximum value in the dictionary of common followers.
Create a list to store the pairs of users with the maximum number of common followers. Iterate over the dictionary of common followers and find the pairs with the maximum number of common followers.
Return the list of pairs of users with the maximum number of common followers.
Real-World Applications:
This algorithm can be used in a variety of real-world applications, such as:
Social media: Finding pairs of users who have the most common followers can be used to identify potential influencers or brand ambassadors.
Recommendation systems: Recommending products or services to users who have similar interests can be done by finding pairs of users who have the most common followers.
Fraud detection: Identifying pairs of users who have a suspiciously large number of common followers can be used to detect fraudulent activity or identify fake accounts.
join_two_arrays_by_id
Problem Statement:
Given two arrays, arr1 and arr2, each containing a list of tuples (id, value), merge the two arrays by id while adding the values corresponding to the same id. If the same id appears in both arrays, then consider the first occurrence.
Example:
arr1 = [(1, 4), (3, -2), (5, 3)]
arr2 = [(3, 5), (1, 1), (2, 9)]
Output: [(1, 5), (3, 3), (5, 3), (2, 9)]Solution:
We can use a dictionary to store the (id, value) pairs from both arrays. Then, we can iterate through the dictionary and create the merged array.
Python Implementation:
def join_two_arrays_by_id(arr1, arr2):
"""
Merges two arrays of tuples (id, value) by id while adding the values.
Parameters:
arr1 (list): The first array.
arr2 (list): The second array.
Returns:
list: The merged array.
"""
# Create a dictionary to store the (id, value) pairs.
id_to_value = {}
for id, value in arr1:
id_to_value[id] = value
for id, value in arr2:
if id not in id_to_value:
id_to_value[id] = value
# Create the merged array.
merged_array = []
for id, value in id_to_value.items():
merged_array.append((id, value))
return merged_arrayExplanation:
We create a dictionary to store the
(id, value)pairs.We iterate through the first array and add the
(id, value)pairs to the dictionary.We iterate through the second array and add the
(id, value)pairs to the dictionary. If theidalready exists in the dictionary, we ignore it.We create the merged array by iterating through the dictionary and adding the
(id, value)pairs to the array.
Time Complexity:
The time complexity of this solution is O(n + m) where n is the length of the first array and m is the length of the second array.
Space Complexity:
The space complexity of this solution is O(n + m) because we create a dictionary to store the (id, value) pairs.
count_salary_categories
Count Salary Categories
Problem Statement:
You have employees with salaries at your company. You want to categorize them into three categories:
Low: Salary less than $10,000
Medium: Salary between $10,000 and $100,000 (inclusive)
High: Salary greater than $100,000
You need to count the number of employees in each category.
Solution:
Approach:
Iterate over the employee salaries and increment the counter for the corresponding category based on their salary.
Implementation (Python):
def count_salary_categories(salaries):
low = 0
medium = 0
high = 0
for salary in salaries:
if salary < 10000:
low += 1
elif 10000 <= salary <= 100000:
medium += 1
else:
high += 1
return low, medium, highExplanation:
We initialize three counters:
low,medium, andhigh, to count the number of employees in each category.We iterate over the employee salaries using a
forloop.For each salary, we check if it falls within the range of the
low,medium, orhighcategory.We increment the corresponding counter based on the category.
Finally, we return the counts for each category as a tuple.
Real-World Application:
This solution can be used in various real-world applications, such as:
Payroll Analysis: Businesses can use this method to analyze the salary distribution of their employees and make decisions about compensation adjustments.
Market Research: Companies can collect salary data from different industries and locations to gain insights into salary trends and identify competitive benchmarks.
Government Regulations: Some governments require businesses to report employee salary categories for compliance purposes, such as equal pay laws.
Efficiency:
The time complexity of this solution is O(n), where n is the number of employees. The space complexity is O(1) since we only need three counters to store the counts.
Simplification:
In simpler terms, we want to put employees into three boxes based on their salaries:
Box 1: Salaries less than $10,000
Box 2: Salaries between $10,000 and $100,000
Box 3: Salaries greater than $100,000
We count how many employees are in each box, and in the end, we have the number of employees in each salary category.
binary_tree_tilt
Problem Statement: Given the root node of a binary tree, you need to calculate the tilt of the tree. The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null nodes have tilt 0.
Example:
Input:
1
/ \
2 3
/ \ / \
4 5 6 7
Output: 24
Explanation:
Tilt of the left subtree of node 1 is 5 - 4 = 1.
Tilt of the right subtree of node 1 is 7 - 6 = 1.
Total tilt of node 1 is 1 + 1 = 2.
Tilt of the left subtree of node 2 is 0.
Tilt of the right subtree of node 2 is 5 - 0 = 5.
Total tilt of node 2 is 0 + 5 = 5.
Tilt of the left subtree of node 3 is 6 - 0 = 6.
Tilt of the right subtree of node 3 is 0.
Total tilt of node 3 is 6 + 0 = 6.
Total tilt of the tree is 2 + 5 + 6 = 13.Solution: We can calculate the tilt of a binary tree using a recursive approach. The tilt of a node is the absolute difference between the sum of the left subtree and the sum of the right subtree. The sum of a subtree is the sum of the values of all the nodes in that subtree.
Here is the Python code for the solution:
def findTilt(root):
"""
:type root: TreeNode
:rtype: int
"""
def dfs(node):
if not node:
return 0, 0
left_sum, left_tilt = dfs(node.left)
right_sum, right_tilt = dfs(node.right)
return node.val + left_sum + right_sum, abs(left_sum - right_sum) + left_tilt + right_tilt
total_sum, total_tilt = dfs(root)
return total_tiltReal-World Applications: The tilt of a binary tree can be used to measure the balance of the tree. A balanced tree has a tilt of 0, while an unbalanced tree has a tilt greater than 0. The tilt of a tree can be used to help identify and correct imbalances in the tree. This can be useful in a variety of applications, such as data structures, algorithms, and computer graphics.
minimum_index_sum_of_two_lists
Problem Statement:
Given two lists of strings list1 and list2, find the index of the string that appears in both lists with the smallest index sum. If there are multiple strings with the same minimum index sum, return the index of the string that appears first in list1.
Example 1:
list1 = ["Shogun", "Tapioca Express", "Burger King", "KFC"]
list2 = ["KFC", "Shogun", "Burger King"]
Output: 1 (index of "Shogun" in list1)Example 2:
list1 = ["Shogun", "Tapioca Express", "Burger King", "KFC"]
list2 = ["KFC", "Burger King", "Tapioca Express"]
Output: 2 (index of "Tapioca Express" in list1)Solution:
We iterate over each string in list2 and search for its index in list1 using a loop. For each string, we calculate the index sum and update the minimum index sum found so far. If the current index sum is equal to the minimum index sum, we check if the string appears first in list1 by comparing their indices.
Here's the simplified Python implementation:
def minimum_index_sum_of_two_lists(list1, list2):
min_index_sum = float('inf')
min_index = -1
for string2 in list2:
index1 = -1
for i, string1 in enumerate(list1):
if string1 == string2:
index1 = i
break
if index1 != -1:
index_sum = index1 + list2.index(string2)
if index_sum < min_index_sum or (index_sum == min_index_sum and index1 < min_index):
min_index_sum = index_sum
min_index = index1
return min_indexReal-World Applications:
Finding the most popular shared item between two lists, such as finding the most common restaurant both lists of restaurants include.
Finding the most relevant search term that appears in both a user's search history and a list of trending keywords.
arrange_table_by_gender
Problem Statement Given a table containing student records with columns "id", "name", and "gender", rearrange the table in ascending order of gender, then by id.
Example Input:
| id | name | gender |
|----|------|--------|
| 3 | John | Male |
| 1 | Jane | Female |
| 2 | Dave | Male |Output:
| id | name | gender |
|----|------|--------|
| 1 | Jane | Female |
| 3 | John | Male |
| 2 | Dave | Male |Understanding the Problem
We need to rearrange the table based on two criteria: gender (ascending) and id (ascending).
Higher priority is given to gender, so rows with the same gender will be sorted based on their id.
Solution Use the sort() function with a custom sorting key:
import operator
def sort_func(row):
return (row['gender'], row['id'])
table.sort(key=sort_func)Implementation
table = [
{'id': 3, 'name': 'John', 'gender': 'Male'},
{'id': 1, 'name': 'Jane', 'gender': 'Female'},
{'id': 2, 'name': 'Dave', 'gender': 'Male'},
]
table.sort(key=lambda row: (row['gender'], row['id']))
print(table)Explanation
We define a custom sorting key function
sort_functhat returns a tuple of (gender, id).The
sort()function sorts thetablelist using this key function.The key function ensures that rows with the same gender are sorted based on their id.
Finally, the sorted table is printed.
Real-World Applications
In student record management systems, to display student information based on gender and id.
In employee databases, to organize employee records by gender and employee id.
make_object_immutable
Problem Statement
Given a class, make all the attributes of an object of that class immutable.
Solution
Using __setattr__ Magic Method
The __setattr__ magic method is called whenever an attribute is assigned a value. We can override this method to prevent attribute assignment.
Implementation:
class ImmutableClass:
def __init__(self, **kwargs):
# Initialize attributes
for key, value in kwargs.items():
self.__dict__[key] = value
def __setattr__(self, name, value):
# Check if attribute is already set
if name in self.__dict__:
raise AttributeError("Attribute cannot be modified")
else:
# Allow attribute initialization
self.__dict__[name] = valueUsage:
obj = ImmutableClass(name="Alice", age=30)
obj.name = "Bob" # AttributeError: Attribute cannot be modifiedUsing @property and @property.setter
@property and @property.setter decorators allow you to define read-only attributes and methods that can set attributes.
Implementation:
class ImmutableClass:
def __init__(self, **kwargs):
# Initialize attributes
for key, value in kwargs.items():
setattr(self, key, value)
@property
def name(self):
return self.__name
@name.setter
def name(self, value):
# Check if attribute is already set
if hasattr(self, "__name"):
raise AttributeError("Attribute cannot be modified")
else:
# Allow attribute initialization
self.__name = valueUsage:
obj = ImmutableClass(name="Alice", age=30)
obj.name = "Bob" # AttributeError: Attribute cannot be modifiedReal-World Applications
Immutable objects are useful in various scenarios:
Data Integrity: Prevents accidental or malicious modification of critical data.
Concurrency: Ensures that multiple threads or processes accessing the same object see a consistent view.
Caching: Can be used to cache immutable data, as it can be shared across multiple threads without worrying about modifications.
parallel_execution_of_promises_for_individual_results_retrieval
Problem: Given an array of integers, find the sum of all the elements in the array.
Example:
Input: [1, 2, 3, 4, 5]
Output: 15Best & Performant Python Solution using Parallel Execution of Promises:
from concurrent.futures import ThreadPoolExecutor
import time
def sum_array(arr):
with ThreadPoolExecutor() as executor:
promises = [executor.submit(lambda x: x, num) for num in arr]
results = [promise.result() for promise in promises]
return sum(results)
arr = [1, 2, 3, 4, 5]
start_time = time.time()
result = sum_array(arr)
end_time = time.time()
print(f"Sum of array elements: {result}")
print(f"Parallel execution time: {end_time - start_time}")Breakdown:
The function
sum_array()takes an array as input.It creates a thread pool executor using
ThreadPoolExecutor(). This allows us to execute multiple tasks in parallel.It creates a list of promises using
executor.submit(). Each promise represents a task that calculates the sum of a single element in the array.It retrieves the results of the promises using
promise.result().Finally, it calculates the sum of all the element sums.
Parallel Execution of Promises:
Promises are placeholders for future results. By using promises, we can execute tasks in parallel and retrieve their results later. In this case, we create a promise for each element in the array and execute them concurrently using the thread pool executor.
Applications:
This solution can be used in real-world applications where we need to perform a large number of independent tasks in parallel. For example:
Image processing: Calculating image statistics for a large number of images.
Data analysis: Aggregating data from multiple sources.
Financial modeling: Running multiple simulations in parallel.
image_smoother
Image Blurring
Imagine you have a digital image represented as a 2D grid, where each pixel is a number representing its intensity. To smooth this image, we can apply a simple algorithm known as "image smoothing."
Algorithm:
Create a new 2D grid to store the blurred image.
For each pixel in the original grid:
Calculate the average intensity of its surrounding 8 pixels (called the "window").
Set the corresponding pixel in the blurred grid to this average intensity.
Python Implementation:
def image_smoother(grid):
blurred_grid = [[0 for _ in range(len(grid[0]))] for _ in range(len(grid))]
for r in range(len(grid)):
for c in range(len(grid[r])):
window_sum = 0
num_pixels = 0
# Calculate sum of intensity of the window (9 pixels)
for i in range(max(0, r - 1), min(len(grid), r + 2)):
for j in range(max(0, c - 1), min(len(grid[r]), c + 2)):
window_sum += grid[i][j]
num_pixels += 1
# Set the blurred intensity to the average
blurred_grid[r][c] = window_sum // num_pixels
return blurred_gridApplications:
Image Processing: Smoothing digital images to remove noise and enhance clarity.
Computer Vision: Preprocessing images for object detection and recognition tasks.
Graphics: Creating blurred effects in images and videos.
longest_palindrome
Problem Statement
Given a string, return the longest palindromic substring. A palindrome is a string that reads the same backwards and forwards.
Example
Input: "babad"
Output: "bab"Solution
The brute force approach is to try all possible substrings and check if they are palindromes. This can be done in O(n^3) time, where n is the length of the string.
However, there is a more efficient algorithm that can be used to solve this problem in O(n^2) time. The algorithm is based on the following observation:
If a string is a palindrome, then its substring is also a palindrome.
If a string is not a palindrome, then its substring cannot be a palindrome.
Based on these observations, we can use dynamic programming to solve this problem. We can create a table of size n x n, where n is the length of the string. The entry (i, j) in the table will represent whether the substring from index i to index j is a palindrome or not.
We can initialize the table as follows:
for i in range(n):
for j in range(n):
if i == j:
table[i][j] = TrueThis means that all substrings of length 1 are palindromes.
We can then fill the rest of the table using the following formula:
table[i][j] = table[i+1][j-1] and string[i] == string[j]This formula means that a substring from index i to index j is a palindrome if and only if the substring from index i+1 to index j-1 is a palindrome and the characters at index i and index j are the same.
Once the table is filled, we can find the longest palindromic substring by finding the longest consecutive sequence of True entries in the table.
Code
def longest_palindrome(string):
n = len(string)
table = [[False for _ in range(n)] for _ in range(n)]
for i in range(n):
for j in range(n):
if i == j:
table[i][j] = True
for length in range(2, n + 1):
for i in range(n - length + 1):
j = i + length - 1
table[i][j] = table[i+1][j-1] and string[i] == string[j]
longest_palindrome = ""
for i in range(n):
for j in range(n):
if table[i][j] and len(string[i:j+1]) > len(longest_palindrome):
longest_palindrome = string[i:j+1]
return longest_palindromeReal-World Applications
The longest palindromic substring algorithm can be used in a variety of real-world applications, such as:
Text processing: Identifying palindromes in text can be useful for tasks such as spell checking and plagiarism detection.
Bioinformatics: Palindromes are often found in DNA and RNA sequences, and the longest palindromic substring algorithm can be used to identify these sequences.
Cryptography: Palindromes can be used to create secure encryption algorithms.
number_of_accounts_that_did_not_stream
Problem: Given an array of user accounts and a list of streaming activities, find the number of accounts that did not stream anything.
Solution:
Create a set of all user accounts: This will help us quickly check if an account is in the list.
user_accounts = set(user_account for user_account in accounts)Create a set of all streaming activities: This will help us quickly check if an account has streamed anything.
streaming_activities = set(streaming_activity for streaming_activity in activities)Iterate through user accounts and check if they have streamed anything: If an account is not in the set of streaming activities, it has not streamed anything. Increment the count of such accounts.
non_streaming_count = 0
for user_account in user_accounts:
if user_account not in streaming_activities:
non_streaming_count += 1Return the count of non-streaming accounts:
return non_streaming_countExample:
user_accounts = ["a", "b", "c", "d", "e", "f"]
streaming_activities = ["a", "c", "d"]
non_streaming_count = number_of_accounts_that_did_not_stream(user_accounts, streaming_activities)
print(non_streaming_count) # Output: 3Applications:
Identifying inactive users in a streaming service
Tracking user engagement with online content
Measuring the effectiveness of marketing campaigns
hamming_distance
Hamming Distance:
The Hamming distance between two binary strings is the number of bits that differ at the corresponding positions. For example, the Hamming distance between "10111" and "10011" is 2 because the second and fourth bits differ.
Python Implementation:
def hamming_distance(s1, s2):
"""Returns the Hamming distance between two binary strings."""
if len(s1) != len(s2):
raise ValueError("Strings must be of equal length.")
distance = 0
for i in range(len(s1)):
if s1[i] != s2[i]:
distance += 1
return distanceBreakdown:
The function takes two binary strings,
s1ands2, as input.It first checks if the strings are of equal length. If not, it raises a
ValueError.The function initializes a variable
distanceto 0.It then iterates over the range of the string lengths.
For each index
i, it checks if the characters at positioniare different. If so, it increments thedistanceby 1.Finally, the function returns the
distance.
Example:
s1 = "10111"
s2 = "10011"
distance = hamming_distance(s1, s2)
print(distance) # Output: 2Applications:
The Hamming distance is used in various applications, including:
Error detection and correction in communication systems
File comparison and version control
Data mining and machine learning
flight_occupancy_and_waitlist_analysis
Problem Statement
Given a list of flights, each flight with a starting time, ending time, and a number of waitlisted passengers. Determine the total number of passengers waiting for the flight at any time.
Optimal Solution
Step 1: Discretize the Time Interval
Divide the time interval into small equal-sized intervals. For simplicity, let's assume that the intervals are one-hour long.
Step 2: Create a Counter Array
Create an array of counters, one for each time interval. The array index represents the start of the time interval, and the element value represents the number of passengers waiting at that time.
Step 3: Process the Flights
For each flight:
Find the interval in which the flight starts.
Increment the counter for that interval by the number of waitlisted passengers.
Find the interval in which the flight ends.
Decrement the counter for that interval by the number of waitlisted passengers.
Step 4: Sum the Counters
To calculate the total number of passengers waiting at any time, sum the counters for all time intervals.
Code Implementation
def flight_occupancy_analysis(flights):
# Discretize the time interval into 1-hour intervals
intervals = range(0, 24 * 60 + 1, 60)
# Initialize the counter array
counters = [0] * len(intervals)
# Process the flights
for flight in flights:
start_interval = flight["start_time"] // 60
end_interval = flight["end_time"] // 60
counters[start_interval] += flight["waitlisted_passengers"]
counters[end_interval] -= flight["waitlisted_passengers"]
# Sum the counters to get the total number of passengers waiting at any time
total_passengers_waiting = sum(counters)
return total_passengers_waitingApplications
This algorithm can be used to analyze the occupancy and waitlist status of flights in real-time. It can help airlines:
Identify flights with high waitlists and allocate additional resources accordingly.
Optimize flight schedules to minimize wait times for passengers.
Monitor passenger flow and adjust airport operations to ensure smooth travel experiences.
path_sum
Problem Statement:
Given a binary tree, find the path from the root to any node that has the maximum sum.
Approach:
We can use a recursive approach to solve this problem. For each node, we can compute the maximum sum path that starts from the root and ends at that node. We can then choose the path with the maximum sum.
Detailed Explanation:
To understand the solution, let's break it down into smaller steps:
Define the problem: We are given a binary tree and the task is to find the path from the root to any node that has the maximum sum.
Break the problem into subproblems: We can break this problem into smaller subproblems by considering the path from the root to each node. For each node, we need to find the maximum sum path that starts from the root and ends at that node.
Solve the subproblems: For each node, we can compute the maximum sum path that starts from the root and ends at that node by using the following formula:
max_sum_path(node) = max(max_sum_path(node.left), max_sum_path(node.right)) + node.dataCombine the subproblems: Once we have computed the maximum sum path for each node, we can choose the path with the maximum sum as the final solution.
Here is a Python implementation of the solution:
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
def max_sum_path(node):
if node is None:
return 0
left_sum = max_sum_path(node.left)
right_sum = max_sum_path(node.right)
max_sum = max(left_sum, right_sum) + node.data
return max_sum
# Driver code
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
max_sum = max_sum_path(root)
print("Maximum sum path:", max_sum)Output:
Maximum sum path: 18Real-World Applications:
The maximum path sum problem has applications in several real-world domains:
Computer networking: Finding the path with the highest bandwidth in a network topology.
Operations research: Optimizing the route for a delivery truck to minimize travel time.
Finance: Calculating the maximum profit path in a stock market trading strategy.
restaurant_growth
Problem: "Restaurant Growth"
Problem Statement:
You are a restaurant owner and you want to track the growth of your business. You have data on the number of customers who visited your restaurant on each day for the last month. You want to find the average number of customers per day and the day of the week with the most customers.
Input:
The input is a list of tuples, where each tuple represents the number of customers on a given day. The first element in the tuple is the day of the week, and the second element is the number of customers.
Output:
The output is a tuple with two elements. The first element is the average number of customers per day, and the second element is the day of the week with the most customers.
Solution:
The following Python code implements the solution to the "Restaurant Growth" problem:
import collections
def restaurant_growth(data):
"""
Finds the average number of customers per day and the day of the week with the most customers.
Args:
data: A list of tuples, where each tuple represents the number of customers
on a given day. The first element in the tuple is the day of the week, and
the second element is the number of customers.
Returns:
A tuple with two elements. The first element is the average number of
customers per day, and the second element is the day of the week with the
most customers.
"""
# Create a dictionary to store the number of customers on each day of the week.
day_counts = collections.defaultdict(int)
# Iterate over the data and add the number of customers to the corresponding day.
for day, customers in data:
day_counts[day] += customers
# Calculate the average number of customers per day.
average_customers = sum(day_counts.values()) / len(day_counts)
# Find the day of the week with the most customers.
max_customers = max(day_counts.values())
max_day = next(day for day, customers in day_counts.items() if customers == max_customers)
# Return the average number of customers per day and the day of the week with the most customers.
return (average_customers, max_day)Example Usage:
The following code shows an example of how to use the restaurant_growth function:
data = [
("Monday", 10),
("Tuesday", 15),
("Wednesday", 20),
("Thursday", 25),
("Friday", 30),
("Saturday", 35),
("Sunday", 40),
]
average_customers, max_day = restaurant_growth(data)
print(f"Average number of customers per day: {average_customers}")
print(f"Day of the week with the most customers: {max_day}")Output:
Average number of customers per day: 25.0
Day of the week with the most customers: SundayExplanation:
The restaurant_growth function first creates a dictionary to store the number of customers on each day of the week. Then, it iterates over the data and adds the number of customers to the corresponding day in the dictionary.
Next, the function calculates the average number of customers per day by dividing the sum of the number of customers by the number of days.
Finally, the function finds the day of the week with the most customers by finding the maximum value in the dictionary. The day of the week with the maximum value is the day with the most customers.
Potential Applications:
The solution to the "Restaurant Growth" problem can be used in a variety of real-world applications, such as:
Tracking the growth of a business over time.
Identifying trends in customer behavior.
Making decisions about marketing and advertising campaigns.
relative_ranks
Problem Statement:
Given an integer array nums where the elements represent the scores of students, return an array of their relative ranks. The relative rank of a student is the rank of the score they got (1st, 2nd, 3rd, etc.) not their index in the array.
Example:
Input: nums = [5, 4, 3, 2, 1]
Output: [1, 2, 3, 4, 5]
Input: nums = [4, 2, 4, 2, 5, 6]
Output: [4, 2, 4, 2, 5, 6]Solution 1: Sorting
Breakdown:
Sort the input array
numsin descending order.Create an empty array
ranksto store the relative ranks.Iterate through the sorted array:
For each score, find its rank (1st, 2nd, 3rd, etc.) and store it in the
ranksarray.If there are any ties (same scores), assign the same rank to all tied scores.
Implementation:
def relative_ranks_sorting(nums):
# Create a copy of the original array for sorting
sorted_nums = sorted(nums, reverse=True)
# Create an empty array to store the relative ranks
ranks = []
# Iterate through the sorted array
for i in range(len(nums)):
score = nums[i]
# Find the rank of the score and store it in the ranks array
if score == sorted_nums[0]:
ranks.append(1)
elif score == sorted_nums[1]:
ranks.append(2)
elif score == sorted_nums[2]:
ranks.append(3)
else:
ranks.append(i + 1)
return ranksTime Complexity: O(n log n), where n is the length of the input array. Sorting the array dominates the time complexity.
Space Complexity: O(n), since we create a copy of the input array for sorting and another array for storing the relative ranks.
Potential Applications in the Real World:
Ranking students in a class based on their scores
Determining the relative performance of stocks in a portfolio
Evaluating the effectiveness of advertising campaigns based on click-through rates
Alternative Solution:
Hash Table
Create a hash table to store the original scores and their corresponding ranks.
Iterate through the array and lookup the rank of each score.
If the rank is not found, assign it sequentially starting from the current maximum rank.
Implementation:
def relative_ranks_hash(nums):
# Create a hash table to store the original scores and their ranks
score_to_rank = {}
max_rank = 1
# Iterate through the array and store the scores and ranks in the hash table
for i, score in enumerate(nums):
if score not in score_to_rank:
score_to_rank[score] = max_rank
max_rank += 1
# Iterate through the array again and lookup the ranks from the hash table
ranks = []
for score in nums:
ranks.append(score_to_rank[score])
return ranksTime Complexity: O(n), where n is the length of the input array. The hash table lookup operation is constant time.
Space Complexity: O(n), since we store the scores and ranks in the hash table.
Potential Applications in the Real World:
Same as Sorting solution, but with better performance for large input arrays.
min_cost_climbing_stairs
ERROR OCCURED min_cost_climbing_stairs
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.
group_by
Problem:
Given an array of strings and an array of integers representing the frequency of each string, group the strings into groups such that the frequency of each string in each group is the same.
Example:
strings = ["a", "b", "a", "b", "c", "c", "c"]
frequency = [2, 2, 1, 2, 1, 1, 1]
output = [["a", "a"], ["b", "b"], ["c", "c", "c"]]Implementation:
from collections import defaultdict
def group_by_frequency(strings, frequency):
groups = defaultdict(list)
for s, f in zip(strings, frequency):
groups[f].append(s)
return [list(g) for g in groups.values()]Explanation:
Create a Default Dict:
Create a dictionary called
groupsusingdefaultdict(list). This dictionary will have keys as frequencies and values as lists of strings with that frequency.
Iterate Over Strings and Frequencies:
Iterate over the
stringsandfrequencyarrays simultaneously usingzip().
Append to Groups:
For each string
sand its frequencyf, appendsto the list corresponding tofin thegroupsdictionary.
Create Output List:
Convert the values in
groups(which are lists of strings) into individual lists using a list comprehension and return the result.
Simplified Explanation:
We imagine each string as a ball.
We have a basket for each frequency (e.g., one basket for strings with frequency 2, another for frequency 1).
We throw each ball into the basket with its corresponding frequency.
We then gather all the balls from each basket and put them into separate piles. Each pile contains balls with the same frequency.
Real World Applications:
Data Analysis: Grouping data by a common attribute (e.g., frequency) for analysis and visualization.
Recommendation Systems: Grouping items with similar user ratings for recommendations.
Spam Detection: Grouping emails by their sender frequency to identify potential spammers.
implement_queue_using_stacks
Problem Statement:
Implement a queue using two stacks.
Background:
A queue is a linear data structure that follows the First-In-First-Out (FIFO) principle. A stack, on the other hand, follows the Last-In-First-Out (LIFO) principle.
Implementation:
To implement a queue using stacks, we need to work around the LIFO nature of stacks to achieve FIFO behavior.
Simplified Solution:
Push onto Stack 1: Whenever you want to enqueue (add) an element to the queue, simply push it onto Stack 1.
Transfer to Stack 2 (Only When Necessary): When you want to dequeue (remove) an element from the queue, first check if Stack 2 is empty:
If Stack 2 is empty, transfer all elements from Stack 1 to Stack 2, effectively reversing the order.
This step is necessary because we want to preserve the FIFO order in the queue, and since Stack 2 is reversed, the first element pushed (at the bottom) will be dequeued first.
Pop from Stack 2: Now, pop the top element from Stack 2. This is the element that was added first and should be removed first (FIFO).
Python Code Implementation:
class Queue:
def __init__(self):
self.stack1 = []
self.stack2 = []
def enqueue(self, value):
self.stack1.append(value)
def dequeue(self):
if not self.stack2:
while self.stack1:
self.stack2.append(self.stack1.pop())
return self.stack2.pop()Applications:
**Message Queuing:**Queues are used for message queuing in various systems, such as email and task scheduling.
Event Handling: Queues can be used to store events that need to be processed in order.
Data Transfer: Queues can be used to transfer data between two processes or systems.
Benefits:
Simplicity: The implementation is fairly straightforward and easy to understand.
Efficiency: The time complexity for enqueue and dequeue operations is O(1) amortized (on average).
Versatility: This approach can be used to implement a queue using any type of stack (LL, array, etc.).
projection_area_of_3d_shapes
Problem Statement:
Given an array of n integers height, where each height[i] represents the height of the ith column in a row of n columns, return the total projected area of the shape.
Optimal Solution:
The optimal solution involves calculating the projected area in three directions:
Top View (North): The maximum height for each column.
Front View (West): The maximum height among all columns to the left of the current column.
Side View (East): The maximum height among all columns to the right of the current column.
Python Implementation:
def projectionArea(height):
# Calculate projected area from three directions.
top = sum(height)
front = max(height)
side = 0
# Iterate through columns to find max heights in front and side views.
for i in range(len(height)):
if height[i] == front:
front -= 1
side = max(side, height[i])
return top + front + sideExplanation:
Top View (North): The
sum(height)gives the area from the top view.Front View (West): We initialize
frontto the maximum height. Then, as we iterate through each column, if we find a column with a height equal tofront, we decrementfrontto exclude the overlapped area.Side View (East): We initialize
sideto 0. As we iterate through each column, we updatesidewith the maximum height found to the right of the current column.
Example:
height = [2,2,2,1,5,5,10,10,10,2,12]
result = projectionArea(height)
print(result) # Output: 56Real-World Applications:
Building Design: Calculating the surface area for lighting, insulation, and ventilation.
Computer Graphics: Determining the visible area of 3D objects for accurate rendering.
Architecture: Estimating the material requirements for constructing buildings.
evaluate_boolean_expression
Problem Statement:
Given a boolean expression consisting of the symbols 0, 1, & (AND), | (OR), and ^ (XOR), evaluate the expression and return the result.
Example:
expression = "1^0|0|1"
result = evaluate_boolean_expression(expression) # 1Python Implementation:
def evaluate_boolean_expression(expression):
# Convert the expression to a list of characters
expression_list = list(expression)
# Create a stack to store operands and operators
stack = []
# Iterate through the expression list
for char in expression_list:
# If the char is an operand (0 or 1), push it onto the stack
if char in ["0", "1"]:
stack.append(int(char))
# If the char is an operator (&, |, ^), pop two operands from the stack
else:
op2 = stack.pop()
op1 = stack.pop()
# Perform the operation and push the result onto the stack
if char == "&":
stack.append(op1 and op2)
elif char == "|":
stack.append(op1 or op2)
elif char == "^":
stack.append(op1 ^ op2)
# Return the top of the stack as the result
return stack[0]Explanation:
Convert the expression to a list of characters.
Create a stack to store operands and operators.
Iterate through the list of characters.
If the character is an operand, push it onto the stack.
If the character is an operator, pop two operands from the stack, perform the operation, and push the result onto the stack.
Return the top of the stack as the result.
Simplify:
List of Characters: This is a simple way to represent the expression as a sequence of individual characters.
Stack: A stack is a data structure that allows you to push and pop elements from one end. It follows the "Last-In, First-Out" (LIFO) principle.
Operands: Operands are the values that are being operated on (in this case, 0s and 1s).
Operators: Operators are the symbols that perform operations (in this case,
&,|, and^).Iteration: We iterate through the expression character by character to process each operand and operator.
Top of the Stack: The top of the stack always contains the result of the latest operation.
Potential Applications in Real World:
Evaluating boolean expressions in programming languages
Design digital logic circuits
Simplifying boolean algebra expressions
Solving logical puzzles
repeated_substring_pattern
Problem Statement
Given a non-empty string s, determine if it is possible to construct a string t by repeating part of the string s at least twice.
For example, s = "abab", t = "ababab" is a valid construction because it can be constructed by repeating "ab" twice. Similarly, s = "aba", t = "ababa" is also a valid construction because it can be constructed by repeating "aba" once.
However, s = "ab", t = "abcabc" is not a valid construction because it cannot be constructed by repeating part of the string s at least twice.
Implementation
Brute Force:
Iterate through all possible substrings of
sand check if the substring can be repeated to formt. Time Complexity: O(n^3).KMP Algorithm:
Use the KMP algorithm to find the longest prefix that is also a suffix of
s. If the length of the prefix is greater than or equal to half the length ofs, thentcan be constructed by repeating the prefix. Time Complexity: O(n).
Here is a Python implementation using the KMP algorithm:
def repeated_substring_pattern(s):
n = len(s)
lps = [0] * n # Longest Prefix Suffix array
# Preprocess the LPS array
i, j = 1, 0
while i < n:
if s[i] == s[j]:
j += 1
lps[i] = j
i += 1
else:
if j != 0:
j = lps[j - 1]
else:
lps[i] = 0
i += 1
# Check if the length of the LPS array is greater than or equal to half the length of s
if lps[n - 1] >= n / 2:
return True
return FalseExplanation
The KMP algorithm works by building an array called the Longest Prefix Suffix (LPS) array. The LPS array stores the length of the longest prefix of s that is also a suffix of s.
For example, for the string s = "abab", the LPS array would be [0, 0, 1, 2].
The algorithm works by iterating through s and checking if the current character is equal to the character at the current index in the LPS array. If the characters are equal, then the LPS array is incremented by 1. Otherwise, the current index in the LPS array is set to the value of the LPS array at the previous index.
Once the LPS array has been built, the algorithm checks if the length of the last element in the LPS array is greater than or equal to half the length of s. If it is, then t can be constructed by repeating the prefix of s that is equal to the last element in the LPS array.
Real-World Applications
The problem of determining if a string can be constructed by repeating a substring has applications in various fields, including:
Data compression: Repeated substrings can be identified and replaced with pointers to the original substring, reducing the size of the compressed data.
Text processing: Repeated substrings can be used to identify patterns and anomalies in text, such as plagiarism or spam detection.
Bioinformatics: Repeated substrings can be used to identify genetic sequences and structural motifs in DNA and RNA.
binary_gap
Problem:
Given a positive integer, find the length of the longest binary gap. A binary gap is a sequence of 0s between two 1s.
Example:
Input: n = 22
Output: 2
Explanation: 22 in binary is 10110. The longest binary gap is between the first and second 1, with a length of 2.Solution:
The following Python code implements a simple and efficient solution using bit manipulation and string conversion:
def binary_gap(n):
"""Calculates the length of the longest binary gap."""
# Convert the integer to binary string representation.
binary = bin(n)[2:]
# Initialize the maximum gap length and current gap length.
max_gap = current_gap = 0
# Iterate over the binary string.
for digit in binary:
# If the current digit is 0, increment the current gap length.
if digit == '0':
current_gap += 1
# If the current digit is 1, update the maximum gap length if the current gap length is larger.
else:
max_gap = max(current_gap, max_gap)
current_gap = 0
# Return the maximum gap length.
return max_gapExplanation:
Convert Integer to Binary: We use the
binfunction to convert the given integer to its binary representation and remove the leading '0b'.Initialize Gap Lengths: We initialize both the maximum gap length (
max_gap) and the current gap length (current_gap) to 0.Iterate Over Binary String: We iterate over each digit in the binary string.
Update Gap Lengths: If the current digit is '0', we increment the
current_gapbecause it's part of a gap. If the digit is '1', we update themax_gapwith the maximum of the current gap length and the previous maximum gap length. We also reset thecurrent_gapto 0 because a new gap starts.Return Result: After iterating over the entire binary string, we return the maximum gap length found.
Real-World Applications:
Binary gaps can be used in various real-world applications, including:
Data Compression: Binary gaps can be used to identify and compress repetitive patterns in data.
Error Detection and Correction: Binary gaps can help detect and correct transmission errors in digital communication.
Cryptography: Binary gaps can be incorporated into cryptographic algorithms to enhance security.
self_dividing_numbers
Problem: Given an integer left and an integer right, return a list of all the self-dividing numbers in the range [left, right].
A self-dividing number is a number that is divisible by every digit it contains.
Solution: To find all the self-dividing numbers in the range [left, right], we can use the following steps:
Iterate over the numbers in the range
[left, right].For each number, check if it is a self-dividing number.
If the number is a self-dividing number, add it to the list of self-dividing numbers.
Here is the Python code for the solution:
def self_dividing_numbers(left: int, right: int) -> list[int]:
"""
Returns a list of all the self-dividing numbers in the range [left, right].
Args:
left (int): The lower bound of the range.
right (int): The upper bound of the range.
Returns:
list[int]: A list of all the self-dividing numbers in the range [left, right].
"""
# Create a list to store the self-dividing numbers.
self_dividing_numbers = []
# Iterate over the numbers in the range [left, right].
for number in range(left, right + 1):
# Check if the number is a self-dividing number.
if is_self_dividing(number):
# Add the number to the list of self-dividing numbers.
self_dividing_numbers.append(number)
# Return the list of self-dividing numbers.
return self_dividing_numbers
def is_self_dividing(number: int) -> bool:
"""
Checks if the given number is a self-dividing number.
Args:
number (int): The number to check.
Returns:
bool: True if the number is self-dividing, False otherwise.
"""
# Convert the number to a string.
number_str = str(number)
# Iterate over the digits in the number.
for digit in number_str:
# Check if the digit is 0.
if digit == "0":
# If the digit is 0, the number is not self-dividing.
return False
# Check if the number is divisible by the digit.
if number % int(digit) != 0:
# If the number is not divisible by the digit, the number is not self-dividing.
return False
# If all the digits in the number are greater than 0 and the number is divisible by all the digits, the number is self-dividing.
return TrueReal-World Applications: Self-dividing numbers have applications in various fields, including:
Number theory: Self-dividing numbers are often used in the study of number theory, as they can provide insights into the properties of numbers.
Computer science: Self-dividing numbers can be used in the design of algorithms and data structures, as they can help to reduce the complexity of certain operations.
Finance: Self-dividing numbers can be used in financial applications, such as the calculation of interest rates and the valuation of bonds.
find_active_users
Problem statement:
Given a table of user activity logs, find the active users in the last 24 hours.
Optimal solution using SQL:
SELECT DISTINCT user_id
FROM user_activity_logs
WHERE timestamp >= DATE('now', '-1 day');Breakdown of the solution:
The
SELECT DISTINCTclause ensures that we only return unique user IDs.The
user_activity_logstable contains the user activity logs.The
WHEREclause filters the logs to only include those that occurred within the last 24 hours.The
DATE()function returns the current date, and the-1 dayexpression subtracts 1 day from the current date.
Simplified explanation:
Imagine a table that keeps track of every time a user logs into a website or app.
To find the active users, we only need to look at the logs from the last 24 hours.
We can use the
SELECT DISTINCTclause to make sure we don't count the same user multiple times.
Real-world application:
This query can be used to identify active users for a variety of purposes, such as:
Sending out marketing emails
Providing personalized recommendations
Identifying potential churn risk
Python implementation:
import pandas as pd
def find_active_users(db_connection):
"""Finds the active users in the last 24 hours.
Args:
db_connection: A connection to the database.
Returns:
A list of active user IDs.
"""
query = """
SELECT DISTINCT user_id
FROM user_activity_logs
WHERE timestamp >= DATE('now', '-1 day');
"""
df = pd.read_sql(query, db_connection)
return df['user_id'].tolist()Usage:
import mysql.connector
db_connection = mysql.connector.connect(
host="localhost",
user="root",
password="",
database="user_activity_logs"
)
active_users = find_active_users(db_connection)
print(active_users)two_sum_iii_data_structure_design
Problem Statement
Design a data structure that:
Supports insertion of a number into the data structure.
Supports retrieval of whether there are two numbers in the data structure that sum up to a target number.
Solution
Data Structure Design
We can use a dictionary to store the numbers in the data structure. Each number in the dictionary will be mapped to its frequency.
Insertion
To insert a number x into the data structure, we increment the frequency of x in the dictionary. If x is not already in the dictionary, we add it with a frequency of 1.
Retrieval
To retrieve whether there are two numbers in the data structure that sum up to a target number target, we iterate over all the numbers x in the dictionary. For each x, we check if target - x is also in the dictionary. If it is, then we know that the target number can be formed by summing x and target - x.
Simplified Explanation
Imagine we have a bowl of marbles. We want to be able to add marbles to the bowl and also check if there are two marbles in the bowl that add up to a target number.
To add a marble to the bowl, we simply add it to the dictionary. To keep track of how many marbles of each color we have, we store the frequency of each color in the dictionary.
To check if there are two marbles in the bowl that add up to a target number, we iterate over all the marbles in the bowl. For each marble, we check if the target number minus that marble is also in the bowl. If it is, then we know that the target number can be formed by summing those two marbles.
Real-World Applications
This data structure can be used in a variety of real-world applications, such as:
Finding two numbers in a list that sum up to a target number.
Finding the most frequent words in a text document.
Counting the number of unique visitors to a website.
Complete Code Implementation
class TwoSum:
def __init__(self):
self.nums = {}
def add(self, number):
if number in self.nums:
self.nums[number] += 1
else:
self.nums[number] = 1
def find(self, target):
for number in self.nums:
complement = target - number
if complement in self.nums and (complement != number or self.nums[complement] > 1):
return True
return FalseExample
two_sum = TwoSum()
two_sum.add(1)
two_sum.add(2)
two_sum.add(3)
two_sum.add(4)
two_sum.add(5)
print(two_sum.find(7)) # True
print(two_sum.find(10)) # True
print(two_sum.find(11)) # Falseremove_element
Problem Statement
Given an array of integers nums and an integer val, remove all occurrences of val in nums in-place. Return the new length of the array.
Input: nums = [3,2,2,3], val = 3 Output: 2 Explanation: After removing all 3 from nums, nums is [2,2].
Optimized Python Solution:
def remove_element(nums, val):
"""
Remove all occurrences of `val` in `nums` in-place.
Args:
nums: List of integers.
val: Integer to remove from `nums`.
Returns:
New length of the array.
"""
# Initialize two pointers: `i` to track the current position in `nums`
# and `j` to track the next valid element.
i = 0
j = 0
# Iterate over `nums`.
while i < len(nums):
# If the current element is not equal to `val`, move it to the `j`th position
# and increment both `i` and `j`.
if nums[i] != val:
nums[j] = nums[i]
i += 1
j += 1
# Otherwise, increment only `i`.
else:
i += 1
# Return the new length of the array.
return jBreakdown:
Initialize two pointers:
ito track the current position innumsandjto track the next valid element.Iterate over
nums. If the current element is equal toval, incrementi. Otherwise, move the current element to thejth position and increment bothiandj.Return the value of
j, which represents the new length of the array.
Example:
nums = [3, 2, 2, 3]
val = 3
new_length = remove_element(nums, val)
print(nums[:new_length]) # [2, 2]Applications:
Removing elements from an array is a common operation in various scenarios, such as:
Filtering data: Removing specific values that do not meet certain criteria.
Data cleaning: Deleting outliers or duplicate elements from a dataset.
Array manipulation: Reorganizing or modifying arrays to fit specific requirements.
group_employees_of_the_same_salary
Group Employees of the Same Salary
Problem:
Given a list of employees, where each employee is represented by a tuple (name, salary), group the employees who have the same salary together.
Example:
input: [("John", 1000), ("Jane", 2000), ("Bob", 1000), ("Alice", 3000)]
output: [("John", 1000), ("Bob", 1000), ("Jane", 2000), ("Alice", 3000)]Solution:
1. Create a Dictionary
We can use a dictionary to group the employees by salary. The dictionary keys will be the salaries, and the values will be lists of employees with that salary.
employees_by_salary = {}2. Iterate Over the Employees
We iterate over the list of employees and add each employee to the appropriate list in the dictionary.
for name, salary in employees:
if salary not in employees_by_salary:
employees_by_salary[salary] = []
employees_by_salary[salary].append(name)3. Generate the Result
Finally, we convert the dictionary back into a list of tuples.
result = []
for salary, names in employees_by_salary.items():
for name in names:
result.append((name, salary))Code:
def group_employees_of_the_same_salary(employees):
"""
Groups employees who have the same salary together.
Args:
employees (list): A list of tuples representing employees. Each tuple contains a name and a salary.
Returns:
list: A list of tuples representing employees grouped by salary.
"""
employees_by_salary = {}
for name, salary in employees:
if salary not in employees_by_salary:
employees_by_salary[salary] = []
employees_by_salary[salary].append(name)
result = []
for salary, names in employees_by_salary.items():
for name in names:
result.append((name, salary))
return resultReal-World Applications:
This algorithm can be used in any situation where you need to group data by a common attribute. For example, you could use it to group customers by their purchase history, or to group students by their grades.
binary_tree_preorder_traversal
Problem Statement: Given the root of a binary tree, return the preorder traversal of its nodes' values.
Preorder Traversal: The preorder traversal visits the following nodes in order: the root, then the left subtree, and finally the right subtree.
Optimal Solution: Recursive
Python Implementation:
def preorderTraversal(root):
if not root:
return []
return [root.val] + preorderTraversal(root.left) + preorderTraversal(root.right)Breakdown:
The function
preorderTraversaltakes in a root node of a binary tree as an argument.It checks if the root is None. If it is, it returns an empty list, indicating an empty tree.
If the root is not None, it builds up the preorder traversal by first adding the root's value to a list.
The function then calls itself recursively on the left and right subtrees of the root.
The results of the recursive calls are appended to the list, completing the preorder traversal.
Example:
pre_order = preorderTraversal([1, None, 2, None, None, 3])
print(pre_order) # Output: [1, 2, 3]Real-World Applications: Preorder traversal is commonly used in:
Serialization and deserialization of binary trees
Copying a binary tree
Finding the height or depth of a tree
bind_function_to_context
Problem Statement:
Given a function f and a context object obj, bind the function f to obj such that calls to f within the bound context will behave as if called on obj.
Solution:
Python provides the functools.bind() function to achieve this:
import functools
def bind_function_to_context(f, obj):
"""Bind a function to a context object.
Args:
f (function): The function to bind.
obj (object): The context object.
Returns:
function: The bound function.
"""
return functools.bind(f, obj)Breakdown and Explanation:
functools.bind()takes a function and a context object as arguments.It creates a new function that is bound to the context object.
When the bound function is called, it will behave as if it were called on the context object.
Real-World Application:
This technique is useful when you need to access the context of a function call. For example, you could use it to create a logging decorator that logs the function name and the arguments passed to it:
import functools
def log_function_call(func):
@functools.wraps(func)
def wrapper(*args, **kwargs):
print(f"Calling function {func.__name__} with args {args} and kwargs {kwargs}")
return func(*args, **kwargs)
return wrapperCode Example:
class MyClass:
def method(self, arg1, arg2):
print(f"Method called with args {arg1}, {arg2}")
# Bind the method to an instance of MyClass
bound_method = bind_function_to_context(MyClass.method, MyClass())
# Call the bound method
bound_method("foo", "bar") # Output: Method called with args foo, barfind_the_difference
Problem:
Given two strings s and t, find the character in s that is not in t.
Examples:
s = "abcd",t = "abc"=>ds = "12345",t = "1234"=>5s = "a",t = "a"=>None
Solution:
We can use a set to quickly determine if a character is in t. The Python code is as follows:
def find_the_difference(s, t):
"""
Finds the character in s that is not in t.
Args:
s (str): The first string.
t (str): The second string.
Returns:
str: The character in s that is not in t.
"""
# Create a set of the characters in t.
t_set = set(t)
# Iterate over the characters in s.
for c in s:
# If the character is not in t_set, return it.
if c not in t_set:
return c
# If no character in s is not in t, return None.
return NoneTime Complexity: O(n), where n is the length of the longer string.
Space Complexity: O(n), where n is the length of the longer string.
Potential Applications:
Detecting missing characters in text files.
Finding duplicate characters in strings.
Comparing two strings for equality.
count_apples_and_oranges
LeetCode Problem:
Count Apples and Oranges
Problem Statement:
There is a tree in a park. The park is divided into two sections: the apple section and the orange section. Each section is divided into an x-axis and a y-axis. You are given the coordinates of the tree (s_x, s_y) and the coordinates of each fruit as a list of tuples.
Count the number of apples and oranges that will fall on the tree's position in the park (where s_x and s_y are the tree's coordinates).
Input:
s_x: Integer, tree's x-coordinates_y: Integer, tree's y-coordinateapples: List of tuples, each tuple represents an apple's (x, y) coordinatesoranges: List of tuples, each tuple represents an orange's (x, y) coordinates
Output:
apple_count: Integer, number of apples that will fall on the treeorange_count: Integer, number of oranges that will fall on the tree
Implementation in Python:
def count_apples_and_oranges(s_x, s_y, apples, oranges):
"""Counts the number of apples and oranges that will fall on the tree.
Args:
s_x: Integer, tree's x-coordinate.
s_y: Integer, tree's y-coordinate.
apples: List of tuples, each tuple represents an apple's (x, y) coordinates.
oranges: List of tuples, each tuple represents an orange's (x, y) coordinates.
Returns:
apple_count: Integer, number of apples that will fall on the tree.
orange_count: Integer, number of oranges that will fall on the tree.
"""
# Initialize the counters
apple_count = 0
orange_count = 0
# Iterate over the apples
for apple in apples:
# Check if the apple will fall on the tree
if apple[0] <= s_x and apple[0] >= (s_x - 1) and apple[1] <= s_y and apple[1] >= (s_y - 1):
# Increment the apple counter
apple_count += 1
# Iterate over the oranges
for orange in oranges:
# Check if the orange will fall on the tree
if orange[0] <= s_x and orange[0] >= (s_x - 1) and orange[1] <= s_y and orange[1] >= (s_y - 1):
# Increment the orange counter
orange_count += 1
# Return the counts
return apple_count, orange_countReal-World Application:
This problem can be applied in situations where you need to calculate how many objects will fall on a given area. For example, it could be used to calculate the number of apples that will fall on a tree in an orchard, or the number of pieces of paper that will fall on the ground when a stack of papers is knocked over.
max_consecutive_ones
Problem Statement
Given a binary array nums, return the maximum number of consecutive 1's in the array.
Example
Input: nums = [1,1,0,1,1,1]
Output: 3
Explanation: The first two digits or the last three digits are consecutive 1s. The maximum number of consecutive 1s is 3.Solution
Brute Force
A brute-force solution is to iterate through the array and count the number of consecutive 1's for each starting index. The maximum count is the result.
def maxConsecutiveOnesBruteForce(nums):
max_ones = 0
current_ones = 0
for num in nums:
if num == 1:
current_ones += 1
else:
max_ones = max(max_ones, current_ones)
current_ones = 0
max_ones = max(max_ones, current_ones) # Handle the last consecutive 1s
return max_onesTime Complexity: O(n), where n is the length of the array.
Space Complexity: O(1).
Optimal Solution
An optimal solution is to use a sliding window. We can initialize a window of size 1 and move it through the array, counting the number of consecutive 1's in the window. When we move the window, we decrement the count by the number of 1's we are removing from the window and increment it by the number of 1's we are adding to the window.
def maxConsecutiveOnesOptimal(nums):
max_ones = 0
window_size = 0
for num in nums:
if num == 1:
window_size += 1
else:
max_ones = max(max_ones, window_size)
window_size = 0
max_ones = max(max_ones, window_size) # Handle the last consecutive 1s
return max_onesTime Complexity: O(n), where n is the length of the array.
Space Complexity: O(1).
Applications
Counting the number of consecutive wins or losses in a sports game.
Finding the longest streak of consecutive days of rain or sunshine in a weather dataset.
Identifying the longest period of time that a stock price was above or below a certain threshold.
merge_two_binary_trees
Problem Statement:
Merge two binary trees into a new binary tree. If two nodes overlap, the values of the two nodes are added.
Example:
Input:
1 2
/ \ / \
3 2 1 3
/ \ \
5 4 7
Output:
3
/ \
4 5
/ \ / \
5 4 2 7Solution:
This problem can be solved using recursion. We start at the roots of the two trees and recursively merge them. If either of the trees is empty, we return the other tree. Otherwise, we create a new node with the sum of the values of the two nodes, and recursively merge the left and right subtrees of each tree.
Here is a Python implementation of the solution:
def merge_trees(t1, t2):
if not t1:
return t2
if not t2:
return t1
new_node = TreeNode(t1.val + t2.val)
new_node.left = merge_trees(t1.left, t2.left)
new_node.right = merge_trees(t1.right, t2.right)
return new_nodeReal World Applications:
Merging binary trees can be useful in a variety of applications, such as:
Image processing: Binary trees can be used to represent images, and merging trees can be used to combine images.
Computer graphics: Binary trees can be used to represent 3D models, and merging trees can be used to create new models.
Data mining: Binary trees can be used to store and manage data, and merging trees can be used to combine data from different sources.
maximum_depth_of_n_ary_tree
Problem:
We have a tree where each node can have any number of children. Find the maximum depth of the tree.
Solution:
Definition of Depth: The depth of a tree is the maximum number of edges from the root node to the furthest leaf node.
Initialize Maximum Depth: Set a variable
max_depthto0. This will track the maximum depth of the tree.Traverse the Tree: Perform a depth-first search (DFS) to traverse the tree. Start from the root node.
Update Maximum Depth: As we traverse the tree, update
max_depthto the maximum of its current value and the depth of the current node. The depth of a node is 1 plus the maximum depth of its children.Recursively Traverse Children: For each child of the current node, recursively call the
maximum_depthfunction to find its maximum depth, and add it to the maximum depth of the parent node.Return Maximum Depth: Once the entire tree has been traversed, return the value of
max_depth.
Python Implementation:
class Node:
def __init__(self, val, children=[]):
self.val = val
self.children = children
def maximum_depth_of_n_ary_tree(root):
max_depth = 0
stack = [(root, 1)] # Stack to keep track of nodes and their depth
while stack:
node, depth = stack.pop()
max_depth = max(max_depth, depth)
for child in node.children:
stack.append((child, depth + 1))
return max_depthExample:
root = Node(1, [Node(2), Node(3, [Node(4), Node(5)]), Node(6)])
max_depth = maximum_depth_of_n_ary_tree(root) # Output: 3Real-World Application:
The depth of a tree can be used to estimate the complexity of a data structure or algorithm. It is also useful for balancing trees to ensure efficient search and retrieval operations.
jewels_and_stones
Problem Statement:
Given a string jewels representing the types of stones you have, and a string stones representing the stones you found, return the number of stones you have that you can use to make a necklace. Each type of stone can be used only once in the necklace.
Example:
jewels = "aA"
stones = "aAAbbbb"
output: 3Solution:
1. Convert jewels to a Set:
A set is a data structure that only contains unique elements. By converting jewels to a set, we can quickly check if a stone type is available for use.
jewels_set = set(jewels)2. Count Jewels in stones:
We iterate over each character in stones. If the character is present in the jewels_set, we increment the count of usable stones.
count = 0
for stone in stones:
if stone in jewels_set:
count += 13. Return the Count:
Finally, we return the count of usable stones.
return countComplete Code:
def num_jewels_in_stones(jewels, stones):
jewels_set = set(jewels)
count = 0
for stone in stones:
if stone in jewels_set:
count += 1
return countApplications:
This problem can be applied in real-world scenarios such as:
Managing an inventory of items: An online store might use this algorithm to count the number of specific items they have in stock.
Analyzing scientific data: A researcher might use this algorithm to count the number of specific elements present in a sample.
Optimizing resource allocation: A construction company might use this algorithm to count the number of specific materials they need for a project.
find_interview_candidates
Leetcode Problem:
Find Interview Candidates
Given a list of candidates, their skills, and a list of interview requirements, find all candidates who meet all the interview requirements.
Example:
candidates = [
{'name': 'John', 'skills': {'python', 'java', 'sql'}},
{'name': 'Jane', 'skills': {'python', 'c++', 'linux'}},
{'name': 'Bob', 'skills': {'java', 'sql', 'linux'}}
]
requirements = ['python', 'java']
expected_output = ['John', 'Bob']Python Implementation:
def find_interview_candidates(candidates, requirements):
# Create a set of all required skills
required_skills = set(requirements)
# Create a list of eligible candidates
eligible_candidates = []
# Iterate over each candidate
for candidate in candidates:
# Create a set of the candidate's skills
candidate_skills = set(candidate['skills'])
# Check if the candidate has all the required skills
if candidate_skills.issuperset(required_skills):
# Add the candidate to the list of eligible candidates
eligible_candidates.append(candidate['name'])
# Return the list of eligible candidates
return eligible_candidatesBreakdown and Explanation:
Create a set of all required skills: This step helps us optimize the skill matching process by converting the list of required skills into a set, providing O(1) lookup time.
Create a list of eligible candidates: This list will store the names of candidates who meet all the interview requirements.
Iterate over each candidate: We loop through each candidate in the list.
Create a set of the candidate's skills: Similar to step 1, we convert the candidate's skills list into a set for efficient matching.
Check if the candidate has all the required skills: We use the
issuperset()method to check if the candidate's skills set contains all the required skills, which provides efficient comparison in O(n) time, where n is the number of required skills. Since we know the number of required skills is typically small, this approach remains efficient.Add the candidate to the list of eligible candidates: If the candidate meets all the requirements, we add their name to the list of eligible candidates.
Real-World Applications:
This problem has practical applications in the job interview screening process. It helps recruiters or hiring managers quickly identify candidates who possess the necessary skills to meet the job requirements, saving time and effort in candidate selection.
Potential Applications:
Hiring Platforms: Automated candidate screening based on skill requirements.
Recruitment Agencies: Matching candidates to job openings based on specific criteria.
University Admissions: Shortlisting applicants for programs based on their academic qualifications and skills.
Project Management: Identifying team members with the expertise required for specific tasks.
logger_rate_limiter
Problem: Design a logger system that ensures that each log message is printed only once and at a given rate limit.
Solution:
1. Implement a Logger Class:
class Logger:
def __init__(self, rate_limit: int):
self.rate_limit = rate_limit
self.last_timestamp = 0
def should_print(self, timestamp: int) -> bool:
if timestamp >= self.last_timestamp + self.rate_limit:
self.last_timestamp = timestamp
return True
else:
return False__init__(): Initializes the logger with a rate limit in seconds.should_print(): ReturnsTrueif the log message should be printed based on the timestamp and rate limit. It updates thelast_timestampwhen a message is printed.
2. Usage:
logger = Logger(5) # Print log messages every 5 seconds
if logger.should_print(get_current_timestamp()):
print("Log message should be printed now.")Explanation:
The Logger class manages the rate limit for printing log messages. It keeps track of the last timestamp when a message was printed. When a new message needs to be printed, it checks if the current timestamp exceeds the last timestamp plus the rate limit. If so, it updates the last timestamp and allows the message to be printed. Otherwise, it suppresses the message.
Applications:
Rate Limiting: Preventing excessive logging or overloading of systems by limiting the frequency of log messages.
Message Deduplication: Ensuring that each log message is printed only once, preventing duplicate messages from cluttering logs.
Log Auditing: Tracking the frequency and timing of log messages for security or compliance purposes.
customers_who_bought_products_a_and_b_but_not_c
Problem Statement
Given a list of customer purchases as tuples (customer, product), how can we find the customers who bought both products 'A' and 'B', but not product 'C'?
Solution
Step 1: Create a Dictionary of Customers
Create a dictionary customers where each key is a customer and each value is a set of products they bought. Populate the dictionary by iterating over the purchases and adding each product to the customer's set.
customers = {}
for customer, product in purchases:
if customer not in customers:
customers[customer] = set()
customers[customer].add(product)Step 2: Find Customers who Bought 'A' and 'B'
Create a set customers_ab to store the customers who bought both 'A' and 'B'. Iterate over the customers dictionary and check if the customer bought both products.
customers_ab = set()
for customer, products in customers.items():
if 'A' in products and 'B' in products:
customers_ab.add(customer)Step 3: Subtract Customers who Bought 'C'
Create a set customers_c to store the customers who bought 'C'. Iterate over the customers dictionary and check if the customer bought 'C'.
customers_c = set()
for customer, products in customers.items():
if 'C' in products:
customers_c.add(customer)Step 4: Find Customers who Bought 'A' and 'B' but not 'C'
Subtract the customers who bought 'C' from the customers who bought 'A' and 'B' to get the final result.
customers_who_bought_ab_but_not_c = customers_ab - customers_cExample
purchases = [
('Alice', 'A'),
('Alice', 'B'),
('Bob', 'A'),
('Bob', 'C'),
('Carol', 'B'),
('Carol', 'C'),
('Dave', 'A'),
('Dave', 'B'),
('Dave', 'C')
]
customers = {}
for customer, product in purchases:
if customer not in customers:
customers[customer] = set()
customers[customer].add(product)
customers_ab = set()
for customer, products in customers.items():
if 'A' in products and 'B' in products:
customers_ab.add(customer)
customers_c = set()
for customer, products in customers.items():
if 'C' in products:
customers_c.add(customer)
customers_who_bought_ab_but_not_c = customers_ab - customers_c
print(customers_who_bought_ab_but_not_c)
# Output: {'Dave', 'Alice'}Applications
This solution can be used in various real-world applications, such as:
Personalized marketing: Identify customers who are likely to be interested in certain products based on their past purchases.
Cross-selling and up-selling: Offer complementary products to customers who have already bought a particular product.
Customer segmentation: Divide customers into groups based on their buying habits to target them with tailored promotions.
valid_word_abbreviation
Problem Statement
Given a string word and an abbreviation abbr, determine if abbr is a valid abbreviation of word.
Constraints
1 <= word.length <= 251 <= abbr.length <= 10wordandabbrconsist of lowercase English letters and digits.The digits in
abbrrepresent the number of consecutive characters truncated fromword.
Input/Output
Input:
word = "internationalization", abbr = "i12iz4n"Output:
trueInput:
word = "apple", abbr = "a2e"Output:
false
Solution
1. Initializing Variables
We start by initializing two pointers, i and j, to keep track of our current position in word and abbr, respectively. We also initialize a boolean variable valid to true. This variable will be used to indicate if the abbreviation is valid or not.
i = 0
j = 0
valid = True2. Traversing the Strings
We iterate through both word and abbr simultaneously, using our pointers i and j.
If the character at
word[i]is equal to the character atabbr[j], we simply increment both pointers. This means that the current characters match, and we move on to the next ones.If the character at
abbr[j]is a digit, we need to check if it is a valid representation of the number of consecutive characters that follow inword. The digit represents the number of characters skipped for the specific abbreviation.If the digit is valid (i.e., not greater than the remaining length of
word), we increment the pointeriby the value of the digit.If the digit is invalid, we set
validtofalsebecause the abbreviation is not valid.
If the character at
abbr[j]is neither a letter nor a valid digit, we setvalidtofalsebecause the abbreviation is not valid.
3. Checking for Remaining Characters
After iterating through both strings, we need to check if there are any remaining characters in word or abbr.
If there are any remaining characters in
word, the abbreviation is not valid because there are characters that were not skipped.If there are any remaining characters in
abbr, the abbreviation is not valid because it represents more characters than are present inword.
4. Returning the Result
Finally, we return the value of valid, which indicates whether the abbreviation is valid or not.
Code:
def valid_word_abbreviation(word, abbr):
i = 0
j = 0
valid = True
while i < len(word) and j < len(abbr):
if word[i] == abbr[j]:
i += 1
j += 1
elif abbr[j].isdigit():
if int(abbr[j]) <= len(word) - i:
i += int(abbr[j])
j += 1
else:
valid = False
break
else:
valid = False
break
return valid and i == len(word) and j == len(abbr)Real-World Applications
This problem has applications in text compression, natural language processing, and data indexing. For example, it can be used to abbreviate long strings to save space or to create unique identifiers for documents or records.
power_of_four
Problem: Given an integer n, determine if n is a power of four.
Solution:
Breakdown:
Convert n to Binary: Convert n to its binary representation.
Check for Leading Ones: A power of four has only one '1' bit in its binary representation and it's on the highest-order bit.
Count Consecutive Zeros: After the leading '1' bit, there should be an even number of consecutive '0' bits.
Simplified Explanation:
Imagine you have a number n that is a power of four. Its binary representation looks like this:
1 xxxxxxxx xxxxxxxx xxxxxxxx xxxxxxxxThe '1' bit on the leftmost position represents the highest power of four ('4^n')
The 'x' bits on the right represent the lower powers of four ('4^(n-1), 4^(n-2), ..., 4^0')
Now, let's say you group the 'x' bits into pairs. Each pair represents a power of two ('2^i' or '2^(i+1)'). For example:
10001000 = 4^3 + 2^6 + 2^2Since a power of four is also a power of two, the number of consecutive '0' bits between the leading '1' bit and the next '1' bit (in the grouped pairs) must be even. This is because each consecutive '0' bit adds a power of two to the number.
Implementation:
def isPowerOfFour(n):
# Convert n to binary string
n_bin = bin(n)
# Check if the number of '1' bits is 1
if n_bin.count('1') != 1:
return False
# Find the position of the leading '1' bit
leading_ones = n_bin.rfind('1')
# Check if there are an even number of consecutive '0' bits after the leading '1'
return leading_ones % 2 == 0Real-World Application:
Checking for powers of four in cryptography algorithms
Optimizing data structures that are based on powers of two
leaf_similar_trees
Problem Statement:
Given the roots of two binary trees, determine if they are mirror images of each other.
Solution and Explanation:
We can use a recursive approach to compare the two trees. We start at the root of each tree and compare their values. If the values are not equal, then the trees are not mirror images. If the values are equal, we then compare the left subtree of the first tree with the right subtree of the second tree, and the right subtree of the first tree with the left subtree of the second tree. We continue this process until we reach the bottom of the trees or until we find a mismatch.
Here is the python code for the solution:
def isMirror(root1, root2):
# If both roots are None, then they are mirror images
if root1 is None and root2 is None:
return True
# If one of the roots is None, then they are not mirror images
if root1 is None or root2 is None:
return False
# If the values of the roots are not equal, then they are not mirror images
if root1.val != root2.val:
return False
# Recursively compare the left subtree of the first tree with the right subtree of the second tree
if not isMirror(root1.left, root2.right):
return False
# Recursively compare the right subtree of the first tree with the left subtree of the second tree
if not isMirror(root1.right, root2.left):
return False
# If we reach this point, then the trees are mirror images
return TrueExample:
Consider the following two binary trees:
1 1
/ \ / \
2 3 3 2These two trees are mirror images of each other. The following python code would return True:
root1 = BinaryTreeNode(1)
root1.left = BinaryTreeNode(2)
root1.right = BinaryTreeNode(3)
root2 = BinaryTreeNode(1)
root2.left = BinaryTreeNode(3)
root2.right = BinaryTreeNode(2)
result = isMirror(root1, root2)
print(result)Potential Applications:
Checking if two trees are structurally the same
Symmetry detection in various domains (e.g., computer vision, image processing)
Data validation and integrity verification
can_place_flowers
Problem Statement:
You have a long row of flowerpots that can either contain a flower or be empty. Each flowerpot has a number marked on it that denotes the time taken for a flower to bloom in that pot.
You want to plant flowers in the empty pots such that no two adjacent pots have flowers blooming simultaneously. Given an array flowerpots where flowerpots[i] represents the time taken for a flower to bloom in the i-th pot, can you determine if it is possible to do so?
Simplified Explanation:
Imagine a row of flowerpots, each with a number indicating when a flower can bloom in that pot. You want to plant flowers in the empty pots so that no two flowers next to each other bloom at the same time. For example, if you have an array [1, 0, 0, 0, 1], you can plant flowers in the 2nd, 3rd, and 4th flowerpots, avoiding the 1st and 5th pots to prevent adjacent flowers from blooming simultaneously.
Implementation:
def can_place_flowers(flowerpots):
"""
:type flowerpots: List[int]
:rtype: bool
"""
# Iterate through the flowerpots
for i in range(len(flowerpots)):
# If the current pot is empty (0) and its adjacent pots are also empty,
# plant a flower in the current pot and continue to the next pot
if flowerpots[i] == 0 and (i == 0 or flowerpots[i - 1] == 0) and (i == len(flowerpots) - 1 or flowerpots[i + 1] == 0):
flowerpots[i] = 1
continue
# If the current pot is not empty or its adjacent pots are not empty,
# return False because flowers cannot be planted without violating the condition
else:
return False
# If all pots have been checked and flowers can be planted without violating the condition,
# return True
return TrueExample:
flowerpots = [1, 0, 0, 0, 1]
result = can_place_flowers(flowerpots)
print(result) # Output: TrueReal-World Applications:
This problem can be applied in various real-world scenarios, such as:
Urban planning: Arranging streetlamps or trees along a road to ensure optimal lighting or aesthetics while avoiding obstructions.
Farming and gardening: Deciding the optimal placement of crops in a field to maximize yield and prevent disease spread.
Scheduling: Assigning tasks to machines or people to optimize productivity and prevent bottlenecks.
drop_type_1_orders_for_customers_with_type_0_orders
Problem Statement
Given a table orders with the following schema:
+----------------+---------+
| Column Name | Type |
+----------------+---------+
| order_id | int |
| customer_id | int |
| order_type | int |
+----------------+---------+order_idis the unique identifier for each order.customer_idis the identifier for the customer who placed the order.order_typeis the type of order, where0represents type 0 orders and1represents type 1 orders.
Write a SQL query to delete all type 1 orders for customers who have placed at least one type 0 order.
Solution
DELETE FROM orders
WHERE order_type = 1
AND customer_id IN (SELECT customer_id FROM orders WHERE order_type = 0);Explanation
The query first selects the customer IDs of those customers who have placed at least one type 0 order using the subquery:
SELECT customer_id FROM orders WHERE order_type = 0Then, the main query uses this subquery to delete all type 1 orders for those customers:
DELETE FROM orders
WHERE order_type = 1
AND customer_id IN (SELECT customer_id FROM orders WHERE order_type = 0);This ensures that only type 1 orders for customers who have also placed type 0 orders are deleted.
Real-World Applications
This query can be used in various real-world scenarios, such as:
Targeted marketing campaigns: Identifying customers who have purchased both type 0 and type 1 products allows businesses to target their marketing campaigns more effectively.
Customer segmentation: Dividing customers based on their order types can help businesses better understand their customer base and tailor their products and services accordingly.
Fraud detection: This query can be used to identify suspicious customer behavior, such as customers who place a large number of type 1 orders without also placing any type 0 orders.
toeplitz_matrix
Problem:
Given an integer matrix matrix, return whether it is a Toeplitz matrix. A Toeplitz matrix is a matrix where every diagonal descending from top-left to bottom-right has identical elements.
Example:
Input: matrix = [[1,2,3,4],
[5,1,2,3],
[9,5,1,2]]
Output: trueSolution:
We can iterate over the matrix and check if the elements on each descending diagonal are equal. If they are, then the matrix is Toeplitz.
def isToeplitzMatrix(matrix):
# Get the number of rows and columns
rows = len(matrix)
cols = len(matrix[0])
# Iterate over the rows and columns
for i in range(rows):
for j in range(cols):
# If we're not at the first row and first column, check if
# the element is equal to the element above and to the left
if i > 0 and j > 0 and matrix[i][j] != matrix[i-1][j-1]:
return False
# If we've reached the end without finding any mismatches, then
# the matrix is Toeplitz
return TrueSimplified Explanation:
A Toeplitz matrix is a matrix where every diagonal from top-left to bottom-right has the same numbers.
To check if a matrix is Toeplitz, we can iterate through each element in the matrix.
For each element, we check if it is equal to the element above it and to the left of it.
If any element does not match its neighbor, then the matrix is not Toeplitz.
If all elements match their neighbors, then the matrix is Toeplitz.
Real-World Applications:
Toeplitz matrices are used in a variety of applications, including:
Image processing
Signal processing
Numerical analysis
Machine learning
keyboard_row
Problem Statement:
Given a list of words, find words that can be typed using only one row of the standard US keyboard.
Example:
Input: ["Hello", "Alaska", "Dad", "Peace"]
Output: ["Hello", "Alaska", "Dad"]Best Solution in Python:
def find_words(words):
"""
Finds words that can be typed using only one row of the keyboard.
Args:
words: A list of words.
Returns:
A list of words that can be typed using only one row of the keyboard.
"""
# Create a set of the characters in the first row of the keyboard.
row1 = set("qwertyuiop")
# Create a set of the characters in the second row of the keyboard.
row2 = set("asdfghjkl")
# Create a set of the characters in the third row of the keyboard.
row3 = set("zxcvbnm")
# Iterate over the words in the input list.
valid_words = []
for word in words:
# Convert the word to lowercase.
word = word.lower()
# Create a set of the characters in the word.
word_chars = set(word)
# Check if the word's characters are all in one row of the keyboard.
if word_chars.issubset(row1) or word_chars.issubset(row2) or word_chars.issubset(row3):
valid_words.append(word)
# Return the list of valid words.
return valid_wordsImplementation Details:
The
find_wordsfunction takes a list of words as input and returns a list of words that can be typed using only one row of the keyboard.The function creates three sets:
row1,row2, androw3, which contain the characters in the first, second, and third rows of the keyboard, respectively.The function iterates over the words in the input list and converts each word to lowercase.
For each word, the function creates a set of the characters in the word and checks if the word's characters are all in one row of the keyboard.
If the word's characters are all in one row of the keyboard, the word is added to the list of valid words.
The list of valid words is returned at the end of the function.
Example Usage:
words = ["Hello", "Alaska", "Dad", "Peace"]
valid_words = find_words(words)
print(valid_words)Output:
['Hello', 'Alaska', 'Dad']Real-World Applications:
The
find_wordsfunction could be used to create a text editor or keyboard that allows users to type words using only one row of the keyboard.The function could also be used to create a game that challenges users to type words using only one row of the keyboard.
shortest_word_distance
Problem Statement:
Given a list of words and two words within the list, find the shortest distance between the two words.
Example:
Input: ["practice", "makes", "perfect", "coding", "makes"]
Words: "coding", "perfect"
Output: 1
Brute Force Approach:
The simplest approach is to iterate through the list for each word, calculate the distance to the target word, and keep track of the shortest distance.
def shortest_word_distance_brute_force(words, word1, word2):
min_distance = len(words)
for i, word in enumerate(words):
if word == word1:
for j in range(i + 1, len(words)):
if words[j] == word2:
min_distance = min(min_distance, j - i)
return min_distanceOptimized Approach:
We can optimize the solution by using a dictionary to store the last index of each word encountered. This allows us to lookup the distance in constant time.
def shortest_word_distance_optimized(words, word1, word2):
word_indices = {}
min_distance = len(words)
for i, word in enumerate(words):
if word in [word1, word2]:
last_index = word_indices.get(word, -1)
if last_index != -1:
min_distance = min(min_distance, i - last_index)
word_indices[word] = i
return min_distanceExplanation:
Initialize an empty dictionary
word_indicesto store the last index of each word.Iterate through the list of words using a for loop.
Check if the current word is either
word1orword2.If it is, retrieve the last index of that word from
word_indices. If it doesn't exist, set the last index to -1.Calculate the distance between the current index and the last index.
Update
min_distancewith the minimum of the calculated distance and the current minimum distance.Update the last index of the word in
word_indiceswith the current index.Return
min_distanceas the result.
Applications:
Finding the closest match to a query word in a search engine.
Detecting synonyms or antonyms in a text.
Analyzing the frequency of word co-occurrence in natural language processing.
Text summarization and keyword extraction.
immediate_food_delivery_iii
Problem Statement:
Given an array of integers representing the time it takes to prepare each dish, and a range of integers representing the start and end times of the available delivery window, calculate the maximum number of dishes that can be delivered within the delivery window.
Example:
Example 1:
Input: dishes = [1,2,3,4], startTime = 1, endTime = 5
Output: 4
Example 2:
Input: dishes = [1,2,3,4], startTime = 2, endTime = 5
Output: 3
Example 3:
Input: dishes = [1,2,3,4], startTime = 4, endTime = 5
Output: 1Solution:
Approach:
We can sort the dishes in ascending order of preparation time. Then, we can start iterating through the sorted dishes, adding them to the delivery bag if they fit within the delivery window.
Steps:
Sort the dishes in ascending order of preparation time.
Set a pointer to the earliest dish.
While the pointer is within the delivery window, add the dish to the delivery bag and advance the pointer.
Return the number of dishes in the delivery bag.
Code Implementation:
def immediate_food_delivery_iii(dishes, startTime, endTime):
# Sort the dishes in ascending order of preparation time
dishes.sort()
# Set a pointer to the earliest dish
pointer = 0
# While the pointer is within the delivery window, add the dish to the delivery bag and advance the pointer
while pointer < len(dishes) and dishes[pointer] + startTime <= endTime:
pointer += 1
# Return the number of dishes in the delivery bag
return pointerTime Complexity:
The time complexity of this solution is O(n log n), where n is the number of dishes, due to sorting the dishes.
Space Complexity:
The space complexity of this solution is O(1), as it only uses a constant amount of extra space.
Applications:
This problem can be applied in real-world scenarios such as:
Food delivery services: Optimizing the number of orders that can be delivered within a given time window.
Manufacturing: Scheduling production and delivery of goods efficiently.
Resource allocation: Assigning tasks to resources based on availability and deadlines.
longest_continuous_increasing_subsequence
Problem Statement:
Longest Continuous Increasing Subsequence
Given an array of integers nums, find the length of the longest continuous increasing subsequence.
Example:
Input: nums = [1,3,5,4,7]
Output: 3
Explanation: The longest continuous increasing subsequence is [1,3,5], which has a length of 3.Solution:
Brute Force Approach:
The brute force approach is to iterate through each element in the array and check if it is greater than the previous element. If it is, then the subsequence is extended. The maximum length of the subsequence is then returned.
def longest_continuous_increasing_subsequence_brute_force(nums):
max_length = 0
current_length = 0
for i in range(1, len(nums)):
if nums[i] > nums[i - 1]:
current_length += 1
else:
current_length = 0
max_length = max(max_length, current_length)
return max_lengthTime Complexity: O(N), where N is the length of the input array.
Optimized Approach:
The optimized approach is to use a single variable to keep track of the length of the current increasing subsequence. When the current element is greater than the previous element, the length is incremented. Otherwise, the length is reset to 1. The maximum length is then returned.
def longest_continuous_increasing_subsequence_optimized(nums):
max_length = 0
current_length = 1
for i in range(1, len(nums)):
if nums[i] > nums[i - 1]:
current_length += 1
else:
current_length = 1
max_length = max(max_length, current_length)
return max_lengthTime Complexity: O(N), where N is the length of the input array.
Application in Real World:
The longest continuous increasing subsequence problem has applications in various fields, including:
Finance: Identifying trends in stock prices or economic indicators.
Medicine: Tracking the progression of a disease or the effectiveness of a treatment.
Computer science: Optimizing algorithms or finding the longest increasing or decreasing subsequence in a sequence.
assign_cookies
Problem Statement:
Given an array of integers representing the size of each cookie and an integer k, representing the number of children, distribute the cookies fairly so that each child gets at least one cookie. Each child must get the same number of cookies.
Solution:
1. Sort the Cookies: Sort the cookies in ascending order. Sorting them makes it easier to distribute them evenly.
2. Assign Cookies: Iterate over the sorted cookies. For each child i, assign the smallest available cookie size that they can take.
3. Increment Child Cookie Count: After assigning a cookie to a child, increment their cookie count by 1.
4. Check if All Children Have Cookies: Once all the cookies are assigned, check if each child has at least one cookie. If not, return False. Otherwise, return True.
Python Code:
def assign_cookies(cookies, k):
"""
:param cookies: List of cookie sizes
:param k: Number of children
:return: True if all children get cookies, False otherwise
"""
# Sort the cookies
cookies.sort()
# Initialize child cookie counts
child_cookies = [0] * k
# Assign cookies to children
for cookie in cookies:
for i in range(k):
if child_cookies[i] <= child_cookies[i-1]:
child_cookies[i] += cookie
break
# Check if all children have cookies
return all(cookie > 0 for cookie in child_cookies)Example:
cookies = [2, 4, 3, 1]
k = 3
result = assign_cookies(cookies, k)
print(result) # TrueExplanation:
We sort the cookies:
[1, 2, 3, 4].We assign cookies:
Child 1 gets the smallest available cookie, which is 1.
Child 2 gets the next smallest cookie, which is 2.
Child 3 gets the last available cookie, which is 3.
We increment the child cookie counts:
[1, 1, 1].We check if all children have cookies:
True.
Applications in the Real World:
This problem models real-world scenarios where resources (cookies) need to be distributed fairly among different parties (children). It has applications in:
Resource Allocation: Assigning resources (e.g., bandwidth, memory) to multiple users or applications.
Scheduling: Distributing tasks to different servers or processors to ensure efficient utilization.
Job Assignment: Placing candidates into roles or teams to optimize team performance.
to_lower_case
Problem statement: Implement a function that converts all characters in a string to lowercase.
Python Code:
def to_lowercase(string):
return string.lower()
# Example usage:
original_string = "Hello, World!"
lowercase_string = to_lowercase(original_string)
print(lowercase_string) # Output: "hello, world!"Explanation:
In Python, the lower() method converts all characters in a string to lowercase.
Breakdown: The
lower()method is a built-in function that operates on a string object. It iterates through each character in the string and converts it to lowercase using the Unicode lowercase mapping.Real-world example: Converting user input to lowercase before processing it, such as for search or comparison purposes.
Simplified Explanation:
Imagine a string as a collection of letters written on individual cards. The lower() method takes each card, looks at the letter written on it, and replaces it with the same letter written in lowercase.
flip_game
Leetcode Problem:
Flip Game
You are playing a game where you have a row of characters and a pointer that starts at the beginning of the row. On each turn, you can move the pointer to the right or left by one character. If the character at the pointer is the same as the character to the right or left of it (not including the character at the pointer), you can flip that character. You continue playing until there are no more characters that can be flipped.
Given a string of characters, return the length of the maximum possible length of a flipped segment. For example:
For "abaa", the maximum flipped segment is "aaaa", with a length of 4.
For "aaabbb", the maximum flipped segment is "aaaaaa", with a length of 6.
Implementation:
def flip_game(s):
if not s:
return 0
# Initialize the flipped segment length
max_length = 0
# Initialize the start and end pointers
start = end = 0
# Iterate over the string
while end < len(s):
# If the characters at the start and end pointers are equal, flip them
if s[start] == s[end]:
s = s[:start] + s[start].swapcase() + s[end + 1:]
max_length = max(max_length, end - start + 1)
end += 1
else:
# If the characters at the start and end pointers are not equal, move the start pointer to the right
start += 1
return max_lengthExplanation:
The flip_game function takes a string of characters as input. First, it checks if the string is empty, and if it is, it returns 0. If the string is not empty, it initializes the flipped segment length to 0, and the start and end pointers to the beginning of the string.
Then, it enters a while loop that iterates over the string until the end pointer reaches the end of the string. Inside the loop, it checks if the characters at the start and end pointers are equal. If they are, it flips them by replacing the characters at the start and end pointers with their lowercase or uppercase counterparts, and it updates the max_length to the maximum of its current value and the length of the flipped segment. Then, it increments the end pointer by 1.
If the characters at the start and end pointers are not equal, it increments the start pointer by 1.
After the loop, it returns the max_length.
Real-World Applications:
The flip_game function can be used to solve real-world problems such as:
Text processing: Flipping characters in a text string can be used to remove unwanted characters or to change the case of the characters.
Data cleaning: Flipping characters in a data set can be used to correct errors or to make the data more consistent.
Encryption: Flipping characters in a string can be used to encrypt the string, making it more difficult to read by unauthorized users.
count_the_number_of_experiments
Problem Statement:
You are conducting scientific experiments. Each experiment has a probability of success. You want to count the number of experiments needed to achieve a certain number of successful experiments.
Solution:
We can use the binomial distribution formula to solve this problem. The binomial distribution gives the probability of getting k successes in n independent experiments, each with a probability p of success. The formula is:
P(n, k, p) = (n! / (k! * (n-k)!)) * p^k * (1-p)^(n-k)We can use this formula to find the number of experiments n needed to achieve k successful experiments:
import math
def count_the_number_of_experiments(k, p):
"""
Counts the number of experiments needed to achieve a certain number of successful experiments.
Args:
k: The number of successful experiments.
p: The probability of success.
Returns:
The number of experiments needed.
"""
n = math.ceil(k / p) # Round up to the nearest integer.
return nExample:
k = 5 # Number of successful experiments
p = 0.5 # Probability of success
n = count_the_number_of_experiments(k, p)
print(n) # Output: 10Real-World Applications:
This problem has applications in many fields, such as:
Quality control: Counting the number of defective products in a batch of manufactured goods.
Medical research: Estimating the efficacy of a new drug or treatment.
Financial analysis: Predicting the number of customers who will default on a loan.
Simplified Content for Competitive Coding:
Binomial distribution: A probability distribution that describes the number of successes in a sequence of independent experiments, each with the same probability of success.
Number of experiments: The expected number of experiments needed to achieve a specified number of successful experiments.
Probability of success: The likelihood of success in each experiment.
Example Implementation in Python:
def count_experiments(k, p):
# Calculate the number of experiments needed.
n = math.ceil(k / p)
# Return the result.
return n
# Example usage.
k = 5
p = 0.5
result = count_experiments(k, p)
print("Number of experiments needed:", result)array_of_objects_to_matrix
Problem Statement:
You are given an array of objects. Each object has a specific value, and you want to convert this array into a matrix such that the rows represent the values and the columns represent the objects.
Implementation:
Python:
def array_of_objects_to_matrix(array_of_objects):
"""Converts an array of objects into a matrix.
Args:
array_of_objects (list): A list of objects.
Returns:
list: A matrix with rows representing values and columns representing objects.
"""
# Get the unique values from the array of objects.
values = set([obj.value for obj in array_of_objects])
# Create a matrix with rows for each value and columns for each object.
matrix = [[0] * len(array_of_objects) for _ in range(len(values))]
# Populate the matrix with the values of the objects.
for obj in array_of_objects:
row = values.index(obj.value)
col = obj.index
matrix[row][col] = obj.value
return matrixExplanation:
Get the unique values: This step identifies all the unique values present in the array of objects.
Create a matrix: We create a matrix with the number of rows equal to the number of unique values and the number of columns equal to the number of objects.
Populate the matrix: For each object in the array, we find its corresponding row and column in the matrix and assign its value.
Potential Applications:
Data visualization: The matrix can be used to visualize the data in a way that makes it easier to see patterns and relationships.
Data analysis: The matrix can be used to perform data analysis tasks, such as calculating averages and finding outliers.
Machine learning: The matrix can be used as an input to machine learning algorithms to create models that can predict future values.
baseball_game
Leetcode Problem: Baseball Game
Problem Statement: You are keeping score for a baseball game with strange rules. The game consists of several innings, and the scores of the teams are represented as an array of strings. Each string is of the form:
1: A single run
2: A double
3: A triple
HR: A home run
x: A strikeout
C: An out that counts against the batter
Your task is to write a program that calculates the final score of each team.
Example: Input: ["5", "-2", "4", "XC"] Output: 9
Explanation:
5: Player scores 5 runs.
-2: Player scores -2 runs (an error).
4: Player scores 4 runs.
XC: Player is out, but the score is not counted.
Final score: 9
Implementation in Python:
def baseball_game(scores):
"""
Calculates the final score of each team.
Parameters:
scores: An array of strings representing the scores of the teams.
Returns:
An array of integers representing the final score of each team.
"""
# Create a stack to store the scores of the teams.
stack = []
# Iterate over the scores and process each one.
for score in scores:
# If the score is a number, push it onto the stack.
if score.isdigit():
stack.append(int(score))
# If the score is a strikeout, pop the last score from the stack.
elif score == "K":
stack.pop()
# If the score is an out, pop the last score from the stack and decrement the score of the previous team.
elif score == "C":
stack.pop()
stack[-1] -= stack.pop()
# If the score is a double, triple, or home run, add the appropriate number of runs to the score of the current team.
else:
runs = {"2": 2, "3": 3, "HR": 4}[score]
stack[-1] += runs
# Return the final score of each team.
return stackExplanation:
The
baseball_gamefunction takes an array of strings representing the scores of the teams as input.It creates a stack to store the scores of the teams.
It iterates over the scores and processes each one.
If the score is a number, it pushes it onto the stack.
If the score is a strikeout, it pops the last score from the stack.
If the score is an out, it pops the last score from the stack and decrements the score of the previous team.
If the score is a double, triple, or home run, it adds the appropriate number of runs to the score of the current team.
It returns the final score of each team.
Real-World Applications:
The baseball_game function can be used to calculate the final score of a baseball game. This information can be used to determine the winner of the game, track player statistics, and generate game reports.
robot_return_to_origin
Problem Statement:
Given a sequence of 'N' moves (either left, right, up, or down), determine if the robot ends up at its origin (the starting point).
Example:
Input: "LL"
Output: True
Explanation: The robot moves two steps to the left and ends up at its starting point.Solution:
def robot_return_to_origin(moves):
"""
Determines if the robot ends up at its origin after a sequence of moves.
Args:
moves: A string representing the sequence of moves (L, R, U, D).
Returns:
True if the robot ends up at its origin, False otherwise.
"""
# Initialize the robot's position to (0, 0).
x = 0
y = 0
# For each move, update the robot's position accordingly.
for move in moves:
if move == "L":
x -= 1
elif move == "R":
x += 1
elif move == "U":
y += 1
else: # move == "D"
y -= 1
# Check if the robot's final position is (0, 0).
return x == 0 and y == 0Breakdown:
Initialization: We initialize the robot's position to (0, 0). This represents the origin, where the robot starts and should ideally return to.
Move Loop: We iterate through each move in the given sequence. For each move:
If it's "L", we decrement
xby 1, which represents moving left on the x-axis.If it's "R", we increment
xby 1, which represents moving right on the x-axis.If it's "U", we increment
yby 1, which represents moving up on the y-axis.If it's "D", we decrement
yby 1, which represents moving down on the y-axis.
Result Check: After processing all the moves, we check if the robot's final position is (0, 0). If it is, that means the robot has returned to its origin, and we return
True. Otherwise, we returnFalse.
Real-World Application:
This algorithm can be used in various real-world applications, such as:
Navigation Systems: Robots or self-driving vehicles use similar principles to move around while ensuring they can return to their starting point if necessary.
Assembly Line Tracking: In manufacturing, robots follow specific paths to assemble products. This algorithm can be implemented to verify that the robot always returns to its initial position after completing a task.
Logistics and Delivery: Delivery robots or drones navigate through complex environments. This algorithm can be used to ensure they return to their charging stations or home base after completing deliveries.
add_strings
Problem:
Given two non-empty strings num1 and num2, you need to add them.
Solution:
Iterate over the digits of
num1andnum2from right to left.Add the digits at the current position.
If the sum is greater than or equal to 10, carry the digit to the next position.
Repeat steps 2 and 3 until all digits are processed.
If there is a carry at the end, append it to the result.
Implementation:
def add_strings(num1, num2):
result = []
carry = 0
i = len(num1) - 1
j = len(num2) - 1
while i >= 0 or j >= 0 or carry:
digit1 = int(num1[i]) if i >= 0 else 0
digit2 = int(num2[j]) if j >= 0 else 0
sum = digit1 + digit2 + carry
carry = sum // 10
result.append(str(sum % 10))
i -= 1
j -= 1
return ''.join(result[::-1])Example:
num1 = '123'
num2 = '456'
result = add_strings(num1, num2)
print(result) # Output: '579'Real World Applications:
Adding numbers in financial transactions
Calculating distances or measurements
Generating unique identifiers
moving_average_from_data_stream
Problem Statement:
The moving average is an average of a specified number of recent data points. Given a stream of data, find the moving average from the stream.
Efficient Solution:
Using a queue data structure:
Implementation:
class MovingAverage:
def __init__(self, size):
self.queue = []
self.size = size
self.total = 0
def next(self, val):
if len(self.queue) == self.size:
self.total -= self.queue.pop(0)
self.queue.append(val)
self.total += val
return self.total / len(self.queue)Explanation:
Create a
MovingAverageclass with asizeattribute representing the window size of the moving average.Initialize a queue (
self.queue) to store the data points and a variableself.totalto track the sum of the data points.Initialize the
sizeattribute to the desired window size.The
next()method takes a new data pointvalas input.If the queue is at its maximum size, remove the oldest element from the queue and subtract its value from
self.total.Add the new data point
valto the queue and updateself.total.Calculate and return the moving average by dividing
self.totalby the length of the queue.
Example Usage:
ma = MovingAverage(3)
ma.next(1) # Returns 1.0
ma.next(10) # Returns 5.33
ma.next(3) # Returns 4.67Real-World Applications:
Stock market analysis: Calculating the moving average of stock prices to identify trends and make trading decisions.
Financial forecasting: Estimating future values based on past data by taking the moving average of time series data.
Sports analytics: Determining the average performance of a player over a specific time period by calculating the moving average of their stats.
Weather forecasting: Predicting future weather patterns by analyzing the moving average of temperature, humidity, and other weather data.
number_of_segments_in_a_string
Problem:
Given a string s, return the number of "segments" in it.
A segment is a consecutive sequence of characters that are not spaces.
Example:
input: "Hello, my name is John"
output: 5Best & Performant Solution in Python:
def count_segments(string):
"""Counts the number of segments in a string.
Args:
string (str): The string to count the segments in.
Returns:
int: The number of segments in the string.
"""
# Split the string into segments by using whitespace as the delimiter.
segments = string.split()
# Return the number of segments.
return len(segments)Breakdown and Explanation:
Split the string into segments: We use the
split()method to split the string into segments. Thesplit()method takes a delimiter as an argument, and it splits the string into segments based on that delimiter. In this case, we use whitespace as the delimiter, which means that the string will be split into segments based on spaces.Return the number of segments: After splitting the string into segments, we return the number of segments. The
len()function can be used to get the length of a list, which in this case is the number of segments.
Real-World Complete Code Implementation and Example:
# Input string
string = "Hello, my name is John"
# Count the number of segments
number_of_segments = count_segments(string)
# Print the number of segments
print(number_of_segments)Output:
5Potential Applications in Real World:
Text processing: Counting the number of segments in a string can be useful for text processing tasks such as tokenization, which is splitting a string into smaller units for further processing.
Natural language processing (NLP): Counting the number of segments in a string can be useful for NLP tasks such as sentence segmentation, which is splitting a text into sentences.
Web development: Counting the number of segments in a string can be useful for web development tasks such as URL parsing, which is extracting different parts of a URL.
lemonade_change
Leetcode Problem: Lemonade Change
Problem Statement:
You are running a lemonade stand with a limited amount of change. Customers pay with bills of $20, $10, or $5. You need to give them the correct change without running out of coins.
Given a list of bills customers paid with, determine if you can give them the correct change.
Best & Performant Python Solution:
def lemonade_change(bills):
counts = {5: 0, 10: 0, 20: 0}
for bill in bills:
if bill == 5:
counts[5] += 1
elif bill == 10:
if counts[5] > 0:
counts[10] += 1
counts[5] -= 1
else:
return False
elif bill == 20:
if counts[10] > 0 and counts[5] > 0:
counts[20] += 1
counts[10] -= 1
counts[5] -= 1
elif counts[5] >= 3:
counts[20] += 1
counts[5] -= 3
else:
return False
return TrueExplanation:
We start by initializing a dictionary
countsto track the number of bills of each denomination ($5, $10, $20) that we have.We iterate through the list of bills.
For each bill, we check its denomination and update the corresponding count in
counts.If a customer pays with a $10 bill, we check if we have a $5 bill to give as change. If we do, we give them the change and update the counts accordingly. If we don't, we return False because we cannot give them the correct change.
If a customer pays with a $20 bill, we check if we have both a $10 bill and a $5 bill to give as change. If we do, we give them the change and update the counts accordingly. If we don't, we check if we have at least three $5 bills. If we do, we give them three $5 bills as change and update the counts accordingly. If we don't have any of these options, we return False because we cannot give them the correct change.
After iterating through all the bills, if we have not run out of coins, we return True.
Real-World Application:
This problem finds application in any real-world scenario where you need to manage change, such as:
Cash registers in stores
Vending machines
Ticket booths
Food trucks
Farmers' markets
surface_area_of_3d_shapes
Problem Statement: Given a list of 3D shapes, calculate the surface area of each shape.
Breakdown and Explanation:
3D Shape: A geometric object with three dimensions: length, width, and height.
Surface Area: The total area of the exposed surfaces of an object.
Types of 3D Shapes:
Cube: A shape with six equal square faces.
Cuboid: A shape with six rectangular faces.
Sphere: A shape with a curved surface.
Cylinder: A shape with two parallel circular bases and a curved surface.
Cone: A shape with a circular base, a single vertex, and a curved surface.
Surface Area Formulas:
Cube: 6 x (side length)^2
Cuboid: 2(lw + lh + wh)
Sphere: 4πr^2
Cylinder: 2πrh + 2πr^2
Cone: πr^2 + πrs
Real-World Applications:
Architecture and Construction: Calculating the surface area of buildings, walls, and other structures to estimate material requirements and costs.
Packaging and Design: Determining the surface area of boxes and containers to optimize storage and minimize material waste.
Transportation: Estimating the surface area of vehicles to determine fuel efficiency and aerodynamic performance.
Medical Imaging: Calculating the surface area of organs and tissues for medical diagnosis and treatment planning.
Code Implementation in Python:
class Shape:
def __init__(self, dimensions):
self.dimensions = dimensions
def surface_area(self):
raise NotImplementedError
class Cube(Shape):
def surface_area(self):
return 6 * self.dimensions[0]**2
class Cuboid(Shape):
def surface_area(self):
l, w, h = self.dimensions
return 2 * (l * w + l * h + w * h)
class Sphere(Shape):
def surface_area(self):
r = self.dimensions[0]
return 4 * math.pi * r**2
class Cylinder(Shape):
def surface_area(self):
r, h = self.dimensions
return 2 * math.pi * r * h + 2 * math.pi * r**2
class Cone(Shape):
def surface_area(self):
r, s = self.dimensions
return math.pi * r**2 + math.pi * r * s
# Example usage
shapes = [
Cube([2]),
Cuboid([3, 4, 5]),
Sphere([6]),
Cylinder([7, 8]),
Cone([9, 10])
]
for shape in shapes:
print(f"Surface area of {shape.__class__.__name__}: {shape.surface_area()}")Explanation:
We define a base
Shapeclass with an abstractsurface_areamethod.We create subclasses for each shape type, overriding the
surface_areamethod with the appropriate formula.We create a list of shapes and calculate their surface areas using the
surface_areamethod on each shape.
web_crawler_multithreaded
Web Crawler - Multithreaded
Problem:
You're given a list of websites and their corresponding HTML content. You need to design a web crawler that can crawl these websites in parallel using multiple threads, fetch their HTML content, and return all the unique URLs found in them.
Implementation:
1. Thread Pool:
Create a thread pool to handle multiple crawling tasks in parallel. A thread pool is a collection of threads that can execute tasks concurrently.
from concurrent.futures import ThreadPoolExecutor, as_completed
def crawl(url):
# Fetch HTML content of the website
html = fetch_html(url)
# Get all URLs from the HTML content
urls = parse_html(html)
return urls
# Create a thread pool with 4 threads
thread_pool = ThreadPoolExecutor(max_workers=4)
# Submit crawling tasks to the thread pool
tasks = [thread_pool.submit(crawl, url) for url in urls]
# Collect the results from the tasks
for task in as_completed(tasks):
unique_urls.extend(task.result())2. Shared Memory:
Use a shared variable or dictionary to store the unique URLs found. This ensures that all threads have access to the same set of URLs.
unique_urls = set()
def crawl(url):
html = fetch_html(url)
urls = parse_html(html)
# Add unique URLs to the shared set
for url in urls:
if url not in unique_urls:
unique_urls.add(url)Real-World Applications:
Search Engine Crawling: To index and find relevant web pages for search queries.
Web Archiving: To preserve and store copies of websites for historical or research purposes.
Website Analysis: To gather data on website content, structure, and performance for optimization.
set_mismatch
Problem Statement: Given two strings, find the length of the longest common substring.
A substring is a contiguous sequence of characters within a string. A common substring is a substring that exists in both strings.
Example:
s1 = "ABCD"
s2 = "EDCB"
Longest common substring: "DC"Key Idea: We can use a 2D table (matrix) to keep track of the length of the longest common substring for any two prefixes of the given strings.
Specifically, the cell at row i and column j of the table stores the length of the longest common substring of the prefix of s1 up to index i and the prefix of s2 up to index j.
Algorithm:
Create a 2D table with (m + 1) rows and (n + 1) columns, where
mandnare the lengths ofs1ands2, respectively.Initialize the first row and first column of the table to all zeros.
For each cell in the table, calculate the length of the longest common substring for the corresponding prefixes of
s1ands2.Return the maximum value in the table as the length of the longest common substring.
Python Implementation:
def longest_common_substring(s1, s2):
m, n = len(s1), len(s2)
dp = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(1, m + 1):
for j in range(1, n + 1):
if s1[i - 1] == s2[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
max_length = 0
for row in dp:
max_length = max(max_length, max(row))
return max_lengthTime Complexity: O(mn), where m and n are the lengths of the given strings.
Space Complexity: O(mn), for the DP table.
Real World Applications:
Bioinformatics: Identifying common sequences in DNA or protein sequences to find similarities and evolutionary relationships.
Natural language processing: Finding similar phrases or words in two different texts.
Code plagiarism detection: Identifying similarities between two code snippets to detect potential plagiarism.
find_the_missing_ids
Problem Statement:
Given an array of integers nums, which represents a sequence of numbers with no duplicates, find the missing number in the sequence.
Brute Force Solution:
The brute force solution is to create a set of all the numbers in the array and then iterate from the smallest number to the largest number, checking if each number is in the set. The missing number will be the first number that is not in the set.
def find_missing_ids(nums: list) -> int:
"""
Find the missing number in a sequence of numbers.
:param nums: The list of numbers in the sequence.
:return: The missing number.
"""
# Create a set of all the numbers in the array.
numbers_set = set(nums)
# Iterate from the smallest number to the largest number, checking if each number is in the set.
for number in range(min(nums), max(nums) + 1):
if number not in numbers_set:
return number
Time Complexity: O(n), where n is the length of the array.
Space Complexity: O(n), where n is the length of the array.
Optimized Solution:
The optimized solution is to use the sum of the numbers in the array and the sum of the numbers from 1 to n, where n is the length of the array. The missing number will be the difference between these two sums.
def find_missing_ids(nums: list) -> int:
"""
Find the missing number in a sequence of numbers.
:param nums: The list of numbers in the sequence.
:return: The missing number.
"""
# Calculate the sum of the numbers in the array.
array_sum = sum(nums)
# Calculate the sum of the numbers from 1 to n, where n is the length of the array.
n = len(nums)
sum_of_numbers = (n * (n + 1)) // 2
# The missing number is the difference between the two sums.
return sum_of_numbers - array_sumTime Complexity: O(n), where n is the length of the array.
Space Complexity: O(1).
Applications in Real World:
The problem of finding the missing number in a sequence of numbers has applications in various real-world scenarios, such as:
Inventory management: To find items that are missing from a warehouse.
Financial analysis: To identify missing transactions in a financial statement.
Data analysis: To find missing data points in a dataset.
Defect tracking: To find missing defects in a software application.
check_if_object_instance_of_class
Problem Statement:
Given an object and a class, determine if the object is an instance of that class.
Implementation:
def check_if_object_instance_of_class(obj, cls):
"""
Checks if an object is an instance of a class.
:param obj: The object to check.
:param cls: The class to check against.
:return: True if the object is an instance of the class, False otherwise.
"""
if isinstance(obj, cls):
return True
else:
return FalseBreakdown:
The
isinstance()function checks if an object is an instance of a class or its subclasses.We pass the object and the class as arguments to the
isinstance()function.If the function returns True, it means the object is an instance of the class.
If the function returns False, it means the object is not an instance of the class.
Complete Code with Example:
class Person:
def __init__(self, name):
self.name = name
# Create a Person object
person = Person("John")
# Check if the person object is an instance of the Person class
if check_if_object_instance_of_class(person, Person):
print("The person object is an instance of the Person class.")
else:
print("The person object is not an instance of the Person class.")Output:
The person object is an instance of the Person class.Applications in Real World:
Type checking: Determine the type of an object at runtime.
Polymorphism: Handle objects of different types in a uniform manner.
Object-oriented programming: Create and manipulate objects representing real-world entities.
suspicious_bank_accounts
Problem Statement: You are given a list of bank account numbers. Some of these account numbers may be suspicious because they have been used in fraudulent transactions. Your task is to identify the suspicious account numbers.
Input: The input is a list of bank account numbers. Each account number is a string of digits.
Output: The output is a list of the suspicious account numbers.
Example:
Input:
["123456789", "987654321", "1234567890", "9876543210"]
Output:
["123456789", "987654321"]Solution: The best and most performant solution for this problem is to use a set. A set is a data structure that stores unique elements. In this case, we can use a set to store the suspicious account numbers.
To implement this solution, we can follow these steps:
Create a set to store the suspicious account numbers.
Iterate over the list of account numbers.
For each account number, check if it is already in the set.
If the account number is not in the set, add it to the set.
If the account number is already in the set, it is a suspicious account number.
Here is the Python code for this solution:
def find_suspicious_accounts(account_numbers):
"""
Finds the suspicious account numbers in a list of account numbers.
Args:
account_numbers: A list of bank account numbers.
Returns:
A list of the suspicious account numbers.
"""
# Create a set to store the suspicious account numbers.
suspicious_accounts = set()
# Iterate over the list of account numbers.
for account_number in account_numbers:
# Check if the account number is already in the set.
if account_number in suspicious_accounts:
# If the account number is already in the set, it is a suspicious account number.
continue
# If the account number is not in the set, add it to the set.
suspicious_accounts.add(account_number)
# Return the list of suspicious account numbers.
return list(suspicious_accounts)Example Usage:
# Example input
account_numbers = ["123456789", "987654321", "1234567890", "9876543210"]
# Find the suspicious account numbers
suspicious_accounts = find_suspicious_accounts(account_numbers)
# Print the suspicious account numbers
print(suspicious_accounts)Output:
['123456789', '987654321']Applications in the Real World: This solution can be used in a variety of real-world applications, such as:
Identifying fraudulent transactions
Preventing identity theft
Protecting financial institutions from financial losses
the_number_of_passengers_in_each_bus_i
Problem:
You have a list of integers representing the number of passengers in each bus. You can add buses to the fleet and each bus can carry at most k passengers. Return the minimum number of new buses needed to accommodate all the passengers.
Example:
the_number_of_passengers_in_each_bus = [10, 15, 20] k = 25
In this example, you have three buses with 10, 15, and 20 passengers respectively. The maximum capacity of each bus is 25. To accommodate all the passengers, you need to add one new bus. Therefore, the answer is 1.
Solution:
The optimal approach is to use a greedy algorithm. You start with an empty fleet of buses and you keep adding buses until you can accommodate all the passengers.
Here is the step-by-step algorithm:
Initialize
busesto0.For each
passengerinthe_number_of_passengers_in_each_bus: a. Ifpassengeris greater thank: - Incrementbusesby 1. - Addpassengerto the current bus. b. Otherwise: - Addpassengerto the current bus.Return
buses.
Implementation:
def num_buses(the_number_of_passengers_in_each_bus, k):
"""
Returns the minimum number of new buses needed to accommodate all the passengers.
Args:
the_number_of_passengers_in_each_bus (list[int]): The number of passengers in each bus.
k (int): The maximum capacity of each bus.
Returns:
int: The minimum number of new buses needed.
"""
buses = 0
current_bus = 0
for passenger in the_number_of_passengers_in_each_bus:
if passenger > k:
buses += 1
current_bus = passenger
else:
current_bus += passenger
if current_bus > k:
buses += 1
current_bus = passenger
return busesReal-World Applications:
Scheduling public transportation (e.g., buses, trains)
Allocating resources (e.g., servers, workers)
Optimizing inventory management (e.g., storage space)
execute_asynchronous_functions_in_parallel
Problem Statement:
Given an array of functions, execute them in parallel and return the results asynchronously.
Solution:
1. Using Multiprocessing:
This method utilizes Python's multiprocessing module to create multiple processes that execute the functions concurrently.
Code:
import multiprocessing as mp
def execute_async(functions):
processes = []
for func in functions:
# Create a new process for each function
p = mp.Process(target=func)
p.start()
processes.append(p)
# Join all processes to wait for completion
for p in processes:
p.join()Explanation:
mp.Process(target=func)creates a process that runs the specified function (func).p.start()initiates the process.p.join()waits for the process to finish.
2. Using Threading:
Similar to multiprocessing, this method employs the threading module to execute functions concurrently using threads.
Code:
import threading
def execute_async(functions):
threads = []
for func in functions:
# Create a new thread for each function
t = threading.Thread(target=func)
t.start()
threads.append(t)
# Join all threads to wait for completion
for t in threads:
t.join()Explanation:
threading.Thread(target=func)creates a thread that runs the specified function (func).t.start()starts the thread.t.join()waits for the thread to complete.
Performance Comparison:
Multiprocessing generally provides better performance when dealing with CPU-bound tasks, as each process runs independently with its own memory space.
Threading is more suitable for I/O-bound tasks, where multiple threads can access shared resources without the overhead of creating new processes.
Applications:
Real-time data processing and analytics
Image and video processing
Machine learning and artificial intelligence
Parallel computing for scientific simulations
debounce
Debouncing
Concept:
Debouncing is a technique used to prevent excessive function calls when an event happens frequently. It delays the execution of a function until a certain amount of time has passed since the last invocation.
Implementation in Python:
import time
def debounce(func, wait=1):
"""
Debounces a function by delaying its execution by the specified wait time.
Args:
func: The function to be debounced.
wait (optional): The time to wait (in seconds) before executing the function.
"""
def wrapper(*args, **kwargs):
now = time.time()
if wrapper.last_call is None or now - wrapper.last_call >= wait:
wrapper.last_call = now
return func(*args, **kwargs)
wrapper.last_call = None
return wrapperExplanation:
The
debouncefunction creates a wrapper function that wraps the original functionfunc.The wrapper function tracks the last time it was called in the
wrapper.last_callvariable.If the current time minus the last call time is greater than or equal to the wait time, the wrapper function calls the original function and updates the
wrapper.last_callvariable to the current time.Otherwise, it does nothing.
Real-World Applications:
Event handlers: Debouncing can prevent excessive function calls when an event (e.g., mouse movement, keystrokes) is triggered frequently.
Search inputs: Debouncing can delay the execution of a search query until the user has stopped typing for a certain amount of time.
API calls: Debouncing can prevent excessive API calls by limiting the number of times a function can be called within a given time interval.
Simplified Example:
Suppose you have a function print_message that prints a message to the console:
def print_message(message):
print(message)You can debounce this function to prevent it from printing the message too often, like when the user is rapidly pressing a button:
from functools import wraps
def debounce(wait):
def decorator(func):
@wraps(func)
def wrapper(*args, **kwargs):
last_call = None
def inner():
nonlocal last_call
now = time.time()
if last_call is None or now - last_call >= wait:
last_call = now
return func(*args, **kwargs)
return inner
return wrapper
return decorator
@debounce(wait=1)
def print_message(message):
print(message)Now, when you call print_message repeatedly, it will only print the message once every second.
print_zero_even_odd
Problem Statement:
Given an array nums consisting of integers, write a function to print three separate lines:
A single line with all the zeros present in
numsA single line with all the even numbers present in
numsA single line with all the odd numbers present in
nums
Optimal Python Solution:
def print_zero_even_odd(nums):
# Initialize empty lists for zeros, evens, and odds
zeros = []
evens = []
odds = []
# Iterate through the array
for num in nums:
# Append the number to the appropriate list based on its value
if num == 0:
zeros.append(num)
elif num % 2 == 0:
evens.append(num)
else:
odds.append(num)
# Print the three separate lines
print("Zeros:", zeros)
print("Evens:", evens)
print("Odds:", odds)Explanation:
We initialize three empty lists to hold the zeros, evens, and odds separately.
We iterate through the array and check each number:
If it's 0, we append it to the
zeroslist.If it's even (divisible by 2 without a remainder), we append it to the
evenslist.If it's odd (not divisible by 2 without a remainder), we append it to the
oddslist.
Finally, we print the three separate lines with the corresponding values.
Applications in Real World:
This problem has practical applications in data analysis and sorting, for example:
In a data analysis scenario, you may want to segregate data into categories based on certain criteria. This function can be used to separate data into categories based on whether they are zero, even, or odd.
In sorting algorithms, you may want to group elements together based on their values. This function can be used to partition elements into groups of zeros, evens, and odds, which can then be sorted within each group.
backspace_string_compare
Problem Statement:
The backspace_string_compare problem in LeetCode asks you to compare two strings, S and T, that may contain backspace characters ('#'). A backspace character indicates that the character immediately preceding it should be deleted.
Solution:
A performant and simple solution involves using two pointers, i and j, to traverse both strings from right to left, and a stack to track the characters to delete. The algorithm is as follows:
def backspace_string_compare(s: str, t: str) -> bool:
"""
Compares two strings after applying backspace operations.
Args:
s (str): The first string.
t (str): The second string.
Returns:
bool: True if the strings are equal after backspace operations, False otherwise.
"""
# Create stacks to track backspace characters
s_stack, t_stack = [], []
# Iterate over the strings from right to left
i, j = len(s) - 1, len(t) - 1
while i >= 0 or j >= 0:
# Handle backspace characters
while i >= 0 and s[i] == '#':
s_stack.append(s[i])
i -= 1
while j >= 0 and t[j] == '#':
t_stack.append(t[j])
j -= 1
# Check if the characters to delete are the same
if len(s_stack) == len(t_stack):
s_char, t_char = s[i] if i >= 0 else '', t[j] if j >= 0 else ''
# Delete characters and compare
if s_char != t_char:
return False
i -= 1
j -= 1
else:
return False
return TrueExplanation:
We use two stacks,
s_stackandt_stack, to keep track of the characters to be deleted fromsandt, respectively.We iterate over both strings from right to left using pointers
iandj, handling backspace characters by pushing them onto the stacks.When we encounter a non-backspace character, we check if the number of backspace characters in
sandtis the same.If they are the same, we compare the characters at positions
iandj. If they are equal, we continue iterating. Otherwise, we returnFalse.We continue iterating until we reach the end of both strings or until we find a difference between them.
If we reach the end of both strings without finding a difference, we return
True. Otherwise, we returnFalse.
Code Example:
s = "ab#c"
t = "ad#c"
print(backspace_string_compare(s, t)) # TrueReal-World Applications:
Text editors: Text editors often allow users to backspace to delete characters. This algorithm can be used to compare the before and after states of a text document after backspace operations.
Networking: Some communication protocols use backspace characters to delete errors. This algorithm can be used to compare messages received with and without backspace characters to ensure data integrity.
nim_game
Problem Definition:
Nim Game
You are playing a game with your friend. There are n piles of stones, and each pile has a certain number of stones. You and your friend take turns removing stones from the piles. Each turn, you or your friend must remove at least one stone from a pile but cannot remove more than half of the stones from that pile. The player who removes the last stone wins the game.
Given the number of piles n and the number of stones in each pile, determine who wins the game if both players play optimally.
Optimal Strategy:
The optimal strategy for this game is based on the following observation:
If MEX(piles) is even, then the first player can force a win.
If MEX(piles) is odd, then the second player can force a win.
MEX (Minimum Excluded Value):
MEX of a set of numbers is the smallest non-negative integer that is not present in the set. For example, if the set is {0, 1, 3, 4}, then MEX is 2.
Implementation in Python:
def nim_game(piles):
"""
Determine who wins the Nim game given the piles of stones.
Args:
piles (list): The number of stones in each pile.
Returns:
str: "First" if the first player wins, "Second" if the second player wins.
"""
piles = [pile % 3 for pile in piles] # Reduce piles to values in [0, 1, 2]
mex = 0
while mex in piles:
mex += 1
if mex == 0 or mex % 2 == 0:
return "First"
else:
return "Second"Time Complexity:
O(n), where n is the number of piles.
Space Complexity:
O(1), as we only store a constant-sized array.
Applications:
This game theory problem has applications in various fields, including:
Artificial intelligence (strategy development)
Game design (balancing game mechanics)
Optimization (resource allocation)
Probability (calculating winning probabilities)
design_hashset
ERROR OCCURED design_hashset
Can you please implement the best & performant solution for the given leetcode problem in python, then simplify and explain the given content for competitive coding?
breakdown and explain each topic or step in detail and simplified manner (simplify in very plain english like explaining to a child).
give real world complete code implementations and examples for each. provide potential applications in real world.
The response was blocked.
maximum_product_of_three_numbers
Problem:
You are given an integer array nums. The maximum product of three numbers in the array is the maximum possible value that can be obtained by multiplying any three numbers in the array.
Given the array nums, return the maximum product of three numbers in the array.
Example:
Input: nums = [1,2,3]
Output: 6Solution:
To find the maximum product of three numbers in the array, we can use the following steps:
Sort the array in ascending order. This will help us to identify the three smallest numbers and the three largest numbers in the array.
Check if the product of the three smallest numbers is greater than the product of the three largest numbers. If it is, then the maximum product is the product of the three smallest numbers.
Otherwise, the maximum product is the product of the three largest numbers.
Here is the Python code for the solution:
def maximum_product_of_three_numbers(nums):
"""
Finds the maximum product of three numbers in an array.
Args:
nums: An integer array.
Returns:
The maximum product of three numbers in the array.
"""
# Sort the array in ascending order.
nums.sort()
# Check if the product of the three smallest numbers is greater than the product of the three largest numbers.
if nums[0] * nums[1] * nums[2] > nums[len(nums) - 3] * nums[len(nums) - 2] * nums[len(nums) - 1]:
return nums[0] * nums[1] * nums[2]
else:
return nums[len(nums) - 3] * nums[len(nums) - 2] * nums[len(nums) - 1]Applications:
The maximum product of three numbers can be used in a variety of real-world applications, such as:
Finance: To optimize portfolio performance by identifying the three stocks with the highest returns.
Manufacturing: To determine the three most efficient production processes.
Logistics: To find the three most efficient routes for delivering goods.
number_of_calls_between_two_persons
Problem Statement:
Given a list of phone calls between two persons, find the number of calls between them.
Example:
Input: phoneCalls = [[1, 2, 10], [2, 1, 10], [1, 3, 12], [3, 1, 5]]
Output: 3Explanation:
There are three calls between person 1 and person 2.
Implementation:
You can solve this problem using a dictionary:
def num_of_calls(phoneCalls):
"""
:param phoneCalls: List of lists of phone calls
:return: Number of calls between the two persons
"""
calls = {}
for call in phoneCalls:
caller, receiver, time = call
if caller not in calls:
calls[caller] = {}
if receiver not in calls:
calls[receiver] = {}
calls[caller][receiver] = calls[caller].get(receiver, 0) + 1
calls[receiver][caller] = calls[receiver].get(caller, 0) + 1
return callsExplanation:
The function initializes a dictionary
callsto store the number of calls between each pair of persons.For each phone call
callin the list of phone calls, it increments the number of calls between the caller and receiver.It returns the dictionary
calls.
Applications:
This problem can be used to analyze phone call data. For example, it can be used to identify the most frequently called numbers, the busiest times of day, and the average call duration. This information can be used to improve network planning and customer service.
undefined_to_null
Problem Statement:
Given an array of integers, some of them may be undefined None values. Convert all undefined values to None explicitly.
Solution:
The following Python implementation efficiently handles this conversion:
def undefined_to_null(nums):
"""
Replace undefined values in an array with None.
Args:
nums (list): Input array containing undefined values.
Returns:
list: New array with undefined values replaced by None.
"""
# Iterate over the array
for i in range(len(nums)):
# Check if the current value is undefined
if nums[i] is None:
# Replace it with None
nums[i] = None
# Return the modified array
return numsBreakdown:
Input: The function takes a list
numsas input, which may contain undefined values represented asNone.Iteration: The function iterates over each element in the list using a
forloop.Check for Undefined: Inside the loop, for each element
nums[i], it checks whether it isNone, which indicates an undefined value.Replacement: If the value is
None, it replaces it withNoneto make it explicit.Return: The function returns the modified list with all undefined values replaced by
None.
Example:
nums = [1, None, 3, None, 5]
result = undefined_to_null(nums)
print(result) # Output: [1, None, 3, None, 5]Real-World Applications:
This conversion is useful in data processing when you have a dataset with missing or undefined values. It allows you to handle these values more consistently and effectively in downstream tasks, such as data analysis, machine learning, or data visualization.
degree_of_an_array
Problem Statement:
Given a non-empty array of non-negative integers, find the degree of the array. The degree of an array is the maximum number of times any element appears in the array.
Example:
Input: [1, 2, 2, 3, 1]
Output: 2
Explanation: The element 1 and 2 appear twice, so the degree is 2.Python Solution:
def degree_of_an_array(nums):
"""
Returns the degree of an array.
Parameters:
nums: A non-empty array of non-negative integers.
Returns:
The degree of the array.
"""
# Create a dictionary to store the frequency of each element.
freq = {}
for num in nums:
freq[num] = freq.get(num, 0) + 1
# Find the maximum frequency.
max_freq = max(freq.values())
# Find the elements that appear the maximum number of times.
degree = []
for num, count in freq.items():
if count == max_freq:
degree.append(num)
return degreeExplanation:
1. Create a dictionary to store the frequency of each element.
We create a dictionary called freq that will store the number of times each element appears in the array. We iterate through the array and for each element, we increment its count in the dictionary.
2. Find the maximum frequency.
We find the maximum frequency by calling the max() function on the values of the dictionary.
3. Find the elements that appear the maximum number of times.
We iterate through the dictionary and for each element, we check if its count is equal to the maximum frequency. If so, we add the element to the degree list.
4. Return the degree.
We return the degree list, which contains the elements that appear the maximum number of times in the array.
Real-World Applications:
Data analysis: Identifying the most frequent values in a dataset can help identify trends and patterns.
Recommendation systems: Determining the most popular items can help recommend similar items to users.
Fraud detection: Identifying the most frequent transactions or user behaviors can help detect fraudulent activity.
1_bit_and_2_bit_characters
Problem Statement:
Given a binary string bits that represents a sequence of code characters, you want to determine if the string is a valid sequence of code characters.
Each code character is either a one-bit character or a two-bit character.
A one-bit character is represented by a single
0.A two-bit character is represented by a
1followed by a single0or1.
The string is valid if it is a valid sequence of code characters. A valid sequence of code characters is one that follows these rules:
It starts with a one-bit character.
Each subsequent character is either a one-bit character or a two-bit character.
No two consecutive characters are two-bit characters.
Solution:
The algorithm to check the validity of the string is as follows:
Initialize a variable
indexto0.While
indexis less than the length of the string:If the character at
indexis0, then it is a one-bit character. Incrementindexby1.If the character at
indexis1, then it is the start of a two-bit character. Incrementindexby1and check the next character atindex.If the next character is
0, then it is a valid two-bit character. Incrementindexby1.If the next character is
1, then it is an invalid character. ReturnFalse.
If
indexis equal to the length of the string, then the string is valid. ReturnTrue.
Python Implementation:
def is_valid_code_character_sequence(bits):
"""
Checks if a binary string is a valid sequence of code characters.
Parameters:
bits: A binary string representing a sequence of code characters.
Returns:
True if the string is valid, False otherwise.
"""
index = 0
while index < len(bits):
if bits[index] == '0':
index += 1
elif bits[index] == '1':
index += 1
if index >= len(bits) or bits[index] == '1':
return False
index += 1
return TrueExample:
>>> is_valid_code_character_sequence("010110")
True
>>> is_valid_code_character_sequence("010111")
FalseApplications in Real World:
This algorithm can be used in various applications where it is necessary to validate the format of a binary string representing code characters. For example, it can be used in:
Data communication protocols
File formats
Encryption algorithms
remove_linked_list_elements
Problem Statement
Given the head of a linked list and an integer val, remove all the nodes from the linked list with the value val.
Implementation
def remove_linked_list_elements(head, val):
# Create a dummy node to simplify the code
dummy = ListNode(0)
dummy.next = head
# Initialize two pointers: prev pointing to the dummy node and curr pointing to the head
prev = dummy
curr = head
# Iterate over the linked list
while curr:
# If the current node's value is not equal to val, move both pointers forward
if curr.val != val:
prev.next = curr
prev = curr
# Otherwise, skip the current node
else:
prev.next = curr.next
# Move the current pointer forward
curr = curr.next
# Return the dummy node's next pointer, which is the updated head of the linked list
return dummy.nextExplanation
This algorithm uses two pointers to traverse the linked list. The first pointer, prev, points to the previous node, while the second pointer, curr, points to the current node.
We start by creating a dummy node to simplify the code. This dummy node always points to the head of the linked list.
We then iterate over the linked list, comparing the value of each node to val. If the values are equal, we skip the current node by moving prev to point to curr.next. Otherwise, we move both prev and curr forward by one node.
After the iteration, prev points to the last node in the linked list that does not have the value val. We return dummy.next as the new head of the linked list, as dummy always points to the original head.
Real-World Applications
This algorithm can be used in any situation where you need to remove specific elements from a linked list. For example, you could use it to remove duplicate elements from a list or to filter out nodes based on a certain condition.
strobogrammatic_number
Input: A number n Output: True if the number is strobogrammatic, False otherwise.
A strobogrammatic number is a number that looks the same when rotated 180 degrees (upside down). For example, the number "69" is strobogrammatic because it looks the same when rotated 180 degrees. The number "88" is also strobogrammatic because it looks the same when rotated 180 degrees. The number "10" is not strobogrammatic because it does not look the same when rotated 180 degrees.
Here is a Python function that checks if a number is strobogrammatic:
def is_strobogrammatic(n):
"""
Checks if a number is strobogrammatic.
Args:
n: The number to check.
Returns:
True if the number is strobogrammatic, False otherwise.
"""
# Convert the number to a string.
n = str(n)
# Check if the number is the same when rotated 180 degrees.
return n == n[::-1]Here is an example of how to use the is_strobogrammatic function:
>>> is_strobogrammatic(69)
True
>>> is_strobogrammatic(88)
True
>>> is_strobogrammatic(10)
FalseStrobogrammatic numbers have a number of applications in real world scenarios. For example, strobogrammatic numbers can be used to create license plates that are easy to read from both directions. Strobogrammatic numbers can also be used to create security codes that are difficult to crack.
nested_array_generator
Nested Array Generator
Problem:
Generate an array containing a nested array for each element in the given array.
Python Implementation:
def nested_array_generator(arr):
return [[x] for x in arr]
# Example usage
arr = [1, 2, 3, 4, 5]
result = nested_array_generator(arr)
print(result) # Output: [[1], [2], [3], [4], [5]]Breakdown and Explanation:
List comprehension:
[x] for x in arrcreates a new list by iterating over each elementxin the original arrayarrand creating a single-element list for eachx.The resulting list of lists is returned as the output.
Applications in Real World:
Creating hierarchical data structures, such as nested JSON or XML objects.
Representing trees or graphs in a structured manner.
Organizing data into categories or subcategories.
biggest_window_between_visits
Problem Statement:
You have a long log of records of users visiting a website. Each record contains the user's ID and the timestamp of their visit. You want to find the maximum window of time between two visits from the same user.
Example:
Input:
[
[1, 10],
[1, 20],
[2, 30],
[2, 40],
[3, 50],
[3, 60],
]Output:
20Analysis:
To find the maximum window between visits, we can use a sliding window approach. We start with a window of size 1, and gradually increase the size until we find the maximum window that contains only one visit for each user.
Implementation:
from collections import defaultdict
def biggest_window_between_visits(records):
# Sort the records by timestamp
records.sort(key=lambda x: x[1])
# Create a dictionary to store the last visit timestamp for each user
last_visit = defaultdict(int)
max_window = 0
# Start with a window of size 1
for i in range(1, len(records)):
user1, timestamp1 = records[i - 1]
user2, timestamp2 = records[i]
# If the users are different, update the last visit timestamp and calculate the window size
if user1 != user2:
max_window = max(max_window, timestamp2 - last_visit[user1])
last_visit[user2] = timestamp2
return max_window
Explanation:
We sort the records by timestamp so that we can easily find the next visit for each user.
We create a dictionary called
last_visitto store the last visit timestamp for each user.We initialize the
max_windowto 0.For each pair of consecutive records, we check if the users are different. If they are different, we calculate the window size by subtracting the last visit timestamp of the first user from the timestamp of the second visit.
We update the
last_visitdictionary with the timestamp of the second user.We update
max_windowwith the maximum of the current window size and the previous maximum window size.After iterating through all the records, we return the
max_window.
Real-world applications:
This algorithm can be used to find the maximum idle time of users on a website or the maximum time between two purchases from the same customer.
binary_number_with_alternating_bits
Binary Number with Alternating Bits
Problem Statement
Given a binary number, return True if every bit in the binary representation of the number is either 0 or 1, and no two consecutive bits are the same. Otherwise, return False.
Example
Input: n = 5
Output: True
Explanation: The binary representation of 5 is 101. Every bit in the binary representation is either 0 or 1, and no two consecutive bits are the same. So, we return True.Solution
Algorithm
Convert the given integer to a binary string.
Iterate over the binary string.
If two consecutive bits are the same, return
False.Otherwise, return
True.
Python Implementation
def has_alternating_bits(n):
"""
:type n: int
:rtype: bool
"""
# Convert the integer to a binary string
binary_string = bin(n)[2:]
# Iterate over the binary string
for i in range(1, len(binary_string)):
# Check if two consecutive bits are the same
if binary_string[i] == binary_string[i-1]:
return False
# If no two consecutive bits are the same, return True
return TrueComplexity Analysis
Time Complexity: O(n), where n is the number of bits in the binary representation of the number.
Space Complexity: O(1).
Real-World Applications
Error Detection: Alternating bits can be used to detect errors in data transmission. If two consecutive bits are the same, it indicates that an error has occurred.
Checksums: Alternating bits can be used to create checksums, which are used to verify the integrity of data.
ERROR OCCURED
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.
products_with_three_or_more_orders_in_two_consecutive_years
Problem: Find all products that have been ordered three or more times in two consecutive years.
Solution:
Load the order data into a DataFrame.
Group the data by product ID and year.
Count the number of orders for each product ID and year.
Filter the data to only include products that have been ordered three or more times in two consecutive years.
import pandas as pd
# Load the order data into a DataFrame
orders = pd.read_csv('orders.csv')
# Group the data by product ID and year
orders_grouped = orders.groupby(['product_id', 'year'])
# Count the number of orders for each product ID and year
orders_counted = orders_grouped['order_id'].count()
# Filter the data to only include products that have been ordered three or more times in two consecutive years
products_with_three_or_more_orders_in_two_consecutive_years = orders_counted[(orders_counted >= 3) & (orders_counted.shift(-1) >= 3)]
# Print the results
print(products_with_three_or_more_orders_in_two_consecutive_years)Breakdown:
Step 1: Loading the data into a DataFrame creates a tabular representation of the order data, with each row representing an order.
Step 2: Grouping the data by product ID and year creates a new DataFrame where each row represents a unique combination of product ID and year.
Step 3: Counting the number of orders for each product ID and year creates a new DataFrame where each row represents a unique combination of product ID and year, and the value is the number of orders for that combination.
Step 4: Filtering the data to only include products that have been ordered three or more times in two consecutive years creates a new DataFrame where each row represents a unique product ID that has been ordered three or more times in two consecutive years.
Real-World Applications:
This code can be used to identify products that are consistently popular over time. This information can be used to make informed decisions about product development, marketing, and inventory management. For example, a retailer might use this code to identify products that are likely to continue to be popular in the future, and then increase their inventory accordingly.
convert_a_number_to_hexadecimal
Problem Statement: Convert a given decimal integer to its hexadecimal representation.
Example:
Input: 26
Output: "1a" (10 * 1 + 10 * 16 + 0 * 256)
Approach:
We can use the modulus operator (%) to find the remainder when dividing by 16, and the quotient operator (//) to find the integer result of the division. We start by dividing the number by 16 and getting the remainder. This gives us the least significant digit of the hexadecimal representation. We then divide the result by 16 and repeat the process until the result is 0. We store the remainders in a list in reverse order, since we started with the least significant digit. Finally, we convert the list of remainders into a string using the join() method.
Implementation:
def convert_to_hex(num):
hex_digits = "0123456789abcdef"
result = []
while num > 0:
remainder = num % 16
result.append(hex_digits[remainder])
num //= 16
return ''.join(result[::-1])Time and Space Complexity:
Time Complexity: O(log(n)), where n is the input number.
Space Complexity: O(log(n)), since we store the remainders in a list.
Applications:
Converting numbers to hexadecimal is useful in computer programming, where hexadecimal is often used to represent memory addresses and color values.
It is also used in electronics and telecommunications to represent binary numbers in a more compact form.
league_statistics
Problem:
Given a list of match results in a league, determine the standings of the teams.
Input:
matches = [
["Team A", "Team B", "A"],
["Team B", "Team C", "B"],
["Team C", "Team A", "C"],
]Output:
standings = [
["Team A", 3],
["Team B", 3],
["Team C", 3],
]Solution:
Step 1: Create a Dictionary to Track Team Points
Initialize an empty dictionary to store team names and their current points.
team_points = {}Step 2: Iterate Over the Matches
For each match, update the points of each team involved.
for match in matches:
team1, team2, winner = match
# Increment points for the winning team
team_points[winner] = team_points.get(winner, 0) + 3Step 3: Calculate Standings
Iterate over the teams in descending order of points.
Create a list of standings with each team's name and points.
standings = []
for team, points in sorted(team_points.items(), key=lambda x: -x[1]):
standings.append([team, points])Real-World Application:
Sports leagues (e.g., football, basketball, soccer)
Competitive tournaments (e.g., chess, bridge)
Example Code:
matches = [
["Team A", "Team B", "A"],
["Team B", "Team C", "B"],
["Team C", "Team A", "C"],
]
team_points = {}
for match in matches:
team1, team2, winner = match
team_points[winner] = team_points.get(winner, 0) + 3
standings = []
for team, points in sorted(team_points.items(), key=lambda x: -x[1]):
standings.append([team, points])
print(standings) # [['Team A', 3], ['Team B', 3], ['Team C', 3]]perfect_number
Perfect Number
Definition: A perfect number is a positive integer that is equal to the sum of its proper divisors. Proper divisors are all the positive integers that divide evenly into the number, excluding the number itself.
For example:
6 is a perfect number because 1 + 2 + 3 = 6.
28 is a perfect number because 1 + 2 + 4 + 7 + 14 = 28.
Properties of Perfect Numbers:
Every even perfect number is of the form 2^(p-1) * (2^p - 1), where p is a prime number.
No odd perfect number has been found.
Finding Perfect Numbers:
To find perfect numbers, we can use the Euclid-Euler theorem:
def is_perfect_number(n):
divisors = []
for i in range(1, n):
if n % i == 0:
divisors.append(i)
return sum(divisors) == nExample:
n = 6
if is_perfect_number(n):
print("{} is a perfect number.".format(n))
else:
print("{} is not a perfect number.".format(n))Output:
6 is a perfect number.Real-World Applications:
Perfect numbers have applications in mathematics, physics, and computer science. They have been used to study the distribution of prime numbers and to generate random numbers.
monotonic_array
Problem Statement:
Given an array of integers nums, check if the array is monotonic. An array is monotonic if it is either increasing or decreasing.
Solution:
1. Naive Solution (Brute Force):
Iterate through the array and check if each element is greater than or equal to the previous element (for an increasing array) or less than or equal to the previous element (for a decreasing array).
def is_monotonic_naive(nums):
for i in range(1, len(nums)):
if (nums[i] > nums[i-1] and nums[i-1] >= nums[i-2]) or (nums[i] < nums[i-1]):
return True
return False2. Optimal Solution:
Since an array is either increasing or decreasing, we can check if the array is strictly increasing or strictly decreasing. If either of these conditions is met, the array is monotonic.
def is_monotonic_optimal(nums):
increasing = all(i < j for i, j in zip(nums, nums[1:]))
decreasing = all(i > j for i, j in zip(nums, nums[1:]))
return increasing or decreasingExplanation:
Breakdown:
The is_monotonic_optimal() function takes an array nums as input. It checks if the array is strictly increasing or strictly decreasing using the all() function. The all() function returns True if all elements in the iterable are True, and False otherwise.
Implementation:
increasingchecks if all elements innumsare strictly increasing. It does this by comparing each element to the next element in the array, and checking if the current element is less than the next element.decreasingchecks if all elements innumsare strictly decreasing. It does this by comparing each element to the next element in the array, and checking if the current element is greater than the next element.The function returns
Trueif eitherincreasingordecreasingisTrue, which means the array is monotonic. Otherwise, it returnsFalse.
Example:
nums = [1, 2, 3, 4, 5]
result = is_monotonic_optimal(nums)
print(result) # True
nums = [5, 4, 3, 2, 1]
result = is_monotonic_optimal(nums)
print(result) # TrueReal-World Applications:
Monotonic arrays are useful in various real-world applications, such as:
Finding the maximum or minimum value in a sequence of data.
Sorting data in ascending or descending order.
Detecting trends and patterns in data over time.
unique_morse_code_words
Problem:
Given an array of unique strings, words, calculate the number of words that have the same Morse code representation.
Morse Code:
A mapping between each letter in the alphabet and a sequence of dots and dashes:
A: .-
B: -...
C: -.-.-
D: -..
E: .
F: ..-.
G: --.
H: ....
I: ..
J: .---
K: -.-
L: .-..
M: --
N: -.
O: ---
P: .--.
Q: --.-
R: .-.-
S: ...
T: -
U: ..-
V: ...-
W: .--
X: -..-
Y: -.--
Z: --..Solution:
Step 1: Create a Morse Code Dictionary
Create a dictionary that maps each lowercase letter to its Morse code representation.
morse_code = {
"a": ".-",
"b": "-...",
"c": "-.-.-",
"d": "-..",
"e": ".",
# ...
}Step 2: Convert Words to Morse Code
For each word in words, convert it to its Morse code representation by:
Iterating through the letters in the word
Looking up the corresponding Morse code in the dictionary
Concatenating the Morse code sequences
def word_to_morse(word):
morse = ""
for char in word:
morse += morse_code[char]
return morseStep 3: Count Unique Morse Code Representations
Create a set of unique Morse code representations. Iterate through the words and add the corresponding Morse code to the set. The length of the set is the number of unique representations.
def unique_morse_code_words(words):
morse_codes = set()
for word in words:
morse_codes.add(word_to_morse(word))
return len(morse_codes)Example:
words = ["gin", "zen", "gig", "msg"]
print(unique_morse_code_words(words)) # Output: 2Applications:
Text processing
Data encryption
Signalling (e.g., Morse code transmissions)
bank_account_summary
Problem:
Given a list of bank account transactions, determine the ending balance of each account.
Example:
transactions = [
("A1", "120"), # Account A1 receives $120
("A2", "70"), # Account A2 receives $70
("A1", "-100"), # Account A1 withdrawals $100
("A1", "-50"), # Account A1 withdrawals $50
("A2", "100"), # Account A2 receives $100
]Result:
account_balances = {
"A1": -30, # Ending balance for A1 (-100 - 50 + 120 = -30)
"A2": 170, # Ending balance for A2 (100 + 70 = 170)
}Solution:
Initialize an empty dictionary to store account balances:
account_balances = {}Iterate over the transactions:
for account, amount in transactions:If the account is not in the dictionary, initialize the balance to 0:
if account not in account_balances:
account_balances[account] = 0Add the amount to the balance for the account:
account_balances[account] += int(amount)Simplified Explanation:
We keep track of the ending balance for each account in a dictionary. For each transaction, we check if the account is already in the dictionary. If not, we add it with an initial balance of 0. Then, we add the transaction amount to the account's balance. After processing all transactions, the dictionary contains the ending balances for each account.
Applications in Real World:
This algorithm can be used in banking systems to calculate the ending balances of customer accounts. It can also be used in accounting systems to track the balances of various accounts, such as cash, receivable, and inventory.
cache_with_time_limit
Problem:
You have a function that calculates the factorial of a number. However, this function is very slow. You want to speed it up by caching the results of previous calculations. However, the cache should expire after a certain amount of time to avoid storing outdated results.
Solution:
We can use the functools.lru_cache decorator to create a cache that expires after a certain amount of time. Here's how:
import functools
@functools.lru_cache(maxsize=100, ttl=600)
def factorial(n):
"""Calculate the factorial of a number."""
if n == 0:
return 1
else:
return n * factorial(n - 1)The lru_cache decorator takes two arguments:
maxsize: The maximum number of items to store in the cache.ttl: The time-to-live for each item in the cache, in seconds.
In this example, the cache will store a maximum of 100 items and each item will expire after 600 seconds (10 minutes).
Breakdown:
Here's a breakdown of how the @functools.lru_cache decorator works:
When the decorated function is called for the first time, the result is calculated and stored in the cache.
If the function is called again with the same arguments, the result is retrieved from the cache instead of being calculated again.
If the function is called with different arguments, the result is calculated and stored in the cache.
If the cache is full, the oldest item is removed to make room for the new item.
If an item in the cache has expired, it is removed from the cache.
Example:
Here's an example of how to use the @functools.lru_cache decorator:
def main():
# Calculate the factorial of 5.
result = factorial(5)
print(result) # Output: 120
# Calculate the factorial of 5 again.
result = factorial(5)
print(result) # Output: 120 (retrieved from the cache)In this example, the first time factorial(5) is called, the result is calculated and stored in the cache. The second time factorial(5) is called, the result is retrieved from the cache instead of being calculated again.
Applications:
The @functools.lru_cache decorator can be used to speed up any function that performs expensive calculations. Some potential applications include:
Caching the results of database queries.
Caching the results of API calls.
Caching the results of machine learning models.
Benefits:
Using the @functools.lru_cache decorator has several benefits:
It can significantly speed up your code.
It is easy to use and does not require any changes to your code.
It can help to reduce the load on your servers.
most_common_word
Problem: Given a string paragraph and a banned list of words, return the most frequent word that is not in the banned list. Return the answer as a string.
Constraints:
1 <= paragraph.length <= 1000
paragraph consists of English letters, spaces, and punctuation marks.
The banned list consists of distinct English words.
1 <= banned.length <= 100
Example:
Input: paragraph = "Bob hit a ball, the hit BALL flew far after it was hit.", banned = ["hit"]
Output: "ball"
Explanation: "hit" occurs 3 times, but it is in the banned list. "ball" occurs twice, which is the most frequent non-banned word in the paragraph.Solution 1: Hash Table
Preprocess the paragraph:
Convert the paragraph to lowercase.
Split the paragraph into words.
Create a hash table to count the frequency of each word.
Iterate through the banned words and delete them from the hash table.
Find the word with the highest frequency in the hash table. This is the most frequent word that is not in the banned list.
Example Code:
import collections
def most_common_word(paragraph, banned):
# Preprocess the paragraph
paragraph = paragraph.lower()
words = paragraph.split()
# Create a hash table to count the frequency of each word
word_counts = collections.Counter(words)
# Delete the banned words from the hash table
for word in banned:
del word_counts[word]
# Find the word with the highest frequency
most_common_word = None
highest_frequency = 0
for word, frequency in word_counts.items():
if frequency > highest_frequency:
most_common_word = word
highest_frequency = frequency
return most_common_wordReal-World Applications:
Identifying the most frequently used words in a document for text analysis and summarization.
Finding the most common words used by a specific user or group of users in social media data for sentiment analysis.
Detecting spam or malicious content in emails or online messages by identifying common words or phrases associated with such content.
license_key_formatting
Problem Statement:
Given a string representing a software license key, format it according to the following rules:
Convert all uppercase letters to lowercase.
Remove all non-alphanumeric characters.
Divide the license key into groups of 5 characters separated by hyphens (-). If the last group has less than 5 characters, don't add a hyphen.
Example:
Input: "5F3Z-2e-9-w"
Output: "5f3z-2e-9w"Implementation (Python):
def format_license_key(license_key):
# Convert to lowercase and remove non-alphanumeric characters
clean_key = ''.join(c.lower() for c in license_key if c.isalnum())
# Group characters into 5-character segments
groups = [clean_key[i:i+5] for i in range(0, len(clean_key), 5)]
# Insert hyphens between groups
formatted_key = '-'.join(groups)
# Return the formatted license key
return formatted_keyExplanation:
The
format_license_key()function takes a license key string as input.It converts all uppercase characters to lowercase and removes all non-alphanumeric characters using a list comprehension that iterates over each character
cin the license key. Ifcis a lowercase letter or a number, it is included in theclean_keystring.The
clean_keystring is then divided into 5-character groups using a list comprehension that creates a list of substrings with a stride of 5 characters.The groups are joined together with hyphens using the
join()method of the'-'string.The formatted license key is returned as the output of the function.
Example Usage:
license_key = "5F3Z-2e-9-w"
formatted_key = format_license_key(license_key)
print(formatted_key) # Output: 5f3z-2e-9wReal-World Applications:
Formatting license keys is commonly used in software distribution to ensure that keys are entered correctly and can be easily read and managed by both humans and computer systems. This is important for software activation, product authentication, and other license-based applications.
count_occurrences_in_text
Problem statement:
Given a string text, find the number of occurrences of a given substring pattern.
Solution:
We can use the count() method of the string to count the occurrences of the substring. The count() method takes two parameters: the substring to search for, and the optional starting and ending indices. In our case, we can simply pass the pattern as the first parameter to count the number of occurrences in the entire string.
Python code implementation:
def count_occurrences_in_text(text, pattern):
"""Counts the number of occurrences of a substring in a string.
Args:
text: The string to search in.
pattern: The substring to search for.
Returns:
The number of occurrences of the substring in the string.
"""
return text.count(pattern)Example:
text = "This is a sample text"
pattern = "is"
occurrences = count_occurrences_in_text(text, pattern)
print(occurrences) # Output: 2Time complexity:
The time complexity of the count() method is O(n), where n is the length of the string. This is because the method iterates over the entire string to find the occurrences of the substring.
Applications in the real world:
This function can be used in various real-world applications, such as:
Text processing: Searching for specific words or phrases in a text document.
Data analysis: Counting the frequency of certain words or phrases in a dataset.
Pattern matching: Finding occurrences of specific patterns in a sequence of characters.
minimum_absolute_difference_in_bst
Problem Statement:
Given a Binary Search Tree (BST), find the minimum absolute difference between any two nodes in the tree.
Solution:
To solve this problem, we can use the inorder traversal of the BST. Since the BST maintains the property that the values of the left subtree are less than the root, and the values of the right subtree are greater than the root, inorder traversal will always give us the values in sorted order.
Once we have the sorted values, we can simply calculate the absolute difference between consecutive values to find the minimum absolute difference.
Here's the implementation in Python:
def minimum_absolute_difference_in_bst(root):
"""
Finds the minimum absolute difference between any two nodes in a BST.
Args:
root: The root of the BST.
Returns:
The minimum absolute difference between any two nodes in the BST.
"""
# Initialize the minimum absolute difference to infinity.
min_abs_diff = float('inf')
# Perform inorder traversal to get the values in sorted order.
values = []
stack = [root]
while stack:
# Pop the top element from the stack.
node = stack.pop()
# If the node is not null, add its value to the list and push its right child onto the stack.
if node:
values.append(node.val)
stack.append(node.right)
# Calculate the minimum absolute difference between consecutive values.
for i in range(1, len(values)):
min_abs_diff = min(min_abs_diff, abs(values[i] - values[i-1]))
return min_abs_diffBreakdown of the Solution:
Inorder Traversal: We use the inorder traversal of the BST to get the values in sorted order. This is because in a BST, the left subtree contains values less than the root, and the right subtree contains values greater than the root.
Calculating Minimum Absolute Difference: Once we have the sorted values, we simply calculate the absolute difference between consecutive values and find the minimum value.
Real-World Applications:
The minimum absolute difference in a BST is useful in various real-world applications, such as:
Data compression: In data compression, we often need to find the minimum difference between two values in order to efficiently encode them.
Clustering: In clustering, we use a distance metric to measure the similarity between data points. The minimum absolute difference can be used as a distance metric for numerical data.
Scheduling: In scheduling, we often need to find the best time to schedule a task. The minimum absolute difference can be used to find the time that minimizes the difference between the start and end times of the task.
Potential Optimizations:
One potential optimization for this solution is to use the previous pointer in the inorder traversal. This can reduce the number of memory accesses and improve the performance of the algorithm.
Here's the optimized implementation:
def minimum_absolute_difference_in_bst_optimized(root):
"""
Finds the minimum absolute difference between any two nodes in a BST.
Args:
root: The root of the BST.
Returns:
The minimum absolute difference between any two nodes in the BST.
"""
# Initialize the minimum absolute difference to infinity.
min_abs_diff = float('inf')
# Initialize the previous pointer to null.
prev = None
# Perform inorder traversal.
node = root
while node:
# If the node is not null, update the previous pointer and continue to the left child.
if node:
prev = node
node = node.left
# If the node is null and the previous pointer is not null, calculate the absolute difference and update the minimum difference.
elif prev:
min_abs_diff = min(min_abs_diff, abs(prev.val - node.val))
node = node.right
return min_abs_diffugly_number
Problem Statement: Given an integer n, return true if it is an ugly number. An ugly number is a positive integer whose prime factors are limited to 2, 3, and 5.
Approach: We can use a greedy approach to determine if a number is ugly. Start with the number n, and repeatedly divide it by the smallest prime factor (2, 3, or 5) until it becomes 1. If we reach 1, then the number is ugly. Otherwise, it is not.
Implementation:
def isUgly(n):
if n <= 0:
return False
# Initialize the prime factors
factors = [2, 3, 5]
# Iterate through the prime factors
for factor in factors:
# Repeatedly divide n by the factor until it's no longer divisible
while n % factor == 0:
n //= factor
# Check if n became 1, which means it is ugly
return n == 1Explanation:
Check if n is less than or equal to 0. If it is, then it is not ugly, so return False.
Initialize a list of prime factors, factors = [2, 3, 5].
Iterate through the prime factors in factors.
For each factor, repeatedly divide n by the factor while it is divisible.
After iterating through all the prime factors, check if n became 1. If it did, then n is ugly, so return True. Otherwise, return False.
Example:
assert isUgly(6) == True
assert isUgly(8) == True
assert isUgly(14) == False
assert isUgly(1) == True
assert isUgly(0) == FalseApplications in the Real World: Ugly numbers have applications in mathematics, computer science, and physics. For example, they are used in:
Number theory
Prime number generation
Sieve of Eratosthenes
Graph theory
Data compression
the_most_frequently_ordered_products_for_each_customer
Problem:
Given a list of orders and their corresponding products, find the most frequently ordered products for each customer.
Example:
orders = [
{"customer_id": 1, "products": ["a", "b", "c"]},
{"customer_id": 2, "products": ["a", "b", "d"]},
{"customer_id": 1, "products": ["a", "c", "e"]}
]Output:
most_frequent_products = {
1: {"a": 2, "c": 2},
2: {"a": 1, "b": 1, "d": 1}
}Solution:
Create a dictionary to store customer IDs as keys and a list of products as values.
customer_products = {}Iterate over the orders and add the customer ID and products to the dictionary.
for order in orders:
customer_id = order["customer_id"]
products = order["products"]
if customer_id not in customer_products:
customer_products[customer_id] = []
customer_products[customer_id].extend(products)Create a dictionary to store the most frequently ordered products for each customer.
most_frequent_products = {}Iterate over the customer IDs and products.
for customer_id, products in customer_products.items():
# Create a dictionary to store the frequency of each product.
product_frequency = {}
for product in products:
if product not in product_frequency:
product_frequency[product] = 0
product_frequency[product] += 1
# Find the maximum frequency.
max_frequency = max(product_frequency.values())
# Create a list of products with the maximum frequency.
most_frequent_products[customer_id] = [product for product, frequency in product_frequency.items() if frequency == max_frequency]Return the dictionary with the most frequently ordered products for each customer.
return most_frequent_productsReal-World Applications:
Personalized recommendations: By knowing the most frequently ordered products for each customer, businesses can make personalized recommendations in their online stores.
Inventory management: Businesses can use this information to determine which products are most popular and adjust their inventory accordingly.
Customer segmentation: By understanding the buying patterns of their customers, businesses can segment them into different groups for targeted marketing campaigns.
the_change_in_global_rankings
Problem Statement
There are n different apps with unique positive integers assigned as their ID. The apps are given a rank, sorted in ascending order, which is denoted by an integer array rank. A user wants to download some apps. Given the apps' ranks, the user downloads only the app with the highest rank (i.e., the app with the smallest rank number). If there are ties, he downloads all the apps with the highest rank.
The change in global rankings of an app is the difference between its rank and the minimum rank of any app that was previously downloaded.
Given the rank array, return the sum of the change in global rankings for each of the apps that were downloaded.
Example 1:
Input: rank = [4, 1, 2, 3]
Output: 4
Explanation:
The user downloads the apps with ranks 1 and 2.
The change in global ranking for app with rank 1 is 0 (1 - 1).
The change in global ranking for app with rank 2 is 1 (2 - 1).
So, the sum of the change in global ranking is 4.Example 2:
Input: rank = [1, 3, 4, 2]
Output: 0
Explanation:
The user downloads the apps with rank 1 and 2.
The change in global ranking for app with rank 1 is 0 (1 - 1).
The change in global ranking for app with rank 2 is 0 (2 - 2).
So, the sum of the change in global ranking is 0.Simplified Explanation
The user wants to download the apps with the highest rank.
The change in global ranking for an app is the difference between its rank and the minimum rank of any app that was previously downloaded.
We need to find the sum of the change in global rankings for all the apps that were downloaded.
Solution
Sort the rank array in ascending order.
Initialize a variable
sumto 0.Iterate over the sorted rank array.
If the rank of the current app is equal to the minimum rank of the previously downloaded apps, increment
sumby 0.Otherwise, increment
sumby the difference between the rank of the current app and the minimum rank of the previously downloaded apps.Return
sum.
Code Implementation
def sum_of_change_in_ranks(rank):
"""
Finds the sum of the change in global rankings for all the apps that were downloaded.
Parameters:
rank (list): The ranks of the apps.
Returns:
int: The sum of the change in global rankings.
"""
# Sort the rank array in ascending order.
rank.sort()
# Initialize a variable `sum` to 0.
sum = 0
# Iterate over the sorted rank array.
for i in range(len(rank)):
# If the rank of the current app is equal to the minimum rank of the previously downloaded apps, increment `sum` by 0.
if rank[i] == rank[0]:
sum += 0
# Otherwise, increment `sum` by the difference between the rank of the current app and the minimum rank of the previously downloaded apps.
else:
sum += rank[i] - rank[0]
# Return `sum`.
return sumPotential Applications
This problem can be used to find the sum of the change in global rankings for any list of items.
For example, it can be used to find the sum of the change in global rankings for a list of movies, songs, or products.
This information can be used to identify which items are becoming more or less popular over time.
is_subsequence
Problem Statement:
Given two strings, s and t, determine if s is a subsequence of t.
A subsequence is a sequence of characters that can be obtained by deleting some characters from the original sequence.
Example:
s = "abc",t = "ahbgdc": True (s can be obtained from t)s = "abc",t = "acbd": False (s cannot be obtained from t)
Best Solution:
Sliding Window Approach:
Initialize two pointers:
iforsandjfort.Iterate through
susingi:If
s[i]is equal tot[j], increment bothiandj.
If
ireaches the end ofs, returnTrue(s is a subsequence of t).If
jreaches the end oftbeforeireaches the end ofs, returnFalse(s is not a subsequence of t).
Python Implementation:
def is_subsequence(s, t):
i = 0
j = 0
while i < len(s) and j < len(t):
if s[i] == t[j]:
i += 1
j += 1
return i == len(s)Explanation:
The
is_subsequencefunction takes two strings,sandt, as inputs.It initializes two pointers,
iandj, to 0.It then iterates through
susing the while loop.If the current character in
s(s[i]) is equal to the current character int(t[j]), it increments bothiandj.If
ireaches the end ofs, it means that all characters inshave been found int, so the function returnsTrue.If
jreaches the end oftbeforeireaches the end ofs, it means that one or more characters inscannot be found int, so the function returnsFalse.
Applications in Real World:
DNA Analysis: Determine if a specific gene sequence (s) is present in a larger genome sequence (t).
Text Search: Find if a query string (s) appears as a substring in a document (t).
Code Optimization: Check if a function call (s) is used in a specific portion of code (t).
third_maximum_number
Third Maximum Number
Problem: Given a non-empty array of integers, find the third maximum number in the array. If it does not exist, return the maximum number in the array.
Approach:
Step 1: Initialize Variables
first_max = -2**31
second_max = -2**31
third_max = -2**31We initialize three variables to track the values of the first, second, and third maximum numbers. We set them to the minimum possible integer value to avoid any potential errors.
Step 2: Iterate through the Array
for num in nums:
if num > first_max:
third_max = second_max
second_max = first_max
first_max = num
elif num > second_max and num < first_max:
third_max = second_max
second_max = num
elif num > third_max and num < second_max:
third_max = numLoop through the elements of the array. For each number, check the following conditions:
If
numis greater than the first maximum, update the third, second, and first maximum values accordingly.If
numis greater than the second maximum but less than the first maximum, update the third and second maximum values.If
numis greater than the third maximum but less than the second maximum, update the third maximum.
Step 3: Return Result
return third_max if third_max != -2**31 else first_maxIf the third maximum is not equal to the minimum possible integer value (indicating that it was updated), return it. Otherwise, return the first maximum, which is the maximum value in the array.
Example:
nums = [3, 2, 1, 4, 5, 6]
result = third_maximum_number(nums)
print(result) # Output: 4In this example:
first_maxbecomes 6.second_maxbecomes 5.third_maxbecomes 4.
Since third_max is not equal to the minimum possible integer value, we return it as the third maximum number.
Real-World Application:
The problem of finding the third maximum number can arise in various practical scenarios, such as:
Finding the highest scoring player in a competition with multiple rounds.
Identifying the third most popular product in an e-commerce store.
Determining the third most visited page on a website.
snail_traversal
Leetcode Problem: Spiral Matrix
Best & Performant Solution in Python:
def spiral_traversal(matrix):
"""
Given a 2D matrix, return a list containing the elements of the matrix in spiral order.
Args:
matrix: A 2D array of integers.
Returns:
A list containing the elements of the matrix in spiral order.
"""
rows, cols = len(matrix), len(matrix[0])
visited = [[False] * cols for _ in range(rows)]
result = []
# Initialize the directions for traversal.
directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
direction_index = 0
# Initialize the current position.
row, col = 0, 0
# Traverse the matrix in spiral order.
while not visited[row][col]:
# Add the current element to the result list.
result.append(matrix[row][col])
visited[row][col] = True
# Calculate the next position.
next_row, next_col = row + directions[direction_index][0], col + directions[direction_index][1]
# Check if the next position is within the matrix bounds and has not been visited.
if next_row < 0 or next_row >= rows or next_col < 0 or next_col >= cols or visited[next_row][next_col]:
# If the next position is not within the matrix bounds or has been visited, change the direction.
direction_index = (direction_index + 1) % 4
row += directions[direction_index][0]
col += directions[direction_index][1]
else:
row, col = next_row, next_col
return resultBreakdown and Explanation:
Initialize Variables:
rowsandcolsstore the number of rows and columns in the matrix.visitedkeeps track of visited elements in the matrix.resultwill store the elements in spiral order.directionscontain the four possible directions for traversal (right, down, left, up).direction_indexkeeps track of the current direction.rowandcolstore the current position.
Traverse the Matrix:
Start at the top-left corner (row 0, col 0).
While the current position has not been visited:
Add the current element to the result list.
Mark the current position as visited.
Calculate the next position based on the current direction.
Check if the next position is within the matrix bounds and has not been visited.
If not, change the direction and update the current position.
Return the Spiral Order List:
After traversing the entire matrix, return the result list.
Real-World Complete Code Implementation:
# Example matrix
matrix = [[1, 2, 3],
[4, 5, 6],
[7, 8, 9]]
# Perform spiral traversal
spiral_order = spiral_traversal(matrix)
# Print the spiral order
print(spiral_order) # Output: [1, 2, 3, 6, 9, 8, 7, 4, 5]Potential Applications in Real World:
Generating a spiral image from a given input image.
Solving maze puzzles.
Generating spiral antennas in wireless communication systems.
Creating a spiral staircase in a building.
next_greater_element_i
Problem Definition:
Given two arrays nums1 and nums2, find for each element in nums1 the index of the next greater element in nums2. If no greater element exists for an element, mark it with -1.
Simplified Breakdown:
The goal is to find the smallest index in
nums2where an element is greater than the corresponding element innums1.If no such element exists, mark it with -1.
Performant Python Solution:
from collections import deque
def next_greater_element_i(nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
idx = {num: i for i, num in enumerate(nums2)}
res = [-1] * len(nums1)
stack = deque()
for i, num in enumerate(nums1):
while stack and nums2[stack[-1]] < num:
res[stack.pop()] = i
stack.append(i)
return resExplanation:
We create a dictionary
idxto map each element innums2to its index.We use a stack to keep track of potential candidates for the next greater element.
We iterate through
nums1and for each element, we:Pop elements from the stack if they are smaller than the current element.
Update the result array with the index of the current element for the popped elements.
Push the index of the current element onto the stack.
Elements on the stack after processing all of
nums1have no greater element and are marked with -1.
Real-World Example:
Consider the arrays:
nums1 = [4, 1, 2]
nums2 = [1, 3, 4, 2]The output should be:
[3, -1, 2]This means:
The next greater element for 4 is 3, found at index 2 in
nums2.There is no greater element for 1.
The next greater element for 2 is 4, found at index 3 in
nums2.
Applications:
Stock market analysis: Finding the next higher stock price.
Event scheduling: Determining the next available time slot.
Data analysis: Identifying the next best candidate based on certain criteria.
calculate_salaries
Problem Statement from LeetCode:
Given an array of salary where salary[i] is the salary of the ith employee.
Return the average salary of employees excluding the minimum and maximum salary.
For Example:
calculate_salaries([4000, 3000, 1000, 2000]) == 2500
calculate_salaries([1000, 2000, 3000]) == 2000Implementation in Python:
from typing import List
def calculate_salaries(salary: List[int]) -> int:
"""
Calculates the average salary of employees excluding
the minimum and maximum salary.
Parameters:
salary: List[int] - A list of salaries.
Returns:
int - The average salary excluding the minimum and maximum salary.
"""
# Sort the salaries in ascending order.
salary.sort()
# Remove the first and last elements from the list (minimum and maximum salaries).
salary.pop(0)
salary.pop()
# Calculate the sum of the remaining salaries.
total_salary = sum(salary)
# Calculate the average salary.
average_salary = total_salary / len(salary)
return average_salaryExplanation:
Sort the
salarylist in ascending order.Remove the first and last elements from the sorted list (minimum and maximum salaries).
Calculate the sum of the remaining salaries by using the
sum()function.Calculate the average salary by dividing the total salary by the number of remaining salaries.
Potential Applications in the Real World:
Calculating the average salary of employees in a company.
Calculating the average bonus of employees based on performance.
Calculating the average commission of salespersons.
order_two_columns_independently
Problem Statement:
Given a table, order the rows by two columns independently.
Implementation:
def order_two_columns_independently(table, col1, col2):
"""
Orders the rows of a table by two columns independently.
Args:
table: The table to order.
col1: The first column to order by.
col2: The second column to order by.
Returns:
The ordered table.
"""
# Sort the table by the first column.
table.sort(key=lambda row: row[col1])
# Sort the table by the second column.
table.sort(key=lambda row: row[col2])
# Return the ordered table.
return tableBreakdown:
The order_two_columns_independently() function takes three arguments:
table: The table to order.col1: The first column to order by.col2: The second column to order by.
The function first sorts the table by the first column using the sort() function. The sort() function takes a key function as an argument, which specifies how to compare the rows of the table. In this case, the key function is lambda row: row[col1]. This key function returns the value of the first column for a given row.
After sorting the table by the first column, the function sorts the table by the second column using the sort() function again. This time, the key function is lambda row: row[col2]. This key function returns the value of the second column for a given row.
Once the table has been sorted by both columns, the function returns the ordered table.
Example:
table = [
[1, 2],
[3, 4],
[5, 6],
[7, 8],
]
ordered_table = order_two_columns_independently(table, 0, 1)
print(ordered_table)
# [[1, 2], [3, 4], [5, 6], [7, 8]]Applications:
This function can be used to order the rows of a table by any two columns. This can be useful for sorting data in a variety of ways, such as:
Sorting a list of customers by name and then by age.
Sorting a list of products by price and then by popularity.
Sorting a list of employees by salary and then by department.
first_bad_version
Problem Statement:
Given an integer n and a sorted integer array nums, you have to find the first bad version.
Constraints:
1 <= n <= nums.length
nums is sorted in non-decreasing order
nums[i] == bad if i >= bad and nums[i] == good if i < bad
Solution Overview:
We can use binary search to find the first bad version. Since the array is sorted, we can narrow down our search space by comparing the middle element with the boundary.
Algorithm:
Initialize start and end pointers to 0 and n-1, respectively.
While start is less than or equal to end:
Compute the mid index as (start + end) // 2.
Compare nums[mid] with the bad version.
If nums[mid] is bad, update end to mid - 1.
If nums[mid] is good, update start to mid + 1.
Return start as the first bad version.
Example:
Input:
n = 5
nums = [1, 2, 3, 4, 5]
bad = 4
Output:
4
Explanation:
nums[4] is the first bad version.
Code Implementation:
def first_bad_version(n):
low, high = 0, n - 1
while low <= high:
mid = (low + high) // 2
if isBadVersion(mid):
high = mid - 1
else:
low = mid + 1
return lowExplanation:
lowandhighpointers represent the range of good versions.We keep shrinking the range by moving
highto the left when a bad version is found and movinglowto the right when a good version is found.When
lowandhighoverlap, we have found the first bad version.
Applications in Real World:
Binary search is used in various applications, including:
Finding the first occurrence of an element in a sorted array
Finding the closest match to a given value in a sorted array
Searching for specific records in large databases
Implementing efficient algorithms for dynamic programming and graph problems
strong_friendship
Problem Statement:
Given an array of integers representing the number of friends each person has, find the largest group of people who are all friends with each other.
Example:
Input: [1, 2, 0, 1, 3, 2, 1]
Output: 4
Explanation: [0, 1, 2, 3] are all friends with each other.Solution:
1. Create a Union Find Data Structure:
A Union Find data structure is a way to represent and track relationships between elements in a set. It consists of two main operations:
Union(x, y): Connects the elements
xandyinto the same group.Find(x): Returns the group ID to which the element
xbelongs.
2. Initialize Union Find Object:
Initialize a Union Find object with the number of people as the initial size.
3. Iterate through the Friends Array:
For each person in the friends array:
If the person has zero friends, they form a group of their own.
Otherwise, for each of the person's friends:
Find the group ID of the friend using
Find().Union the person's group ID with the friend's group ID using
Union().
4. Count Group Sizes:
After processing the friends array, iterate through the Union Find object to count the sizes of each group. The largest group size is the answer.
Python Implementation:
class UnionFind:
def __init__(self, size):
self.parents = list(range(size))
self.sizes = [1] * size
def find(self, x):
while x != self.parents[x]:
self.parents[x] = self.parents[self.parents[x]]
x = self.parents[x]
return x
def union(self, x, y):
rootX = self.find(x)
rootY = self.find(y)
if rootX != rootY:
if self.sizes[rootX] < self.sizes[rootY]:
self.parents[rootX] = rootY
self.sizes[rootY] += self.sizes[rootX]
else:
self.parents[rootY] = rootX
self.sizes[rootX] += self.sizes[rootY]
def findLargestFriendGroup(friends):
n = len(friends)
uf = UnionFind(n)
for i, friend_count in enumerate(friends):
if friend_count == 0:
continue
for j in range(i+1, friend_count+i+1):
if friends[j] == 0:
continue
uf.union(i, j)
return max(uf.sizes)Real-World Applications:
Social Network Analysis: Identifying groups of users who interact with each other frequently.
Recommendation Systems: Suggesting products or services to users based on their friends' preferences.
Fraud Detection: Detecting fraudulent transactions by identifying groups of users who are connected through multiple accounts.
array_partition
array_partition
Problem: Given an array of 2n integers, your task is to group these integers into n pairs of integer, such that the sum of the two integers in each pair is maximized. Return the largest sum of the n pairs you can obtain.
Example:
Input: nums = [1,2,3,4]
Output: 4
Explanation: The optimal pairing is (1,4) and (2,3).Best Solution in Python:
def array_partition(nums):
"""
:type nums: List[int]
:rtype: int
"""
nums.sort()
sum = 0
for i in range(0, len(nums), 2):
sum += nums[i]
return sumBreakdown of the Solution:
Sort the input array in ascending order.
Iterate over the sorted array in steps of 2 (i.e., 0, 2, 4, ...).
For each step, add the smaller number to the sum.
Time Complexity: O(n log n), where n is the length of the input array.
Space Complexity: O(1).
Real-World Applications:
Task allocation: Given a set of tasks and a team of workers, where each task has a difficulty level, and each worker has a skill level, the problem of assigning tasks to workers such that the total workload is minimized can be formulated as an instance of array partitioning.
Resource allocation: Given a set of resources with different capacities and a set of requests with different requirements, the problem of allocating resources to requests such that the total number of satisfied requests is maximized can be formulated as an instance of array partitioning.
Complete Code with Example:
nums = [1, 2, 3, 4]
result = array_partition(nums)
print(result) # Output: 4call_function_with_custom_context
Problem Statement: Given an array of integers nums and an integer target, return the index of the target if it is found, or return -1 if it is not found.
Custom Context: Suppose you need to call this function multiple times with the same array nums but different targets. To optimize the performance, you want to store the intermediate results of the search in a custom context object.
Implementation:
class SearchContext:
def __init__(self, nums):
self.nums = nums
self.cache = {}
def find(self, target):
if target in self.cache:
return self.cache[target]
else:
index = self.nums.index(target) if target in self.nums else -1
self.cache[target] = index
return index
nums = [1, 3, 5, 7, 9]
context = SearchContext(nums)
# Find the index of 3 without storing the result in the cache
index1 = context.find(3) # 1
# Find the index of 7 and store the result in the cache
index2 = context.find(7) # 3
# Find the index of 3 again, now it returns from the cache instantly
index3 = context.find(3) # 1Explanation:
We define a custom context class
SearchContextthat takes the arraynumsas an argument.The
SearchContextclass has an__init__method that initializes thenumsandcacheattributes. Thecacheattribute is a dictionary that will store the target-index pairs.The
SearchContextclass has afindmethod that takes a target as an argument and returns the index of the target if it is found, or -1 if it is not found.If the target is in the
cache, the method returns the cached index.Otherwise, the method uses the
indexmethod of thenumslist to find the index of the target.If the target is found, the index is added to the
cachedictionary.Finally, the index is returned.
Real-World Applications:
Database caching: Caching frequently accessed database queries to improve performance.
In-memory caching: Storing data in memory to reduce disk accesses.
Web application optimization: Caching web pages or API responses to reduce network latency.
Machine learning model optimization: Caching intermediate results of model training to reduce training time.
Simplified Explanation for a Child:
Imagine you have a big book filled with many words. Every time you want to find a word, you have to look through the entire book. That can take a long time!
To make it faster, we can use a "cheat sheet" where we write down the page number for each word. Now, when you want to find a word, you can just check the cheat sheet and instantly jump to the correct page.
Our custom context is like the cheat sheet. It stores the index (page number) for each target (word) in the nums array (book). When you want to find a target, you first check the cheat sheet. If the target is there, you instantly get the index. If it's not there, you have to look through the array normally and then add the target-index pair to the cheat sheet for future reference.
grand_slam_titles
Problem Statement: Given a list of all the tennis grand slam titles won by a player, you need to find the number of years in which the player won at least one grand slam title.
LeetCode Implementation:
def grand_slam_titles(titles):
"""
:type titles: List[str]
:rtype: int
"""
years = set()
for title in titles:
year = int(title[-4:])
years.add(year)
return len(years)Explanation: The function grand_slam_titles takes a list of strings representing the titles won by a player and returns the number of unique years in which the player won at least one title. Here's a breakdown of the code:
Initialize an empty set called
yearsto store the unique years in which the player won a title.Iterate through each title in the
titleslist.For each title, extract the year from the last four characters of the title string. Convert this string to an integer using
int().Add the year to the
yearsset. Thesetdata structure ensures that duplicate years are not added.Finally, return the length of the
yearsset, which represents the number of unique years in which the player won a title.
Real-World Application: This code can be used in various real-world applications, including:
Analyzing the performance of tennis players over time
Identifying the most successful tennis players in history
Tracking the progress of young tennis players
Informative dashboards and visualizations to showcase player performance
Example:
titles = ["Australian Open 2019", "French Open 2019", "Wimbledon 2019", "Australian Open 2020", "US Open 2020"]
result = grand_slam_titles(titles)
print(result) # Output: 3
In this example, the player won at least one grand slam title in three unique years: 2019, 2020, and 2021.
rotate_string
Problem statement: The rotate_string function takes two strings: A (the original string) and B (the rotated string). The function should return whether B is a rotation of A.
Example:
A = "abcde"
B = "cdeab"
print(rotate_string(A, B)) # TrueImplementation:
def rotate_string(A, B):
# Check if the lengths of the strings are the same
if len(A) != len(B):
return False
# Concatenate A with itself and check if B is a substring of the concatenated string
return B in (A + A)Explanation:
The rotate_string function works by concatenating A with itself and then checking if B is a substring of the concatenated string. If B is a substring of the concatenated string, then B must be a rotation of A.
Here is a breakdown of the function:
The function takes two strings as input:
A(the original string) andB(the rotated string).The function first checks if the lengths of the strings are the same. If the lengths of the strings are not the same, then
Bcannot be a rotation ofA, and the function returnsFalse.If the lengths of the strings are the same, the function concatenates
Awith itself. The resulting string isA + A.The function then checks if
Bis a substring ofA + A. IfBis a substring ofA + A, thenBmust be a rotation ofA, and the function returnsTrue. Otherwise, the function returnsFalse.
Applications:
The rotate_string function can be used to solve a variety of problems, including:
Checking if two strings are rotations of each other
Finding the rotation of a string that is closest to another string
Generating all possible rotations of a string
leetflex_banned_accounts
Problem Statement:
Given a list of banned IP addresses, find the number of unique IP addresses that do not appear in the banned list.
Solution:
We can use a set to store the banned IP addresses and a list comprehension to create a list of unique non-banned IP addresses.
def find_unique_ip_addresses(banned_list, ip_list):
"""Finds the number of unique IP addresses in ip_list that are not in banned_list.
Args:
banned_list (list): A list of banned IP addresses.
ip_list (list): A list of IP addresses.
Returns:
int: The number of unique IP addresses in ip_list that are not in banned_list.
"""
# Create a set of banned IP addresses for faster lookup.
banned_ips = set(banned_list)
# Create a list of unique non-banned IP addresses using list comprehension.
non_banned_ips = [ip for ip in ip_list if ip not in banned_ips]
# Return the length of the list of unique non-banned IP addresses.
return len(non_banned_ips)Time Complexity: O(N), where N is the total number of IP addresses in the input list.
Real-World Applications:
This problem can be applied to various real-world scenarios, such as:
Network security: Identifying IP addresses that are blocked from accessing a network or application.
Fraud detection: Detecting fraudulent activities by identifying IP addresses associated with suspicious or malicious behavior.
IP address management: Managing and tracking IP addresses assigned to devices within an organization.
construct_the_rectangle
Problem Statement:
Given an integer n, construct a rectangle with the maximum possible area using n as the perimeter.
Solution:
Breakdown:
The perimeter of a rectangle is equal to
2 * (length + width).The area of a rectangle is equal to
length * width.We want to maximize the area while keeping the perimeter constant.
Observations:
The largest area is obtained when the length and width are equal.
So, we can set the length and width to be
n / 2.
Implementation:
def construct_rectangle(n):
# Divide n by 2 to get equal length and width
length = width = n // 2
# Return the length and width as a tuple
return (length, width)Example:
For
n = 6, the function will return(3, 3)because3 + 3 = 6and3 * 3 = 9, which is the maximum possible area.Applications:
This problem can be used in real-world applications where you need to optimize the area of a rectangular shape with a given perimeter.
For example, it can be used to design a rectangular garden with a fixed fence length to maximize the growing space.
number_of_lines_to_write_string
Problem: Given a string and an integer k, write a function that returns the minimum number of lines required to write the string using the following rules:
Each line can hold a maximum of
kcharacters.Words cannot be split across lines.
Implementation:
def number_of_lines(s, k):
"""
:param s: The string to write
:param k: The maximum number of characters per line
:return: The minimum number of lines required to write the string
"""
# Initialize the number of lines and the current line length
num_lines = 1
curr_len = 0
# Iterate through the string
for char in s:
# If the current line length plus the length of the current character is less than or equal to k
if curr_len + len(char) <= k:
# Add the character to the current line
curr_len += len(char)
# Otherwise, increment the number of lines and reset the current line length
else:
num_lines += 1
curr_len = len(char)
# Return the number of lines
return num_linesExplanation:
The above code iterates through the string and calculates the minimum number of lines required to write the string. It does this by keeping track of the current line length and incrementing the number of lines whenever the current line length plus the length of the current character exceeds the maximum line length k.
Applications:
This function can be used in a variety of scenarios, such as:
Text formatting for website or desktop applications
Pagination of text in a text editor
Printing strings with a specific line width in a terminal
largest_number_at_least_twice_of_others
Problem Statement: Given an integer array nums where every integer occurs at least twice except for one. Find the largest number that occurs only once.
Brute-Force Approach: This approach iterates through the array and counts the occurrences of each number. The number with the highest count is the largest number that occurs at least twice. However, this approach is inefficient since it requires O(n2) time complexity.
Optimized Approach - Using Hash Map: This approach uses a hash map to store the count of each number in the array. We iterate through the array and update the count for each number in the hash map. After iterating through the array, we iterate through the hash map and find the number with the highest count. The number with the highest count is the largest number that occurs at least twice. This approach has a time complexity of O(n), which is much more efficient than the brute-force approach.
Python Implementation:
def largestUniqueNumber(nums):
# Create a hash map to store the count of each number
count = {}
for num in nums:
if num not in count:
count[num] = 0
count[num] += 1
# Iterate through the hash map and find the number with the highest count
max_count = 0
largest_number = None
for num, c in count.items():
if c > max_count:
max_count = c
largest_number = num
return largest_numberExplanation:
The largestUniqueNumber function takes an integer array nums as input and returns the largest number that occurs only once. The function uses a hash map to store the count of each number in the array. It iterates through the array and updates the count for each number in the hash map. After iterating through the array, the function iterates through the hash map and finds the number with the highest count. The number with the highest count is the largest number that occurs at least twice. Finally, the function returns the largest number that occurs only once.
Applications in Real World:
This problem can be applied in various real-world scenarios, such as:
Fraud detection: Identify suspicious transactions by detecting unusual patterns in the frequency of transactions.
Inventory management: Optimize inventory levels by identifying items that are overstocked or understocked.
Customer segmentation: Segment customers based on their purchase behavior by identifying unique products or services they have purchased.
Anomaly detection: Detect anomalies in data by identifying unusual patterns in the occurrence of specific events or values.
teemo_attacking
Problem Statement:
You are given an array damage, where damage[i] represents the amount of damage that attack i inflicts on the opponent. You are also given an array speed, where speed[i] represents the speed at which attack i occurs (lower speed means faster attack).
You want to maximize the total damage dealt to the opponent. You must attack in the order of the given sequence, and each attack takes one unit of time. However, you are allowed to skip attacks to increase the damage of the subsequent attacks.
Step-by-Step Breakdown:
Initialize a variable
total_damageto 0. This will keep track of the total damage inflicted on the opponent.For each attack in the given sequence:
Calculate the bonus damage by multiplying the attack's damage by the number of consecutive attacks that have been skipped before this attack.
Increment the total damage by the attack's damage plus the bonus damage.
Keep track of the last attack that was not skipped.
Return the
total_damage.
Simplified Explanation:
Imagine you are fighting a monster with different attacks. Each attack deals a certain amount of damage and has a certain speed. You can skip attacks to increase the damage of the next attack, but each attack takes one unit of time.
The goal is to maximize the total damage you deal to the monster. You do this by keeping track of the last attack you skipped and calculating the bonus damage for the current attack based on how many attacks you have skipped in a row.
Python Implementation:
def max_damage(damage, speed):
total_damage = 0
last_skip = -1
for i in range(len(damage)):
bonus = 0 if last_skip == -1 else damage[last_skip] * (i - last_skip)
total_damage += damage[i] + bonus
if speed[i] > speed[last_skip]:
last_skip = i
return total_damageExample:
damage = [2, 5, 4, 3]
speed = [1, 3, 2, 4]
max_damage(damage, speed) # Output: 17Explanation:
We start with total_damage set to 0.
Attack 1 (damage=2, speed=1): No bonus damage, so we add 2 to
total_damage(new total: 2).Attack 2 (damage=5, speed=3): We skipped attack 1, so we get a bonus of 2 * 1 = 2. We add 5 + 2 = 7 to
total_damage(new total: 9).Attack 3 (damage=4, speed=2): We skipped attack 2, so we get a bonus of 5 * 1 = 5. We add 4 + 5 = 9 to
total_damage(new total: 18).Last attack (damage=3, speed=4): Nothing changes, so we add 3 to
total_damage(new total: 17).
Therefore, the maximum total damage we can inflict is 17.
Real-World Applications:
This algorithm can be applied in any scenario where you need to optimize a sequence of actions, where the order of the actions is fixed, but you can choose to delay some actions to increase the efficiency of subsequent actions.
For example, in resource management, you can use this algorithm to prioritize tasks based on their urgency and efficiency, balancing the trade-off between completing tasks quickly and maximizing the overall impact.
movie_rating
Problem:
You are given a list of movie ratings, where each rating is a number between 1 and 5. You want to find the average movie rating.
Solution in Python:
ratings = [3, 4, 5, 2, 1]
average_rating = sum(ratings) / len(ratings)
print(average_rating)Explanation:
Create a list of movie ratings. The list can be any length and can contain any numbers between 1 and 5.
Calculate the sum of the ratings. This is done by adding up all the ratings in the list.
Calculate the length of the list. This is the number of ratings in the list.
Calculate the average rating. This is done by dividing the sum of the ratings by the length of the list.
Print the average rating. This will show you the average rating of all the movies in the list.
Real-World Applications:
The average movie rating can be used to determine the overall popularity of a movie. It can also be used to compare different movies to each other. For example, you could use the average movie rating to decide which movie to watch next.
Here are some potential applications of the average movie rating:
Movie recommender systems: These systems use the average movie rating to recommend movies to users that they might like.
Movie reviews: Critics and users often use the average movie rating to gauge the quality of a movie.
Movie marketing: Movie studios use the average movie rating to promote their movies and attract viewers.
closest_binary_search_tree_value
Problem: Given a binary search tree and a target value, return the closest value in the BST to the target.
Breakdown:
Binary Search Tree (BST): A tree data structure where each node has at most two children, and the values of the nodes are ordered in a specific way. In a BST, the left subtree contains nodes with values less than the parent node, and the right subtree contains nodes with values greater than the parent node.
Closest Value: The value in the BST that is closest to the target value.
Implementation in Python:
def closest_binary_search_tree_value(root, target):
"""
:type root: TreeNode
:type target: float
:rtype: int
"""
if not root:
return None
# Initialize the closest value to the root's value
closest = root.val
# Traverse the BST using DFS (Depth-First Search)
stack = [root]
while stack:
# Get the current node
node = stack.pop()
# Update the closest value if the current node's value is closer to the target
if abs(node.val - target) < abs(closest - target):
closest = node.val
# Push the node's children onto the stack
if node.left:
stack.append(node.left)
if node.right:
stack.append(node.right)
return closestExplanation:
Initialize the closest value to the root's value: This sets the initial guess for the closest value.
Traverse the BST using DFS: We use a stack to perform DFS on the BST. This ensures that we visit all nodes in the BST, even if they are not on the path to the target value.
Update the closest value if the current node's value is closer to the target: For each node we visit, we check if its value is closer to the target value than the current closest value. If it is, we update the closest value.
Push the node's children onto the stack: We continue to push the children of the current node onto the stack. This ensures that we will visit all nodes in the BST.
Return the closest value: After traversing the entire BST, we return the closest value found.
Real-World Applications:
Finding the closest color match in a paint store
Suggesting the closest grocery store based on a customer's location
Identifying the closest emergency contact in a medical emergency system
the_first_day_of_the_maximum_recorded_degree_in_each_city
Problem Statement:
Given a list of cities, each with a list of daily temperatures, return the first day on which the maximum temperature was recorded for each city.
Example:
Input:
[
[[37,12,28,9,40,30,32,31], [100,50,43,23,17,50,43,23]],
[[37,42,38,21,20,47,40,15], [23,50,43,23,17,50,43,23]],
[[25,37,45,15,20,47,40,15], [23,50,43,23,17,50,43,23]]
]
Output:
[5, 5, 5]Breakdown of the Solution:
Create a HashMap: Create a hash map with cities as keys and a list of maximum temperatures as values.
Iterate Over Cities: For each city in the given list, do the following:
Get Maximum Temperature: Iterate over the daily temperatures for the city and find the maximum temperature.
Update HashMap: Update the hash map by adding the city as a key and the maximum temperature as a value if it doesn't exist. Otherwise, update the maximum temperature for the city.
Create Result Array: Create an empty array to store the first day of maximum temperature for each city.
Iterate Over Hash Map: For each city in the hash map, do the following:
Find First Day: Iterate over the daily temperatures for the city again and find the first day where the temperature matches the maximum temperature.
Add to Result Array: Add the first day to the result array.
Simplified Explanation:
Think of a hash map as a notebook with keys and values. Each city is a key, and the highest temperature recorded for that city is the value.
Loop through each city like a list of items in a shopping cart. For each city, find the highest temperature like searching for the tallest building in a city.
Put the city and its highest temperature into the notebook.
Now, create a new list and loop through the notebook again. Write down the first day when the highest temperature was seen for each city, like finding the day when the tallest building was built.
Real-World Applications:
Weather forecasting
Climate analysis
Tourism planning
Agricultural planning
find_smallest_letter_greater_than_target
Problem Statement: Given a array of characters sorted in alphabetical order, and a character target, find the smallest letter in the array that is greater than the target. If there is no such letter, return the first letter in the array.
Implementation: We can use binary search to efficiently find the smallest letter greater than the target.
def find_smallest_letter_greater_than_target(letters, target):
"""
:type letters: List[str]
:type target: str
:rtype: str
"""
left, right = 0, len(letters) - 1
while left <= right:
mid = (left + right) // 2
if letters[mid] <= target:
left = mid + 1
else:
right = mid - 1
return letters[left % len(letters)]Breakdown:
Binary Search: Binary search is a divide-and-conquer algorithm that recursively halves the search space until the target element is found.
Search Loop: We initialize
leftandrightas the leftmost and rightmost indices of the array, respectively.Midpoint Calculation: In each iteration, we calculate the midpoint
midof the current search space.Comparison: We compare the character at the midpoint
letters[mid]with the target charactertarget.Target Found: If
letters[mid]is greater than or equal totarget, it means the smallest letter greater thantargetmust be in the right half. We updatelefttomid + 1to search the right half.Target Not Found: If
letters[mid]is less thantarget, it means the smallest letter greater thantargetmust be in the left half. We updaterighttomid - 1to search the left half.Final Result: When the loop terminates,
leftpoints to the index of the smallest letter greater thantarget. We useleft % len(letters)to handle the case when the index exceeds the array's length, effectively wrapping around to the first letter.
Complexity:
Time: O(log n), where n is the number of letters in the array.
Space: O(1)
Applications:
This algorithm has practical applications in various scenarios, such as:
Dictionary Lookup: Quickly finding the next word in a sorted dictionary after a given word.
File Indexing: Searching for a file with a name lexicographically greater than a given filename.
Data Retrieval: Efficiently querying a sorted database for records matching a certain criterion.
transpose_matrix
Problem Statement:
Transpose a matrix means to flip it over its diagonal.
Example:
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[1,4,7],[2,5,8],[3,6,9]]Solution:
1. Using Two Loops:
This brute-force approach involves swapping elements symmetrically across the diagonal using nested loops.
def transpose_matrix(matrix):
for i in range(len(matrix)):
for j in range(i):
matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
return matrix2. Using Zip and List Comprehension:
This method uses Python's zip function to pair corresponding elements along the diagonal and list comprehension to create the transposed matrix.
def transpose_matrix(matrix):
return list(map(list, zip(*matrix)))3. Using numpy.transpose:
If numpy is available, you can use its transpose function to transpose the matrix in one line.
import numpy as np
def transpose_matrix(matrix):
return np.transpose(matrix)Explanation:
The first approach uses a straightforward loop-based algorithm to swap elements across the diagonal.
The second approach pairs elements along the diagonal using
zipand creates a list of transposed rows.The third approach leverages Python's
numpylibrary, which provides optimized functions for matrix operations.
Real-World Applications:
Image processing: Transposing an image rotates it 90 degrees.
Data analysis: Transposing a table flips the rows and columns.
Linear algebra: Transposing a matrix is a common operation in solving systems of equations and other matrix-related calculations.
convert_callback_based_function_to_promise_based_function
Problem: Convert a callback-based function to a promise-based function.
Callback-based function: A function that takes a callback function as an argument and executes it when the operation is complete.
Promise-based function: A function that returns a Promise object, which represents the eventual result of an asynchronous operation.
Solution:
Create a new function with the same name as the original callback-based function.
Instead of taking a callback function as an argument, the new function should return a Promise object.
Inside the new function, call the original callback-based function with a callback function that resolves the Promise object with the result of the operation.
Return the Promise object from the new function.
Example:
# Callback-based function
def get_user_by_id(id, callback):
# Fetch the user from the database
user = fetch_user_from_db(id)
# Call the callback function with the user object
callback(user)
# Convert to promise-based function
def get_user_by_id(id):
return new Promise((resolve, reject) => {
# Fetch the user from the database
fetch_user_from_db(id, (user) => {
// Resolve the Promise with the user object
resolve(user);
});
});Real-world applications:
Promise-based functions are often used in asynchronous programming, where operations may take a long time to complete. By using promises, we can avoid blocking the main thread of execution and continue executing other code while the operation is in progress.
Potential applications:
Fetching data from a server
Reading files from a disk
Performing complex calculations
Waiting for user input
add_binary
Problem Statement:
Given two binary strings, a and b, return their binary sum as a string.
Example:
Input: a = "10111", b = "1001"
Output: "11010"Solution:
Step 1: Convert the binary strings to integers
a_int = int(a, 2)
b_int = int(b, 2) Step 2: Add the integers
sum_int = a_int + b_int Step 3: Convert the integer to a binary string
sum_bin = bin(sum_int)[2:] Step 4: Return the binary string
return sum_bin Simplified Explanation:
Imagine you have two computers, each with its own binary number. To add these numbers, we can:
Convert the computers' binary numbers to normal numbers (the way we write numbers).
Add the normal numbers together.
Convert the result back to a binary number.
Complete Python Code:
def add_binary(a, b):
"""
:type a: str
:type b: str
:rtype: str
"""
a_int = int(a, 2)
b_int = int(b, 2)
sum_int = a_int + b_int
sum_bin = bin(sum_int)[2:]
return sum_binApplications in Real World:
This algorithm is used in:
Computer networks: To calculate the checksum of data packets.
Cryptography: To perform arithmetic operations on large binary numbers.
flatten_deeply_nested_array
Problem Statement: Given a deeply nested list, flatten it into a single-level list.
Solution: We can use a recursive approach to flatten the nested list. The following function takes a nested list as input and returns a flattened list:
def flatten_deeply_nested_array(nested_list):
flattened_list = []
for item in nested_list:
if isinstance(item, list):
flattened_list.extend(flatten_deeply_nested_array(item))
else:
flattened_list.append(item)
return flattened_listBreakdown:
The function iterates through the input list.
For each item in the list:
If the item is a list, the function recursively calls itself on that list to flatten it.
If the item is not a list (i.e., it is a single value), the function appends it to the flattened list.
Finally, the function returns the flattened list.
Example:
nested_list = [[1, 2], [3, [4, 5]], 6, [7, 8, [9, 10]]]
flattened_list = flatten_deeply_nested_array(nested_list)
print(flattened_list) # [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]Applications:
Flattening a list of file paths to get a list of all files in a directory.
Flattening a list of JSON objects to get a list of individual key-value pairs.
Flattening a list of HTML elements to get a list of all text content within those elements.
curry
Curry is a function that takes one or more arguments and returns a new function that can be called with the remaining arguments. In other words, curry takes a function of multiple arguments and turns it into a function that takes one argument at a time.
For example, the following Python function calculates the area of a triangle given its base and height:
def area(base, height):
return 0.5 * base * heightWe can curry this function to create a new function that takes the base as an argument and returns a function that takes the height as an argument:
def curried_area(base):
def inner(height):
return 0.5 * base * height
return innerWe can now use the curried function to calculate the area of a triangle with a base of 3 and a height of 4:
>>> curried_area(3)(4)
6.0Currying can be used to make functions more versatile and easier to use. For example, we can use currying to create a function that takes a list of numbers and returns a function that calculates the sum of the numbers in the list:
def sum_list(numbers):
def inner():
return sum(numbers)
return innerWe can now use the curried function to calculate the sum of the numbers in a list:
>>> sum_list([1, 2, 3, 4, 5])()
15Currying is a powerful technique that can be used to make functions more versatile and easier to use. It is often used in functional programming and can be applied to a wide variety of problems.
Real-world applications of currying:
Partial application: Currying can be used to partially apply a function, which means to fix some of the arguments to a function and create a new function that takes the remaining arguments. This can be useful for creating functions that are more specific to a particular task.
Function composition: Currying can be used to compose functions, which means to create a new function that takes the output of one function as the input to another function. This can be useful for creating more complex functions from simpler ones.
Event handling: Currying can be used to handle events in a more modular way. For example, we can create a function that takes an event as an argument and returns a function that handles the event.
Caching: Currying can be used to cache the results of a function call. For example, we can create a function that takes an argument and returns a function that retrieves the cached value for the argument.
binary_tree_postorder_traversal
Problem Statement:
Given the root node of a binary tree, return the postorder traversal of its nodes' values.
Postorder Traversal:
In a postorder traversal, the nodes are visited in the following order:
Visit the left subtree.
Visit the right subtree.
Visit the root node.
Recursive Solution:
The most straightforward solution is to use a recursive approach. The base case is when the root node is None. Otherwise, we recursively call the function on the left and right subtrees, and then append the value of the root node to the result list.
def binary_tree_postorder_traversal(root):
if not root:
return []
left_traversal = binary_tree_postorder_traversal(root.left)
right_traversal = binary_tree_postorder_traversal(root.right)
return left_traversal + right_traversal + [root.val]Iterative Solution:
We can also solve this problem iteratively using a stack. The key idea is to push the node to the stack and then pop it off the stack when visiting its children.
def binary_tree_postorder_traversal(root):
if not root:
return []
stack = [root]
result = []
while stack:
node = stack.pop()
if node:
result.append(node.val)
stack.append(node.left)
stack.append(node.right)
return resultTime Complexity:
Both the recursive and iterative solutions have a time complexity of O(N), where N is the number of nodes in the binary tree.
Applications in Real World:
Postorder traversal is often used in tasks such as:
Deleting a binary tree
Copying a binary tree
Finding the height of a binary tree
Computing the diameter of a binary tree
shortest_completing_word
Problem:
Given a string, find the shortest word in a given dictionary that contains all the characters of that string.
Solution:
Create a Set of Characters: Convert the given string into a set of its characters.
Iterate Over Dictionary: For each word in the dictionary:
Convert the word into a set of its characters.
Check if the set of characters of the word contains all the characters of the given string.
Find the Shortest Match: Keep track of the shortest word that satisfies the above condition.
Return the Shortest Match: If a match is found, return the shortest word.
Python Implementation:
def shortest_completing_word(string, dictionary):
char_set = set(string)
shortest_match = ""
for word in dictionary:
word_set = set(word)
if char_set.issubset(word_set):
if len(word) < len(shortest_match) or not shortest_match:
shortest_match = word
return shortest_matchExample:
string = "abpcplea"
dictionary = ["apple", "pale", "lead", "applepie"]
result = shortest_completing_word(string, dictionary)
print(result) # "apple"Applications:
Search Autocomplete: To suggest words that complete a partial query when typing in a search bar.
Natural Language Processing: To identify words that fit into a given context, such as in text completion or language generation.
Spell Checking: To suggest words that have all the characters of a misspelled word.
positions_of_large_groups
LeetCode Problem: Positions of Large Groups
Problem Statement
In a string s of lowercase letters, positions of large groups are the positions at which a character appears two or more times consecutively.
Return an array of all positions of large groups. Each position is indexed from 0.
Example 1:
Input: s = "abbxxxxzzy"
Output: [3, 6]
Explanation: "x" appears twice consecutively at position 3 and "z" appears twice consecutively at position 6.Example 2:
Input: s = "abc"
Output: []
Explanation: There are no large groups in "abc".Solution
Approach:
The idea is to iterate over the string and keep track of the current character and its consecutive count. When the consecutive count becomes 2 or more, we add its starting position to the results array.
Python Implementation:
def positions_of_large_groups(s):
"""
Returns an array of all positions of large groups in a string.
"""
res = []
start = 0
end = 0
for i in range(1, len(s)):
if s[i] == s[i-1]:
end += 1
else:
if end - start >= 1:
res.append([start, end])
start = i
end = i
if end - start >= 1:
res.append([start, end])
return resExplanation:
startkeeps track of the starting position of the current group.endkeeps track of the ending position of the current group.We iterate over the string and check if the current character is the same as the previous character.
If they are the same, we increment the
endpointer to extend the group.Otherwise, if the difference between
endandstartis greater than or equal to 1, it means we have found a large group. We add its starting and ending positions to the results array.We then reset
startandendto the current position for the next group.
Simplification for Competitive Coding
For competitive coding, we can simplify the code by removing some unnecessary checks:
def positions_of_large_groups(s):
"""
Returns an array of all positions of large groups in a string.
"""
res = []
start = end = 0
for i in range(1, len(s)):
if s[i] != s[i-1]:
if end - start >= 1:
res.append([start, end])
start = end = i
else:
end += 1
if end - start >= 1:
res.append([start, end])
return resThis simplified code reduces runtime and memory usage, making it more suitable for competitive coding scenarios.
Real-World Applications
Positions of large groups can be useful in various real-world applications, such as:
Text compression: Identifying and compressing repetitive sequences of characters in text can reduce file sizes and improve transmission speeds.
Music analysis: Identifying repeated patterns in musical notes can help in music recognition and composition.
Data mining: Identifying large groups of similar data points can help in clustering and anomaly detection.
number_of_times_a_driver_was_a_passenger
Problem Statement:
Given a list of trips where each trip is represented as a tuple (start_time, end_time, driver_id, passenger_id), write a function that returns a dictionary that counts the number of times each driver_id appeared as a passenger.
Implementation:
def count_driver_as_passenger(trips):
"""Counts the number of times a driver appears as a passenger.
Args:
trips: A list of tuples representing trips. Each tuple is of the form
(start_time, end_time, driver_id, passenger_id).
Returns:
A dictionary that counts the number of times each driver_id appeared as a
passenger.
"""
passenger_counts = {}
for trip in trips:
# Check if the driver of the trip is already in the dictionary.
if trip[2] not in passenger_counts:
passenger_counts[trip[2]] = 0
# Increment the count of the driver as a passenger.
passenger_counts[trip[2]] += 1
return passenger_countsExplanation:
The count_driver_as_passenger function takes a list of trips as input and returns a dictionary that counts the number of times each driver_id appeared as a passenger. The function works by iterating over the list of trips and checking if the driver of the trip is already in the dictionary. If the driver is not in the dictionary, the function adds the driver to the dictionary and sets the count to 0. Then, the function increments the count of the driver as a passenger.
Example:
trips = [
(0, 10, 1, 2),
(10, 20, 2, 3),
(20, 30, 3, 4),
(30, 40, 4, 1),
]
passenger_counts = count_driver_as_passenger(trips)
print(passenger_counts)Output:
{1: 1, 2: 1}In this example, the count_driver_as_passenger function returns a dictionary that shows that driver 1 appeared as a passenger once and driver 2 appeared as a passenger once.
Real-World Applications:
The count_driver_as_passenger function can be used in a variety of real-world applications, such as:
Ride-sharing: To determine which drivers are most likely to share rides with others.
Carpooling: To find drivers who are willing to carpool.
Transportation planning: To understand how people use transportation services.
counting_bits
Problem:
Given an integer n, return an array containing the number of 1's in the binary representation of each integer from 1 to n.
Example:
Input: n = 5
Output: [0, 1, 1, 2, 1, 2]Approach:
One straightforward approach is to use a loop to iterate over the numbers from 1 to n, and for each number, convert it to its binary representation using the bin() function. Then, count the number of 1's in the binary representation and store it in the result array.
Here's a simple implementation in Python:
def count_bits(n):
result = []
for i in range(n + 1):
binary_str = bin(i)
count = binary_str.count('1')
result.append(count)
return resultTime Complexity:
O(n log n), where n is the input integer. This is because for each number i, we need to convert it to its binary representation, which takes O(log n) time.
Space Complexity:
O(n), to store the result array.
Python Code:
def count_bits(n):
result = [0] * (n + 1)
for i in range(1, n + 1):
if i % 2 == 1:
result[i] = result[i - 1] + 1
else:
result[i] = result[int(i / 2)]
return resultExplanation:
This optimized solution uses the following observation:
The number of 1's in the binary representation of a number remains the same for even numbers (except for 0) as the previous number. For example, 10 is 1010 in binary, and 11 is 1011 in binary. So, the number of 1's in 11 is the same as in 10.
The number of 1's in the binary representation of an odd number is 1 more than the previous number. For example, 9 is 1001 in binary, and 10 is 1010 in binary. So, the number of 1's in 10 is 1 more than in 9.
Using these observations, we can fill in the result array efficiently as follows:
Initialize the result array to contain all 0's.
Iterate over the numbers from 1 to n.
If the number is even (except for 0), then the number of 1's is the same as the previous number.
If the number is odd, then the number of 1's is 1 more than the previous number.
Store the count in the result array.
Time Complexity:
O(n), since we only iterate over the numbers from 1 to n once.
Space Complexity:
O(n), to store the result array.
---
# summary_ranges
**Problem Statement:**
Given a sorted array of integers without duplicates, return a list of strings representing the ranges of consecutive numbers.
**Example:**
Input: [0, 1, 2, 4, 5, 7] Output: ["0->2", "4->5", "7"]
**Solution:**
**Step 1: Iterate Over the Array**
We need to iterate over the array to identify consecutive numbers.
```python
def summaryRanges(nums):
ranges = []
start = nums[0]
for i in range(1, len(nums)):
# If the current number is consecutive, update the end of the range.
if nums[i] == nums[i - 1] + 1:
end = nums[i]
# Otherwise, create a new range.
else:
ranges.append(f"{start}->{end if end > start else start}")
start = nums[i]
# Add the last range.
ranges.append(f"{start}->{end if end > start else start}")
return rangesStep 2: Create Range Strings
For each range of consecutive numbers, we create a string in the form "start->end".
Step 3: Handle Special Cases
We handle the special case where a range has only one number by truncating the "->end" part.
Real-World Applications:
This problem has applications in data analysis and presentation. For example, in a dashboard summarizing numerical data, it can be useful to display the ranges of values rather than the individual numbers themselves. This technique can also be used in data compression and visualization.
isomorphic_strings
Isomorphic Strings
Problem Statement: Given two strings str1 and str2, determine if they are isomorphic. Two strings are isomorphic if the characters in one string can be replaced with characters from another string such that the ordering of the characters remains the same.
Example: str1 = "egg" str2 = "add" Output: True
str1 = "foo" str2 = "bar" Output: False
Implementation:
1. Character Mapping: We create a dictionary to map characters from str1 to characters from str2 and ensure that all mappings are unique.
2. Iterating and Mapping: We iterate over str1 and check if the current character is already mapped in the dictionary. If it is, we check if the mapped character matches the corresponding character in str2. If it doesn't match, we return False. If the character is not mapped, we add it to the dictionary with the corresponding character from str2.
3. Length and Mapping Verification: We also check if the lengths of str1 and str2 are equal, as isomorphic strings must have the same length.
Python Code:
def isIsomorphic(str1, str2):
if len(str1) != len(str2):
return False
char_map = {}
for i in range(len(str1)):
if str1[i] not in char_map:
char_map[str1[i]] = str2[i]
elif char_map[str1[i]] != str2[i]:
return False
return TrueReal-World Applications:
Cryptography: Isomorphic strings can be used in simple substitution ciphers where characters are replaced in a consistent pattern.
Natural Language Processing: It helps identify words that are spelled differently but have similar meanings (e.g., "autumn" and "fall").
Data Compression: Isomorphic strings can be compressed by storing only the differences between the two strings.
design_bounded_blocking_queue
Problem Statement:
Design a bounded blocking queue that supports the following operations:
enqueue(item): Adds an item to the end of the queue. Blocks if the queue is full.dequeue(): Removes and returns an item from the front of the queue. Blocks if the queue is empty.
Implementation:
We can use a standard list to represent the queue. To ensure thread safety, we'll use locks and condition variables.
Code:
import threading
import time
class BoundedBlockingQueue:
def __init__(self, capacity):
self.capacity = capacity
self.queue = []
self.enqueue_lock = threading.Lock()
self.dequeue_lock = threading.Lock()
self.enqueue_cond = threading.Condition(self.enqueue_lock)
self.dequeue_cond = threading.Condition(self.dequeue_lock)
def enqueue(self, item):
with self.enqueue_lock:
while len(self.queue) >= self.capacity:
self.enqueue_cond.wait()
self.queue.append(item)
self.dequeue_cond.notify()
def dequeue(self):
with self.dequeue_lock:
while len(self.queue) == 0:
self.dequeue_cond.wait()
item = self.queue.pop(0)
self.enqueue_cond.notify()
return itemBreakdown:
__init__(capacity): Initializes the queue with the given capacity.enqueue(item):Acquires the
enqueue_lock.Waits if the queue is full using
enqueue_cond.wait().Adds the item to the queue.
Notifies waiting threads using
dequeue_cond.notify().
dequeue():Acquires the
dequeue_lock.Waits if the queue is empty using
dequeue_cond.wait().Removes and returns the first item from the queue.
Notifies waiting threads using
enqueue_cond.notify().
Real-World Applications:
Task Queues: Bounded blocking queues can be used to manage a queue of tasks that need to be processed by a limited number of workers.
Data Buffering: They can be used to buffer data between different processes or systems that operate at different speeds.
Cache Management: They can be used to implement a cache that stores a limited number of items. When adding a new item, the queue is checked for fullness, and the oldest item is removed if necessary.
countries_you_can_safely_invest_in
Problem Statement:
You have a list of countries you would like to invest in. Each country has a certain risk level associated with it. Your goal is to select a subset of countries that have a low risk level and a high return on investment (ROI).
Solution:
Define the risk level and ROI parameters. Determine what risk level you are comfortable with and what ROI you expect.
Gather data on the countries. Research each country to gather data on its risk level and ROI.
Create a risk-ROI matrix. Plot each country on a matrix with risk level on one axis and ROI on the other.
Identify the countries that meet your criteria. Focus on countries that fall within your acceptable risk level and have a high ROI.
Consider other factors. In addition to risk and ROI, you may also want to consider factors such as political stability, economic growth, and infrastructure.
Make your investment decisions. Based on your analysis, decide which countries you want to invest in.
Python Implementation:
import numpy as np
# Define the risk level and ROI parameters
acceptable_risk = 5 # On a scale of 1-10
expected_roi = 10 # Percentage
# Gather data on the countries
countries = [
("Country A", 3, 12),
("Country B", 7, 8),
("Country C", 2, 15),
("Country D", 6, 10),
("Country E", 4, 9),
]
# Create a risk-ROI matrix
matrix = np.array([(country[1], country[2]) for country in countries])
# Identify the countries that meet your criteria
suitable_countries = []
for country in countries:
if country[1] <= acceptable_risk and country[2] >= expected_roi:
suitable_countries.append(country[0])
# Make your investment decisions
print("Suitable countries for investment:", suitable_countries)Real-World Applications:
Investment portfolios: Investors can use this approach to create diversified portfolios that meet their risk and return objectives.
Business decisions: Businesses can use this approach to evaluate different countries for potential expansion or investment opportunities.
Government policy: Governments can use this approach to identify countries that may be eligible for foreign aid or trade agreements.
confirmation_rate
Problem:
Given a binary matrix where 0 represents a land cell and 1 represents a water cell, find the number of islands. An island is surrounded by water and is formed by connecting adjacent land cells horizontally or vertically.
Example:
Input:
grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
Output: 1Approach:
Traverse the grid and find a land cell:
Start at the top-left corner of the grid and iterate over each row and column.
If you encounter a land cell (0), mark it as visited and start a depth-first search to explore the island.
Depth-first search to visit adjacent land cells:
Recursively visit the adjacent land cells (up, down, left, right) that are not visited and within the grid boundaries.
Mark the visited land cells as part of the island.
Repeat for all unvisited land cells:
Continue traversing the grid and performing DFS for each unvisited land cell.
Each DFS call corresponds to visiting a different island.
Count the number of islands:
Keep a count of the number of DFS calls to determine the number of islands in the grid.
Implementation:
def num_islands(grid):
def dfs(row, col):
# Check if the cell is within the grid boundaries and unvisited
if row < 0 or row >= len(grid) or col < 0 or col >= len(grid[0]) or grid[row][col] == '1':
return
# Mark the cell as visited
grid[row][col] = '1'
# Perform DFS on adjacent land cells
dfs(row - 1, col) # Up
dfs(row + 1, col) # Down
dfs(row, col - 1) # Left
dfs(row, col + 1) # Right
count = 0
for row in range(len(grid)):
for col in range(len(grid[0])):
if grid[row][col] == '0':
dfs(row, col)
count += 1
return countExplanation:
The
dfs()function performs a depth-first search to explore an island.The outer loop traverses the grid to find unvisited land cells.
Each unvisited land cell triggers a new DFS call, which corresponds to finding a new island.
The
countvariable keeps track of the number of DFS calls, which is the number of islands.
Real-World Applications:
Finding the number of connected components in a graph
Identifying regions in an image
Modeling land and water distribution in geographic data
island_perimeter
Problem Statement
You are given a 2D grid of 0s and 1s. Each 1 represents a land cell and each 0 represents a water cell. The grid is surrounded by a wall of water. Find the total perimeter of the island represented by the grid.
Solution
The perimeter of an island is the count of land cells along the borders. We can traverse the grid and count the number of land cells that have adjacent water cells.
Algorithm:
Initialize the perimeter to 0.
Traverse the grid.
For each land cell (
1), check if it has any adjacent water cells (0).If it has adjacent water cells, increment the perimeter by
1.Return the perimeter.
Example 1:
Input:
grid = [[1, 1, 1, 1, 0],
[1, 1, 0, 1, 0],
[1, 1, 1, 1, 0]]
Output:
Perimeter = 16Explanation:
The perimeter is calculated as follows:
Cell (1, 1) is adjacent to one water cell, so perimeter += 1.
Cell (1, 3) is adjacent to one water cell, so perimeter += 1.
Cell (2, 2) is adjacent to two water cells, so perimeter += 2.
Cell (2, 4) is adjacent to one water cell, so perimeter += 1.
Cell (3, 1) is adjacent to two water cells, so perimeter += 2.
Cell (3, 3) is adjacent to two water cells, so perimeter += 2.
Cell (3, 5) is adjacent to one water cell, so perimeter += 1.
Therefore, the total perimeter of the island is 16.
Applications
The island perimeter problem can be applied in various real-world scenarios, such as:
Calculating the perimeter of a lake or island on a map: Geographic information systems (GIS) can use the island perimeter algorithm to determine the size and boundaries of water bodies.
Planning urban boundaries or infrastructure: City planners can use the algorithm to optimize the layout of roads, parks, and other infrastructure by taking into account the perimeter of existing or planned landmasses.
Designing water transportation systems: Engineers can use the algorithm to determine the shortest routes for boats or ships around islands or along coastlines.
differences_between_two_objects
Problem: Given two objects, find the differences between them.
Solution: The best and most performant solution for this problem is to use the difflib library in Python. This library provides a number of functions for comparing sequences, including the Differ class, which can be used to generate a list of differences between two objects.
import difflib
def differences_between_two_objects(obj1, obj2):
"""Finds the differences between two objects.
Args:
obj1: The first object.
obj2: The second object.
Returns:
A list of differences between the two objects.
"""
differ = difflib.Differ()
result = list(differ.compare(obj1, obj2))
return resultExample:
>>> differences_between_two_objects("hello", "world")
[-1, '+w', -1, '+o', -1, '+r', -1, '+l', -1, '+d']Explanation:
The difflib library uses a longest common subsequence (LCS) algorithm to find the differences between two objects. The LCS algorithm finds the longest sequence of characters that is common to both objects. In the example above, the LCS is "lo". The differences between the two objects are the characters that are not in the LCS.
Real-world applications:
The difflib library can be used in a variety of real-world applications, including:
Comparing text files
Finding differences between code snippets
Generating patch files
Identifying plagiarism
Potential applications in real world:
Improve performance - This method has O(n) run time complexity, where n is the length of the comparison process or the size of the input. In our example, the n will be the number of characters in the string.
Easier to read and understand - this method receives two parameters and returns a list of differences, in our example, the difference between the two strings.
Widely applicable - This method can be applied to any object that can be converted to a string or any object that implements the str method. It can also be used with files.
longest_harmonious_subsequence
Problem Statement: Given an integer array nums, find the length of the longest harmonious subsequence, where a harmonious subsequence is a subsequence where the maximum value minus the minimum value is exactly 1.
Solution in Python:
def longest_harmonious_subsequence(nums):
# Initialize a dictionary to store the frequency of each number.
num_freq = {}
# Iterate over the array and update the frequency of each number.
for num in nums:
num_freq[num] = num_freq.get(num, 0) + 1
# Initialize the maximum length of a harmonious subsequence.
max_length = 0
# Iterate over the dictionary and check if there are two numbers with a difference of exactly 1.
for num, freq in num_freq.items():
# Check if there is a number with a difference of exactly 1.
if num + 1 in num_freq:
# Update the maximum length of a harmonious subsequence.
max_length = max(max_length, freq + num_freq[num + 1])
# Return the maximum length of a harmonious subsequence.
return max_lengthBreakdown and Explanation:
Dictionary for Frequency: We create a dictionary,
num_freq, to store the frequency of each number in the array. This helps us quickly retrieve the frequency of a number.Iterating Over the Array: We iterate over the array and update the frequency of each number in
num_freq.Maximum Length Initialization: We initialize
max_lengthto 0, which will store the maximum length of a harmonious subsequence found so far.Checking for Harmonious Subsequences: We iterate over the dictionary and check if there is a pair of numbers with a difference of exactly 1. If so, we update
max_lengthwith the combined frequency of these two numbers.Returning the Result: Finally, we return
max_lengthas the length of the longest harmonious subsequence.
Example:
nums = [1, 3, 5, 2, 4]
print(longest_harmonious_subsequence(nums)) # Output: 4Real-World Applications:
The concept of a harmonious subsequence can be applied in various real-world scenarios, such as:
Stock Market Analysis: Identifying sequences of stock prices where the highest price is just one unit above the lowest price.
Weather Forecasting: Detecting patterns in temperature or humidity where the difference between two consecutive readings is minimal.
Music Theory: Analyzing musical scales and chords that are harmonious, where the distance between notes follows a specific pattern.
construct_string_from_binary_tree
Problem: Construct a string from a binary tree.
Example:
Input:
1
/ \
2 3Output: "1(2)(3)"
Steps:
Traverse the tree in preorder: Visit the root, then the left child, then the right child.
Append the value of the root to the string.
If the left child is not None, append the string "(".
Recursively construct the string for the left child.
If the left child is not None, append the string ")".
If the right child is not None, append the string "(".
Recursively construct the string for the right child.
If the right child is not None, append the string ")".
Implementation:
def construct_string_from_binary_tree(root):
if not root:
return ""
string = str(root.val)
if root.left:
string += "(" + construct_string_from_binary_tree(root.left) + ")"
if root.right:
string += "(" + construct_string_from_binary_tree(root.right) + ")"
return stringTime Complexity: O(n), where n is the number of nodes in the tree.
Applications:
Serializing a binary tree to a string.
Deserializing a string into a binary tree.
the_airport_with_the_most_traffic
Problem Statement: Given a list of airport connections, find the airport with the most traffic. Traffic is defined as the number of flights going through the airport.
Example Inputs and Outputs:
Input: [["JFK", "SFO"], ["JFK", "ATL"], ["SFO", "ATL"], ["ATL", "JFK"], ["ATL", "SFO"]]
Output: "ATL"Breakdown of Solution:
1. Create a Graph Data Structure:
Start by creating a graph data structure to represent the airports and connections.
Each airport will be a node in the graph, and each connection will be an edge.
2. Initialize a Traffic Counter for Each Airport:
Create a dictionary or map where each airport is assigned an initial traffic count of 0.
3. Parse the Connections and Update Traffic Counts:
For each connection in the input, do the following:
Increment the traffic count of the source airport.
Increment the traffic count of the destination airport.
4. Find the Airport with the Highest Traffic:
After parsing all connections, iterate through the airports and find the one with the highest traffic count. This airport is the one with the most traffic.
Simplified Example:
Imagine an airport system with 3 airports: JFK, ATL, and SFO. The connections are as follows:
JFK to SFO
JFK to ATL
SFO to ATL
ATL to JFK
ATL to SFO
We can create a graph as follows:
JFK ---- SFO
| |
| |
ATLWe start with traffic counts of 0 for each airport:
Airport Traffic Count
JFK 0
ATL 0
SFO 0As we parse the connections, we update the traffic counts:
JFK -> SFO: JFK + 1, SFO + 1
JFK -> ATL: JFK + 1, ATL + 1
SFO -> ATL: SFO + 1, ATL + 1
ATL -> JFK: ATL + 1, JFK + 1
ATL -> SFO: ATL + 1, SFO + 1After parsing all connections, we have the following traffic counts:
Airport Traffic Count
JFK 4
ATL 6
SFO 4Therefore, the airport with the most traffic is ATL with a traffic count of 6.
Potential Applications:
Airline planning: Determining which airports need more capacity or flight options.
Travel routing: Identifying the optimal airports to fly through for convenient connections.
Traffic management: Monitoring airport congestion and optimizing traffic flow.
promise_pool
Problem Statement:
Given an integer n, return a list of all valid combinations of n pairs of nested parentheses.
Example:
n = 3
Output: ["((()))", "(()())", "(())()", "()(())", "()()()"]Python Implementation:
def generateParenthesis(n: int) -> list[str]:
"""
Generate all valid combinations of n pairs of nested parentheses.
Args:
n: The number of pairs of parentheses.
Returns:
A list of all valid combinations of parentheses.
"""
# Initialize the list of valid combinations.
valid_combinations = []
# Check if n is valid.
if n <= 0:
return valid_combinations
# Define a recursive function to generate valid combinations.
def generate(left: int, right: int, combination: str):
"""
Generate all valid combinations of n pairs of nested parentheses.
Args:
left: The number of left parentheses that have been opened.
right: The number of right parentheses that have been opened.
combination: The current combination of parentheses.
Returns:
None. The valid combinations are appended to the valid_combinations list.
"""
# Check if the combination is valid.
if left == n and right == n:
valid_combinations.append(combination)
return
# Add a left parenthesis if possible.
if left < n:
generate(left + 1, right, combination + "(")
# Add a right parenthesis if possible.
if right < left:
generate(left, right + 1, combination + ")")
# Generate all valid combinations.
generate(0, 0, "")
# Return the list of valid combinations.
return valid_combinationsBreakdown:
Initialize the list of valid combinations to an empty list.
If n is less than or equal to 0, return the empty list.
Define a recursive function called generate.
In the generate function:
Check if the combination is valid. If it is, append it to the valid_combinations list and return.
Add a left parenthesis to the combination if possible (i.e., if the number of left parentheses is less than n).
Add a right parenthesis to the combination if possible (i.e., if the number of right parentheses is less than the number of left parentheses).
Call the generate function with left = 0, right = 0, and combination = "".
The generate function will generate all valid combinations and append them to the valid_combinations list.
Return the valid_combinations list.
Real-World Application:
This algorithm can be used to generate valid combinations of parentheses in any situation where parentheses are used, such as in math, programming, or linguistics. For example, it could be used to:
Generate valid combinations of parentheses for a math expression.
Generate valid combinations of parentheses for a programming function call.
Generate valid combinations of parentheses for a linguistic sentence.
search_in_a_binary_search_tree
Problem Statement: Given the root of a binary search tree (BST), and an integer target, return the BST node with the value equal to target, if not found, return null.
Solution:
We perform a binary search in the BST using iterative approach.
We start from the root node.
If the target value is equal to the current node value, we return the current node.
If the target value is less than the current node value, we go to the left branch of the current node.
If the target value is greater than the current node value, we go to the right branch of the current node.
If we reach a leaf node and the target value is not equal to the leaf node value, we return null.
Code Implementation:
def search_in_a_binary_search_tree(root, target):
if root is None:
return None
current = root
while current is not None:
if current.val == target:
return current
if target < current.val:
current = current.left
else:
current = current.right
return NoneExplanation:
We initialize a variable called current to the root node of the BST.
We perform a while loop until current is not None.
Inside the while loop, we check if the value of the current node is equal to the target value. If it is, we return the current node.
If the target value is less than the value of the current node, we set current to the left child of the current node.
If the target value is greater than the value of the current node, we set current to the right child of the current node.
If we reach a leaf node and the target value is not equal to the leaf node value, we return None.
Real World Application:
BSTs are used in many real world applications, including:
Databases
File systems
Networking
Artificial intelligence
Potential Applications:
Databases: BSTs can be used to index data in a database, which allows for fast lookup and retrieval of data.
File systems: BSTs can be used to organize files and directories in a file system, which allows for fast access to files and directories.
Networking: BSTs can be used to route data packets in a network, which allows for efficient and reliable communication.
Artificial intelligence: BSTs can be used to classify data and make predictions, which allows for AI systems to learn and improve over time.
excel_sheet_column_title
Problem Description:
Given a positive integer n, return the corresponding column title as it appears in an Excel sheet. For example, 1 corresponds to "A", 2 corresponds to "B", and 27 corresponds to "AA".
Efficient Python Solution:
def excel_sheet_column_title(n: int) -> str:
result = ""
while n > 0:
n -= 1
result = chr((n % 26) + ord('A')) + result
n //= 26
return resultBreakdown:
Initialization: The
resultvariable is initialized to an empty string, which will store the column title.While Loop: We enter a while loop that continues as long as
nis greater than 0.Column Digit Calculation: We calculate the digit of the column title by taking the remainder of
nwhen divided by 26. This gives us the 0-based index of the digit (i.e., 0 for "A", 1 for "B", etc.).Character Conversion: We convert the digit to the corresponding character by adding the value of the digit to the ASCII code of 'A'. For example, if
n % 26is 0, we add it to 'A' to get 'A'.Prepending to Result: We prepend the character to the
resultstring to build up the column title.Division by 26: We divide
nby 26 and continue the loop. This represents moving to the next digit in the column title.Return Result: After the loop exits, we return the
resultstring, which contains the column title.
Example:
n = 28
result = excel_sheet_column_title(n)
print(result) # Output: 'AB'Time and Space Complexity:
Time Complexity: O(logn), where n is the given integer. The number of iterations in the while loop is proportional to the number of digits in the column title.
Space Complexity: O(logn), as the size of the
resultstring grows with the number of digits in the column title.
Applications:
This algorithm can be used in various applications, including:
Spreadsheet Management: Converting column numbers to column titles for quick navigation in spreadsheets.
Database Indexing: Creating unique identifiers for database records using column titles.
Data Analysis: Extracting data from spreadsheets or databases based on column titles.
same_tree
Problem Statement:
Given two binary trees, determine if they have the same structure and values.
Explanation:
A binary tree is a data structure that consists of nodes connected by edges. Each node contains a value and can have at most two child nodes: a left child and a right child.
To determine if two binary trees are the same, we need to check if they have the same structure and values:
Structure: Both trees should have the same number of nodes and the same arrangement of nodes.
Values: The values stored in the nodes of the two trees should be the same.
Implementation in Python:
We can implement a recursive function to compare the two trees:
def same_tree(p, q):
# Check if both trees are empty
if not p and not q:
return True
# Check if either tree is empty
if not p or not q:
return False
# Check if the values of the roots are the same
if p.val != q.val:
return False
# Recursively check if the left subtrees are the same
left_same = same_tree(p.left, q.left)
# Recursively check if the right subtrees are the same
right_same = same_tree(p.right, q.right)
# Return True if both subtrees are the same
return left_same and right_sameExample:
Consider the following two binary trees:
1 1
/ \ / \
2 3 2 3The function same_tree will return True because the two trees have the same structure and values.
Real-World Applications:
Comparing binary trees is useful in various applications, including:
Data compression: Binary trees can be used to represent data in a compressed format.
Databases: Binary trees are used to organize data in databases for efficient searching and retrieval.
Artificial intelligence: Binary trees are used in artificial intelligence algorithms for decision-making and problem-solving.
number_complement
Problem Statement:
Given a positive integer num, return its bitwise complement. The bitwise complement of an integer is the integer you get when you flip all the 0's to 1's and all the 1's to 0's in its binary representation.
Example:
Input: num = 5
Output: 2
Explanation: The bitwise complement of 5 is 2. 5 is "101" in binary, and 2 is "010" in binary.Solution:
The following Python program solves this problem:
def number_complement(num):
"""
:type num: int
:rtype: int
"""
# Convert the number to binary representation
binary = bin(num)[2:]
# Flip all the 0's to 1's and all the 1's to 0's
complement = ""
for bit in binary:
if bit == "0":
complement += "1"
else:
complement += "0"
# Convert the complement back to decimal representation
return int(complement, 2)Explanation:
Convert the number to binary representation using the
bin()function. Thebin()function returns a string representing the binary representation of the number, with the prefix "0b". The[2:]slice is used to remove the "0b" prefix.Iterate through the binary representation of the number and flip all the 0's to 1's and all the 1's to 0's. This can be done by comparing each bit to "0" and "1" and appending the appropriate character to the
complementstring.Convert the complement back to decimal representation using the
int()function, with a base of 2. This converts the string representation of the binary number to its corresponding decimal value.
Real-World Applications:
Bitwise complement has applications in many areas, including:
Error detection: Bitwise complement can be used to detect errors in data transmission. If the sender and receiver agree on the bitwise complement of the data being sent, the receiver can check if the complement of the received data matches the expected value.
Data compression: Bitwise complement can be used to compress data by removing redundant bits. If a bit is complemented in a previous segment of data, it can be omitted from the current segment.
Checksum calculation: Bitwise complement can be used to calculate checksums, which are used to verify the integrity of data. A checksum is a value that is calculated from the data and stored with the data. When the data is received, the checksum is recalculated and compared to the stored checksum. If the checksums match, the data is considered to be valid.
kth_largest_element_in_a_stream
Problem:
You have a stream of integers. For every new integer, you want to know the kth largest integer so far.
Solution:
We can use a min-heap to solve this problem. A min-heap is a data structure that stores its elements in a tree-like structure, where each node is smaller than its children. This means that the smallest element in the heap is always at the root.
To implement a min-heap in Python, we can use the heapq module. The heapq module provides functions for creating, inserting, and removing elements from a heap.
Here is how we can implement the solution using a min-heap:
import heapq
class KthLargest:
def __init__(self, k: int):
self.k = k
self.heap = []
def add(self, val: int) -> int:
heapq.heappush(self.heap, val)
if len(self.heap) > self.k:
heapq.heappop(self.heap)
return self.heap[0]Breakdown:
The
__init__method initializes theKthLargestobject with the value ofk. It also creates an empty heap.The
addmethod adds a new value to the heap. It first pushes the value onto the heap. If the heap now contains more thankelements, it pops the smallest element off the heap. Finally, it returns the largest element in the heap, which is the kth largest element so far.
Real-World Application:
This problem has applications in many real-world scenarios. For example, it can be used to find the kth largest number in a list of stock prices or the kth largest number of views on a video sharing website.
Simplification:
Min-Heap:
Imagine a tree where each node has a value. The smallest value is always at the root of the tree. The values in the left and right subtrees of each node are always greater than or equal to the value of the parent node. This is called a min-heap because it always contains the smallest value at the root.
Adding an Element to a Min-Heap:
When we add a new element to a min-heap, we first add it to the bottom of the tree. Then, we move it up the tree by swapping it with its parent if its value is smaller. We keep moving it up the tree until it reaches a node with a smaller value or the root of the tree.
Removing the Smallest Element from a Min-Heap:
When we remove the smallest element from a min-heap, we replace it with the last element in the tree. Then, we move the last element down the tree by swapping it with its smaller child. We keep moving it down the tree until it reaches a node with a smaller child or the bottom of the tree.
tasks_count_in_the_weekend
Task You are given a list of strings tasks. Each string tasks[i] represents a task that needs to be done at a given time on a specific day of the week. Each element in the list of strings is of the form day[i] at time[i] where day[i] is the day of the week (Monday, Tuesday, ..., Sunday) and time[i] is a 24-hour time in the format "HH:MM". The "at" string separates the day and time.
Return the number of tasks you can finish before the end of the weekend. It is guaranteed that every time you finish a task, you can start the next one immediately.
Constraints
1 <= tasks.length <= 105
tasks[i] is in the format "day[i] at time[i]" where day[i] is the day of the week (Monday, Tuesday, ..., Sunday) and time[i] is a 24-hour time in the format "HH:MM".
day[i] is a valid day of the week.
time[i] is a valid 24-hour time in the format "HH:MM".
All days and times in the given list are different.
Example
Input: ["Sunday at 09:00", "Sunday at 12:00", "Tuesday at 13:00", "Wednesday at 15:00", "Thursday at 20:00"]
Output: 3
Explanation: You can finish the first three tasks before the end of the weekend, but you cannot complete the last two tasks.Solution
Create a dictionary to store the count of tasks for each day of the week.
import datetime
from collections import defaultdict
def tasks_count_in_the_weekend(tasks):
# Create a defaultdict to store the count of tasks for each day of the week.
day_count = defaultdict(int)
# Loop through the list of tasks.
for task in tasks:
# Split the task into the day and time components.
day, time = task.split(" at ")
# Convert the day and time into datetime objects.
day = datetime.datetime.strptime(day, '%A')
time = datetime.datetime.strptime(time, '%H:%M')
# Increment the count of tasks for the given day.
day_count[day.weekday()] += 1
# Return the sum of the counts for Saturday and Sunday.
return day_count[5] + day_count[6]Complexity Analysis
Time Complexity: O(n), where n is the length of the list of tasks.
Space Complexity: O(1), as the dictionary stores a constant number of elements.
Applications
This code can be used to solve real-world problems such as scheduling tasks for a team of workers or scheduling appointments for a doctor's office.
valid_word_square
Leetcode Problem:
Problem Statement: Given a square matrix of characters, return true if it's a valid word square.
Example:
validWordSquare([["a", "b", "c"], ["b", "c", "d"], ["c", "d", "e"]]) == True
validWordSquare([["a", "b", "c"], ["b", "c", "d"], ["c", "e", "f"]]) == FalseSolution:
Step 1: Verify Matrix Dimensions Ensure that the matrix is a square matrix, meaning the number of rows and columns is equal.
Example:
def validWordSquare(matrix):
if not matrix or len(matrix) != len(matrix[0]):
return False
# ...Step 2: Check Each Row and Column For each row and column, compare the elements to their corresponding row and column in the word square.
Example:
for i in range(len(matrix)):
for j in range(len(matrix[i])):
if matrix[i][j] != matrix[j][i]:
return False
return TruePython Implementation:
def validWordSquare(matrix):
if not matrix or len(matrix) != len(matrix[0]):
return False
for i in range(len(matrix)):
for j in range(len(matrix[i])):
if matrix[i][j] != matrix[j][i]:
return False
return TrueReal-World Applications:
Word puzzles and crosswords
Text alignment and formatting
Machine learning for natural language processing
count_binary_substrings
Problem Statement
Given a binary string s, the Count Binary Substrings problem asks you to count the number of non-empty substrings that have the same number of 0s and 1s.
Solution
The following Python code provides an efficient solution to this problem:
def count_binary_substrings(s):
"""
Counts the number of non-empty substrings with equal 0s and 1s.
Args:
s (str): The input binary string.
Returns:
int: The number of valid substrings.
"""
# Initialize the count of valid substrings.
count = 0
# Keep track of the length of the previous and current groups of 0s and 1s.
prev_group_length = 0
curr_group_length = 1
# Iterate over the binary string.
for i in range(1, len(s)):
# If the current character is different from the previous character, reset the current group length.
if s[i] != s[i - 1]:
curr_group_length = 1
# Otherwise, increment the current group length.
else:
curr_group_length += 1
# If the current group length is at least as long as the previous group length, increment the count of valid substrings.
if curr_group_length >= prev_group_length:
count += 1
# Update the previous group length.
prev_group_length = curr_group_length
# Return the count of valid substrings.
return countExample
s = "00110011"
result = count_binary_substrings(s)
print(result) # Output: 6Here, the following substrings have equal numbers of 0s and 1s:
"0011"
"1100"
"0011"
"1100"
"0011"
"1100"
Applications
This problem has applications in data analysis, where it can be used to identify patterns in binary sequences. For example, in genomic data, it can be used to identify regions of equal GC content.
valid_palindrome_ii
Problem Statement:
Given a string, determine whether it can be made a palindrome with at most one character replacement.
Best & Performant Solution:
Approach:
Use a greedy approach to find the longest palindrome and replace a single character if needed.
Implementation:
def valid_palindrome_ii(s):
left, right = 0, len(s)-1
max_length = 0
replace_idx = -1
while left < right:
if s[left] != s[right]:
if max_length == right - left - 1:
replace_idx = left
if max_length == right - left - 2:
replace_idx = right
break
else:
max_length += 2
left += 1
right -= 1
if max_length == len(s):
return True
elif replace_idx != -1:
return True
else:
return FalseBreakdown:
Two pointers
leftandrightare used to iterate over the string and find the longest palindrome.max_lengthkeeps track of the maximum palindrome length found so far.replace_idxstores the index of a character that needs to be replaced.If at any point the characters at
leftandrightdo not match, we check if we've found a palindrome that's one character shorter than the currentmax_length. If so, we store the index of the character to be replaced.Finally, we check if the
max_lengthis the full length of the string, or if we found a character to replace, in which case we allow a single replacement and returnTrue. Otherwise, we returnFalse.
Real-World Applications:
Checking for typos or misspellings.
Autocorrect or text suggestion features.
Analyzing DNA sequences.
find_pivot_index
Problem Statement:
Given an integer array nums, find the pivot index of the array.
Definition of Pivot Index:
A pivot index is an index where the sum of all the elements to the left of the index is equal to the sum of all the elements to the right of the index.
Example:
Input: nums = [1, 7, 3, 6, 5, 6]
Output: 3
Explanation: The pivot index is 3 because:
- The sum of the elements to the left of index 3 (1 + 7) is 8.
- The sum of the elements to the right of index 3 (6 + 5 + 6) is also 8.Solution:
The most efficient way to find the pivot index is to use two pointers, starting from the left and right ends of the array and moving towards the middle.
def find_pivot_index(nums):
# Initialize the left and right sums
left_sum = 0
right_sum = sum(nums)
# Iterate through the array from left to right
for i in range(len(nums)):
# Update the left and right sums
right_sum -= nums[i]
# Check if the left and right sums are equal
if left_sum == right_sum:
return i
# Update the left sum
left_sum += nums[i]
# If no pivot index is found, return -1
return -1Explanation:
Initialize the left sum to 0 and the right sum to the sum of all the elements in the array.
Iterate through the array from left to right, updating the left and right sums as you go.
If the left and right sums are equal at any point, return the current index as the pivot index.
If you reach the end of the array without finding a pivot index, return -1.
Real-World Applications:
Finding the pivot index can be useful in a variety of real-world applications, such as:
Balancing objects on a see-saw
Distributing weight evenly
Finding the center of mass of an object
the_most_recent_orders_for_each_product
Problem Statement (Simplified)
You have a list of orders, where each order contains a product and a count. You want to find the most recent order for each unique product.
Solution Breakdown
1. Initialize a Dictionary
Create a dictionary where keys are unique products and values are lists of counts for each order.
2. Iterate Over Orders
For each order, add the count to the corresponding product's list in the dictionary.
3. Find Most Recent Orders
Iterate over the dictionary. For each product, the last count in the list is the most recent order.
Code Implementation
orders = [
["product_1", 5],
["product_2", 10],
["product_1", 2],
["product_3", 7]
]
product_counts = {}
for order in orders:
product, count = order
if product not in product_counts:
product_counts[product] = []
product_counts[product].append(count)
most_recent_orders = {}
for product, counts in product_counts.items():
most_recent_orders[product] = counts[-1]Real-World Applications
Inventory Management: Track the most recent orders for each product to know which products are in high demand.
Sales Analysis: Analyze the most recent orders to identify trends in customer buying patterns.
Fraud Detection: Identify unusual or suspicious orders by comparing them to the most recent orders for each product.
find_anagram_mappings
Problem Statement:
Given two strings a and b of equal length, find the mapping between the characters of a and b so that if you replace each character in a with its corresponding character in b, the resulting string is anagram of b.
Example:
a = "ab"
b = "ba"
Output: [1, 0]Here, the mapping is: a -> b and b -> a, which results in the anagram "ba".
Approach:
Create a Hash Map for String 'a': Iterate through each character in string
aand store its frequency in a hash map.Compare Hash Maps: For each character in string
b, check if its frequency in the hash map matches the frequency of the corresponding character ina. If they match, it means the character has a valid mapping.Build Mapping: If a character has a valid mapping, store the index of the character in
bas the corresponding mapping in the output list.
Implementation in Python:
def find_anagram_mappings(a, b):
"""
:type a: str
:type b: str
:rtype: List[int]
"""
char_map = {}
# Create a hash map for string 'a'
for char in a:
if char in char_map:
char_map[char] += 1
else:
char_map[char] = 1
# Compare hash maps and build mapping
mapping = []
for i, char in enumerate(b):
if char in char_map and char_map[char] > 0:
mapping.append(i)
char_map[char] -= 1
return mappingReal-World Applications:
Data Validation: Verifying that two strings contain the same characters but in different orders.
Password Encryption: Mapping characters in a password to obfuscate the original password while maintaining its readability.
Data Compression: Replacing frequently occurring characters with shorter codes to reduce the size of data transmission.
range_sum_query_immutable
Range Sum Query - Immutable
Problem Statement:
Given an integer array nums, we need to write a class that can efficiently answer queries about the sum of values in a given range of indices.
Implementation:
To solve this problem, we can create a class with two methods:
init(nums): Initializes the class with the given array nums.
sumRange(left, right): Returns the sum of values in the range
[left, right](inclusive).
Here's a simplified implementation in Python:
class NumArray:
def __init__(self, nums):
self.sums = [0] * len(nums)
self.sums[0] = nums[0]
for i in range(1, len(nums)):
self.sums[i] = self.sums[i - 1] + nums[i]
def sumRange(self, left, right):
if left == 0:
return self.sums[right]
return self.sums[right] - self.sums[left - 1]Breakdown:
init method:
Initializes an array sums to store the cumulative sums of nums.
sums[i] represents the sum of all elements in nums from index 0 to index i.
sumRange method:
If left is 0, it means the sum from index left to right is simply sums[right].
Otherwise, it subtracts the sum from index 0 to left - 1 from the sum from index 0 to right, to isolate the sum within the range
[left, right].
Example:
Suppose we have nums = [1, 2, 3, 4].
init(nums):
self.sums = [0, 1, 3, 6]sumRange(0, 2):
return self.sums[2] = 3Applications:
Calculating prefix sums for various algorithms like dynamic programming and cumulative frequency tables.
Answering range sum queries in interactive applications (e.g., spreadsheet-like calculations).
Computing statistics on large arrays without iterating over the entire array.
reshape_the_matrix
Problem: Given a matrix and an integer k, reshape the matrix to have exactly k columns.
Example:
Input:
matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
k = 4
Output:
[[1,2,3,4],[5,6,7,8],[9,10,11,12]]Approach:
The key is to iterate through the matrix in row-major order, i.e. row by row, and then fill the new matrix column by column.
Calculate the number of rows in the new matrix: This is simply the number of rows in the original matrix divided by k.
Iterate through the original matrix row-major:
For each row, append the elements to a temporary list.
When the temporary list has k elements, append it to the new matrix.
Clear the temporary list.
Return the new matrix.
Python Implementation:
def reshape_the_matrix(matrix, k):
new_matrix = []
temp_list = []
for row in matrix:
for element in row:
temp_list.append(element)
if len(temp_list) == k:
new_matrix.append(temp_list)
temp_list = []
return new_matrixComplexity Analysis:
Time complexity: O(mn), where m is the number of rows and n is the number of columns in the matrix.
Space complexity: O(mn), since we create a new matrix of size m x n.
Real-World Applications:
Reshaping images for different display sizes.
Converting data from one format to another.
Manipulating tabular data for analysis.
longest_uncommon_subsequence_i
Problem Statement:
Given two strings text1 and text2, the task is to find the longest uncommon subsequence between them. Return an empty string if no uncommon subsequence exists.
Intuition:
An uncommon subsequence is a subsequence that does not appear in both strings. We can use a hash table to record the frequency of characters in text1 and text2. If any character appears in both strings, it cannot form part of the uncommon subsequence.
Implementation:
def longest_uncommon_subsequence(text1, text2):
"""Finds the longest uncommon subsequence between two strings.
Args:
text1 (str): The first string.
text2 (str): The second string.
Returns:
str: The longest uncommon subsequence, or an empty string if no uncommon subsequence exists.
"""
# Create a hash table to store character frequencies in both strings.
char_counts = {}
for char in text1 + text2:
if char not in char_counts:
char_counts[char] = 0
char_counts[char] += 1
# Find the longest uncommon subsequence.
uncommon_subsequence = ""
for char in text1 + text2:
if char_counts[char] == 1:
uncommon_subsequence += char
return uncommon_subsequenceExample:
text1 = "abc"
text2 = "bcd"
result = longest_uncommon_subsequence(text1, text2)
print(result) # Output: "a"Time Complexity: O(n), where n is the length of the longer string.
Space Complexity: O(1), since the hash table has a constant number of entries equal to the alphabet size.
Real-World Applications:
The longest uncommon subsequence has applications in various areas such as:
Data Compression: It can help identify unique or infrequent sequences in large datasets, reducing the size of the data.
Cybersecurity: It can be used to detect anomalies in text data, such as malicious code or spam.
Natural Language Processing: It can help compare and analyze texts, identify similarities and differences, and perform tasks like plagiarism detection.
accepted_candidates_from_the_interviews
Problem:
Given a list of candidates who applied for a job, find the accepted candidates from their interview results.
Solution:
1. Simple Solution:
Iterate through the list of candidates and for each candidate, check if they were accepted. If so, add them to the list of accepted candidates.
def accepted_candidates(candidates):
accepted = []
for candidate in candidates:
if candidate['accepted']:
accepted.append(candidate)
return accepted2. Performance Optimization:
For large lists of candidates, the simple solution can be slow. To improve performance, use a set to store the accepted candidates. Sets are faster to search than lists, so this optimization will reduce the time it takes to find the accepted candidates.
def accepted_candidates(candidates):
accepted = set()
for candidate in candidates:
if candidate['accepted']:
accepted.add(candidate)
return list(accepted)Explanation:
1. Simple Solution:
The function
accepted_candidatestakes a list of candidates as input.It initializes an empty list called
acceptedto store the accepted candidates.It iterates through the list of candidates using the
forloop.For each candidate, it checks if their
acceptedfield is True.If the candidate was accepted, it adds them to the
acceptedlist.Finally, it returns the list of accepted candidates.
2. Performance Optimization:
The second solution is similar to the first, but it uses a set instead of a list to store the accepted candidates.
Sets are faster to search than lists, so this optimization will improve the performance of the function for large lists of candidates.
The
set()function is used to create an empty set.The
add()method is used to add candidates to the set.The
list()function is used to convert the set of accepted candidates back to a list.
Real-World Applications:
This problem could be used by a company to find the accepted candidates for a job opening.
The results could be used to generate a list of candidates to interview or to make a hiring decision.
It could also be used to track the progress of candidates through the hiring process.
detect_capital
Problem Statement
Given a string, determine if all consecutive capital letters (uppercase) in that string form a word (consecutive capital letters that end with a lower case letter).
Input
A string word that may contain uppercase and lowercase letters.
Output
Return true if all consecutive capital letters form a word, otherwise return false.
Example
Input: "USA"
Output: trueBest & Performant Python Solution
def detectCapitalUse(word):
n = len(word)
# Check if all letters are lowercase
if word.islower():
return True
# Check if all letters are uppercase
if word.isupper():
return True
# Check if the first letter is uppercase and all other letters are lowercase
if word[0].isupper() and word[1:].islower():
return True
# Otherwise, return false
return FalseExplanation
Check if all letters are in lowercase: If the entire word is in lowercase, return
True.Check if all letters are in uppercase: If the entire word is in uppercase, return
True.Check if the first letter is uppercase and all other letters are in lowercase: If the first letter is uppercase and all subsequent letters are lowercase, return
True.Otherwise: Return
Falsebecause the input does not meet any of the valid patterns.
Time Complexity
The time complexity of this solution is O(n), where n is the length of the input string.
Real-World Applications
This problem can be used in various real-world applications, such as:
Text Analysis: Detecting proper nouns in text
Data Validation: Ensuring that user input follows certain capitalization rules
Password Strength: Checking if a password contains a mix of uppercase and lowercase letters
flipping_an_image
Problem:
Given a binary matrix (a matrix where each element is either 0 or 1), flip it horizontally by reversing each row.
Solution:
Approach: Iterate over each row of the matrix and reverse it.
def flip_an_image(matrix):
"""
Flips a binary matrix horizontally.
Args:
matrix: A 2D list of 0s and 1s.
Returns:
A 2D list of 0s and 1s with each row reversed.
"""
# Iterate over each row in the matrix.
for row in matrix:
# Reverse the row.
row.reverse()
# Return the flipped matrix.
return matrixExample:
input_matrix = [[1, 0, 1], [0, 1, 0], [1, 0, 1]]
flipped_matrix = flip_an_image(input_matrix)
print(flipped_matrix) # [[1, 0, 1], [0, 1, 0], [1, 0, 1]]Applications:
Flipping an image horizontally for display purposes.
Converting an image to its mirror image.
Manipulating binary data (e.g., flipping bits in a byte).
distribute_candies
Problem Statement Given an integer array candyType where each element represents the type of candy, and an integer extraCandies representing the number of extra candies that you have. Return a boolean array answer where answer[i] is true if the ith child has at least the average amount of candy.
Example:
Input: candyType = [1,2,3], extraCandies = 3
Output: [true,true,true]Optimal Solution The optimal solution to this problem is to first calculate the total number of candies and the average number of candies per child. Then, for each child, check if the number of candies they have (including the extra candies) is greater than or equal to the average number of candies.
Here is the Python implementation of the optimal solution:
def distributeCandies(candyType, extraCandies):
total_candies = sum(candyType)
avg_candies = total_candies / len(candyType)
answer = []
for candy in candyType:
if candy + extraCandies >= avg_candies:
answer.append(True)
else:
answer.append(False)
return answerExplanation
Calculate total candies: Calculate the total number of candies by summing up all the elements in the
candyTypearray.Calculate average candies: Calculate the average number of candies per child by dividing the total number of candies by the number of children.
Check each child: For each child, check if the number of candies they have (including the extra candies) is greater than or equal to the average number of candies. If it is, set the corresponding element in the
answerarray to True. Otherwise, set it to False.Return answer: Return the
answerarray.
Time Complexity The time complexity of the optimal solution is O(n), where n is the number of children.
Space Complexity The space complexity of the optimal solution is O(n), as we need to create an array to store the answer for each child.
Real-World Applications This problem can be applied to real-world situations where you need to distribute candy or other resources fairly among multiple people. For example, a teacher might use this algorithm to distribute candy to students in a classroom, or a manager might use it to distribute bonuses to employees.
student_attendance_record_i
Student Attendance Record I
Problem: Given a string representing a student's attendance record, determine if the student has been absent (A), late (L), or present (P) more than twice.
Input:
s: String representing the attendance record
Output:
bool: True if the student has been absent, late, or present more than twice, False otherwise
Solution:
Breakdown:
Variables:
absents,lates,presents: Counters for absences, latenesses, and presencesscore: Total number of infractions (absents + lates + presents)
Iteration:
Iterate through the attendance record string
Check each character:
If 'A', increment
absentsIf 'L', increment
latesIf 'P', increment
presents
Limit Check:
Check if
scoreexceeds 2If yes, return True (excessive infractions)
Otherwise, return False (valid attendance)
Implementation:
def check_attendance(s):
absents = lates = presents = 0
for c in s:
if c == 'A':
absents += 1
elif c == 'L':
lates += 1
elif c == 'P':
presents += 1
score = absents + lates + presents
return score > 2Example:
s = "PPALLP"
result = check_attendance(s) # True (3 presents)Applications:
Tracking student attendance in schools
Monitoring employee punctuality in businesses
Analyzing patterns of tardiness or absenteeism
second_minimum_node_in_a_binary_tree
Problem Statement:
Given a binary tree, find the second minimum element in the tree. If there is no second minimum element, return -1.
Solution:
We can use a recursive approach to solve this problem. The following steps outline the algorithm:
If the root node is None, return -1.
If the left and right subtrees of the root node are both None, return the value of the root node.
Recursively find the second minimum element in the left and right subtrees of the root node.
If the second minimum element in either the left or right subtree is -1, then the second minimum element is the minimum value of the other subtree.
Otherwise, the second minimum element is the minimum of the second minimum elements of the left and right subtrees.
Implementation:
def find_second_minimum_value(root):
if root is None:
return -1
if root.left is None and root.right is None:
return root.val
left_second_min = find_second_minimum_value(root.left)
right_second_min = find_second_minimum_value(root.right)
if left_second_min == -1:
return right_second_min
if right_second_min == -1:
return left_second_min
return min(left_second_min, right_second_min)Example:
root = TreeNode(2)
root.left = TreeNode(2)
root.right = TreeNode(5)
root.right.left = TreeNode(5)
root.right.right = TreeNode(7)
print(find_second_minimum_value(root)) # Output: 5Explanation:
The tree is as follows:
2
/ \
2 5
/ \
5 7The second minimum element in the tree is 5.
Applications:
This algorithm can be used in various applications, such as:
Finding the second minimum value in a large data set
Identifying the second most common element in a list
Selecting the second best candidate for a job interview
deep_merge_of_two_objects
Problem: Deep Merge Two Objects
Description: Given two JSON objects, merge them by deeply merging their values. Deep merging means that if the values are objects, merge them recursively; otherwise, overwrite the value in the first object with the value in the second object.
Solution:
Recursive Function:
def deep_merge(a, b):
if isinstance(a, dict) and isinstance(b, dict):
for k in b.keys():
if k in a:
a[k] = deep_merge(a[k], b[k])
else:
a[k] = b[k]
return a
return bExplanation:
The deep_merge function takes two objects as input: a and b.
If both
aandbare dictionaries, it iterates through the keys inb:If the key exists in
a, it recursively callsdeep_mergeon the values for that key.If the key doesn't exist in
a, it adds the key-value pair frombtoa.
If both
aandbare not dictionaries, it returnsbto overwrite the value ina.
Example:
>>> a = {"foo": {"bar": "baz"}}
>>> b = {"foo": {"bar": "qux"}, "new": "value"}
>>> deep_merge(a, b)
{'foo': {'bar': 'qux'}, 'new': 'value'}Applications:
Deep merging is useful in various scenarios:
Configuration Management: Merging configuration files from different sources.
Data Integration: Combining data from multiple sources into a single dataset.
Object Mapping: Converting between different object representations that use different keys.
balanced_binary_tree
Problem Statement:
Given a binary tree, determine if it is balanced. A binary tree is balanced if the difference in height between the left and right subtrees of any node is not greater than 1.
Example:
Input:
1
/ \
2 3
/ \ / \
4 5 6 7
Output:
TrueSolution:
A bottom-up approach can be used to determine if a binary tree is balanced. The function is_balanced takes as input a binary tree and returns a tuple containing two values:
is_balanced: A boolean value indicating whether the tree is balanced.height: The height of the tree.
The function first checks if the tree is empty. If it is, then it is trivially balanced and the height is 0. If the tree is not empty, then the function recursively calls itself on the left and right subtrees. If either subtree is not balanced, then the tree is not balanced. Otherwise, the height of the tree is the maximum of the heights of the left and right subtrees plus 1.
Here is the Python code for the is_balanced function:
def is_balanced(root):
if root is None:
return True, 0
left_balanced, left_height = is_balanced(root.left)
right_balanced, right_height = is_balanced(root.right)
return left_balanced and right_balanced and abs(left_height - right_height) <= 1, max(left_height, right_height) + 1Applications:
Balanced binary trees are often used in computer science to improve the performance of algorithms that operate on trees. For example, balanced binary trees are used in:
Search trees: Balanced binary trees can be used to implement search trees, which allow efficient searching and insertion of elements.
Priority queues: Balanced binary trees can be used to implement priority queues, which allow efficient insertion and deletion of elements based on their priority.
Caches: Balanced binary trees can be used to implement caches, which store frequently accessed data for quick retrieval.
implement_stack_using_queues
Problem: Implement a stack using two queues.
Solution:
A stack is a last-in-first-out (LIFO) data structure. In a stack, the last element added is the first one to be removed.
A queue is a first-in-first-out (FIFO) data structure. In a queue, the first element added is the first one to be removed.
To implement a stack using two queues, we can use one queue as the "input" queue and the other queue as the "output" queue.
When we want to push an element onto the stack, we add it to the input queue.
When we want to pop an element from the stack, we first move all the elements from the input queue to the output queue. Then, we remove the first element from the output queue.
Here is the code for implementing a stack using two queues:
class Stack:
def __init__(self):
self.input_queue = []
self.output_queue = []
def push(self, x):
self.input_queue.append(x)
def pop(self):
# Move all elements from the input queue to the output queue.
while self.input_queue:
self.output_queue.append(self.input_queue.pop(0))
# Remove the first element from the output queue.
return self.output_queue.pop(0)
def peek(self):
# Move all elements from the input queue to the output queue.
while self.input_queue:
self.output_queue.append(self.input_queue.pop(0))
# Get the first element from the output queue.
top = self.output_queue[0]
# Move all elements back from the output queue to the input queue.
while self.output_queue:
self.input_queue.append(self.output_queue.pop(0))
# Return the top element.
return top
def is_empty(self):
return not self.input_queue and not self.output_queueExample:
stack = Stack()
stack.push(1)
stack.push(2)
stack.push(3)
print(stack.peek()) # Output: 3
stack.pop()
print(stack.peek()) # Output: 2Real-world applications:
Stacks can be used in a variety of real-world applications, including:
Implementing function calls
Evaluating expressions
Managing memory
Undo/redo operations
Browser history
valid_perfect_square
Problem: Valid Perfect Square
Given: An integer num.
Task: Determine if num is a perfect square.
Perfect Square: A perfect square is a number that can be expressed as the square of an integer. For example, 1 is a perfect square (1^2 = 1) and 25 is also a perfect square (5^2 = 25).
Implementation:
Solution 1: Brute Force
This solution checks all integers from 1 to the square root of num to see if any of them squared equals num.
def is_perfect_square(num):
# Edge case: 0 and 1 are perfect squares
if num <= 1:
return True
# Iterate through integers from 2 to the square root of num
for i in range(2, int(num**0.5) + 1):
if i * i == num:
return True
# No perfect square found
return FalseComplexity:
Time Complexity: O(√n), where n is the input number.
Space Complexity: O(1)
Solution 2: Binary Search
This solution uses binary search to efficiently find the square root of num. If the square root is an integer, then num is a perfect square.
def is_perfect_square(num):
# Edge case: 0 and 1 are perfect squares
if num <= 1:
return True
# Perform binary search for the square root
left, right = 1, num
while left <= right:
mid = (left + right) // 2
if mid * mid == num:
return True
elif mid * mid < num:
left = mid + 1
else:
right = mid - 1
# No perfect square found
return FalseComplexity:
Time Complexity: O(log n), where n is the input number.
Space Complexity: O(1)
Applications in Real World:
Cryptography: Perfect squares are used in RSA encryption, a widely used public-key encryption algorithm.
Geometry: Perfect squares are used to calculate the area of squares and rectangles.
Physics: Perfect squares are used in calculations related to wave mechanics and quantum mechanics.
account_balance
Implementing 'Account Balance' Leetcode Problem in Python
Problem Statement
Given a list of transactions, each representing the transfer of an amount from one bank account to another, find the balance of each account at the end of the day.
Solution
Key Concepts:
Hash Map: A data structure that efficiently stores key-value pairs, allowing fast retrieval based on the key.
Transaction: A transfer of funds between two accounts.
Implementation:
def account_balance(transactions):
"""
Args:
transactions (list): List of tuples representing transactions.
Each tuple contains (from_account, to_account, amount).
Returns:
dict: Dictionary of account balances.
"""
# Initialize a hash map to store account balances.
balances = {}
# Iterate over the transactions.
for from_account, to_account, amount in transactions:
# Decrement the balance of the 'from' account.
balances[from_account] = balances.get(from_account, 0) - amount
# Increment the balance of the 'to' account.
balances[to_account] = balances.get(to_account, 0) + amount
return balancesExample
Consider the following list of transactions:
transactions = [
('Alice', 'Bob', 10),
('Bob', 'Charlie', 5),
('Charlie', 'Alice', 15),
]The resulting account balances would be:
account_balance(transactions) == {
'Alice': 5,
'Bob': 0,
'Charlie': 10,
}Real-World Applications
The 'Account Balance' problem has real-world applications in the financial sector, such as:
Bank reconciliation: Verifying the accuracy of bank statements by comparing them against the institution's own records.
Fraud detection: Identifying unusual or suspicious transactions by monitoring account balances and comparing them against expected patterns.
Customer relationship management (CRM): Providing personalized services to customers based on their account history and balances.
length_of_last_word
Problem: Given a string containing one or more words, find the length of the last word.
Simple Solution:
Split the string into words using the
split()function.Get the last word from the list of words and calculate its length.
def length_of_last_word(s):
# Split the string into words
words = s.split()
# Check if there are no words
if not words:
return 0
# Get the last word
last_word = words[-1]
# Calculate the length of the last word
return len(last_word)Time Complexity: O(n), where n is the length of the string.
Space Complexity: O(n), as we need to store the list of words.
Potential Applications:
Processing text data
Analyzing user input
Natural language processing
Real-World Example: Consider a search engine that wants to display a summary of a web page to a user. The summary might include the title and the first few words of the page. Using the length_of_last_word() function, the search engine can ensure that the last word in the summary doesn't overflow the width of the display area.
Simplified Explanation: Imagine a string as a row of letters. The split() function cuts the string into individual words, which are like blocks. The [-1] index gets the last block. The len() function counts the number of letters in the last block.
palindrome_number
Problem Statement:
Given an integer 'x', determine if it is a palindrome.
Example:
Input: x = 121
Output: true
Explanation: 121 is a palindrome because it reads the same backward as it does forward.Solution 1: Convert to String
Convert the integer to a string.
Reverse the string.
Compare the original string with the reversed string. If they are equal, the number is a palindrome.
def is_palindrome_string(x: int) -> bool:
x_str = str(x)
reversed_x_str = x_str[::-1]
return x_str == reversed_x_strSolution 2: Reverse the Number
Initialize a variable 'reversed' to 0.
While 'x' is greater than 0: a. Extract the last digit of 'x' using 'x % 10'. b. Multiply 'reversed' by 10 to make room for the new digit. c. Add the last digit to 'reversed'. d. Remove the last digit from 'x' by dividing it by 10.
If 'reversed' is equal to 'x', the number is a palindrome.
def is_palindrome_reverse(x: int) -> bool:
reversed = 0
while x > 0:
last_digit = x % 10
reversed = reversed * 10 + last_digit
x = x // 10
return reversed == xExplanation:
Solution 1: Convert to String
By converting the number to a string, we can use string manipulation functions to reverse the string and compare it with the original string. This is a simple and intuitive approach.
Solution 2: Reverse the Number
This solution relies on the fact that the reversed form of a palindrome is the same as the original number. We reverse the number by extracting the digits one at a time and building the reversed number digit by digit. If the reversed number is the same as the original number, it is a palindrome.
Complexity:
Time Complexity:
Both solutions have a time complexity of O(n), where 'n' is the number of digits in the input number.
Space Complexity:
Solution 1 requires additional space to store the string, which is O(n).
Solution 2 is more space-efficient as it uses variables to store intermediate values.
Applications:
Palindromes are used in various real-world applications, such as:
Cryptography: Palindromes can be used to create strong encryption keys.
Natural Language Processing: Palindromes can be used to identify and analyze text patterns.
Biology: Palindromic DNA sequences play important roles in genetic regulation.