ltc7

subtract_the_product_and_sum_of_digits_of_an_integer

Problem Statement:

Given an integer n, find the difference between its product and sum of digits.

Example:

Input: n = 234
Output: 15
Explanation: Product of digits = 2 * 3 * 4 = 24. Sum of digits = 2 + 3 + 4 = 9. Difference = 24 - 9 = 15.

Implementation in Python:

def subtract_product_and_sum(n):
  """
  Calculates the difference between the product and sum of digits of an integer.

  Args:
    n: The input integer.

  Returns:
    The difference between the product and sum of digits.
  """

  # Convert the integer to a string.
  n_str = str(n)

  # Initialize the product and sum of digits.
  product = 1
  sum = 0

  # Iterate over the digits of the string.
  for digit in n_str:
    # Convert the digit to an integer.
    int_digit = int(digit)

    # Multiply the product by the digit.
    product *= int_digit

    # Add the digit to the sum.
    sum += int_digit

  # Return the difference between the product and sum.
  return product - sum

Breakdown and Explanation:

Convert the integer to a string: This is necessary because we need to access the individual digits of the integer.
Initialize the product and sum of digits: We start with a product of 1 and a sum of 0.
Iterate over the digits of the string: We use a for loop to iterate over each character in the string.
Convert the digit to an integer: We convert the character to an integer using the int() function.
Multiply the product by the digit: We update the product by multiplying it with the current digit.
Add the digit to the sum: We update the sum by adding the current digit.
Return the difference between the product and sum: Finally, we return the difference between the product and sum.

Real-World Applications:

This algorithm can be used in various real-world applications, such as:

Checksum generation: Checksums are used to verify data integrity. The product and sum of digits can be used as a simple checksum.
Password validation: Some password validation rules may require that the password contains both digits and non-digits. The product and sum of digits can be used to check this.
Mathematical modeling: This algorithm can be used to solve mathematical problems involving the product and sum of digits.

final_prices_with_a_special_discount_in_a_shop

Problem Statement

You are given the final prices of some items in a shop. Each final price is discounted by the value of the next item in the array toward the right. However, if there is no next item to the right, the final price remains the same. Calculate the final prices for all items in the shop.

Example

Input: prices = [8, 4, 6, 2, 3] Output: [4, 2, 4, 2, 3]

Explanation:

The first item has no next item, so the final price remains 8.
The second item is discounted by the value of the third item (6), so the final price is 4.
The third item is discounted by the value of the fourth item (2), so the final price is 4.
The fourth item is discounted by the value of the fifth item (3), so the final price is 2.
The fifth item has no next item, so the final price remains 3.

Simplified Explanation

Imagine you are shopping in a grocery store. Each item on the shelf has a price tag. The store offers a special discount where each item's price is reduced by the amount of the next item's price.

For example, if you buy a loaf of bread that costs $5, and the next item on the shelf is a bag of chips that costs $2, then you will get a discount of $2 on the bread. So, you will only pay $3 for the bread.

Implementation

def final_prices_with_a_special_discount_in_a_shop(prices):
  """
  :type prices: List[int]
  :rtype: List[int]
  """
  # Initialize a stack to store the indices of the items.
  stack = []

  # Iterate over the prices.
  for i in range(len(prices)):
    # While the stack is not empty and the price of the current item is greater than or equal to the price of the item at the top of the stack, pop the top item.
    while stack and prices[i] <= prices[stack[-1]]:
      stack.pop()

    # If the stack is not empty, the price of the current item is discounted by the price of the item at the top of the stack.
    if stack:
      prices[i] -= prices[stack[-1]]

    # Push the current item's index onto the stack.
    stack.append(i)

  # Return the final prices.
  return prices

Real-World Applications

This problem has applications in the retail industry, where it can be used to calculate discounts on products based on the prices of other products in the store.

For example, a grocery store might offer a discount on bananas if you also buy apples. The store could use this problem to calculate the discounted price of the bananas based on the price of the apples.

shuffle_the_array

Problem: Shuffle the Array

Prompt:

Given an array of integers nums and an integer n, shuffle the array such that the first n elements of the shuffled array are the first n elements of nums in random order. The remaining n elements of the shuffled array are the last n elements of nums in random order.

Constraints:

n will be between 1 and nums.length.

Example 1:

Input: nums = [2,5,1,3,4,7], n = 3
Output: [2,3,5,4,1,7]
Explanation: The first `n` elements of the shuffled array are [2,5,1], which is the first `n` elements of `nums` in random order. The remaining `n` elements of the shuffled array are [4,1,7], which is the last `n` elements of `nums` in random order.

Example 2:

Input: nums = [1,2,3,4,4,3,2,1], n = 4
Output: [1,4,2,3,3,2,4,1]
Explanation: The first `n` elements of the shuffled array are [1,4,2,3], which is the first `n` elements of `nums` in random order. The remaining `n` elements of the shuffled array are [3,2,4,1], which is the last `n` elements of `nums` in random order.

Implementation in Python:

import random

def shuffle_the_array(nums, n):
  """
  Shuffles the given array `nums` such that the first `n` elements of the 
  shuffled array are the first `n` elements of `nums` in random order. The 
  remaining `n` elements of the shuffled array are the last `n` elements of 
  `nums` in random order.

  Parameters:
    nums: The array to shuffle.
    n: The number of elements to shuffle from the front of the array.

  Returns:
    The shuffled array.
  """

  # Shuffle the first `n` elements of the array.
  random.shuffle(nums[:n])

  # Shuffle the last `n` elements of the array.
  random.shuffle(nums[n:])

  # Return the shuffled array.
  return nums

Explanation:

The shuffle_the_array() function takes two arguments: an array nums and an integer n. It first shuffles the first n elements of the array using random.shuffle(nums[:n]). Then, it shuffles the last n elements of the array using random.shuffle(nums[n:]). Finally, it returns the shuffled array.

Real-World Applications:

Shuffling arrays is useful in various real-world applications, such as:

Random sampling: Shuffling an array allows you to randomly select a subset of elements from the array.
Randomized algorithms: Many randomized algorithms rely on shuffling data to ensure fairness or prevent bias.
Games and simulations: Shuffling arrays is used to create random scenarios in games and simulations, such as card games or dice rolls.

confusing_number

Problem Statement:

Given a number, find the largest number that can be formed by rearranging its digits.

Example:

Input: 1234
Output: 4321

Explanation:

The largest number that can be formed by rearranging the digits of 1234 is 4321.

Solution:

The best and most performant solution for this problem is to use a greedy approach. This approach involves sorting the digits of the number in descending order and then concatenating them to form the largest number.

Python Implementation:

def largest_number(num):
    """
    Returns the largest number that can be formed by rearranging the digits of a given number.

    Args:
        num: The number to rearrange.

    Returns:
        The largest number that can be formed by rearranging the digits of the given number.
    """

    # Convert the number to a list of digits.
    digits = list(str(num))

    # Sort the digits in descending order.
    digits.sort(reverse=True)

    # Concatenate the digits to form the largest number.
    return int(''.join(digits))

Time Complexity:

The time complexity of this solution is O(n log n), where n is the number of digits in the given number. This is because sorting the digits takes O(n log n) time.

Space Complexity:

The space complexity of this solution is O(n), where n is the number of digits in the given number. This is because we need to store the list of digits in memory.

Real-World Applications:

This problem has applications in several real-world scenarios, such as:

Lottery: When drawing lottery numbers, it is important to ensure that the numbers are random and cannot be predicted. This greedy approach can be used to generate a random number that is difficult to guess.
Password Generation: When creating passwords, it is important to use strong passwords that are difficult to crack. This greedy approach can be used to generate a strong password that is easy to remember but difficult to guess.

find_the_distance_value_between_two_arrays

Problem Statement:

Given two integer arrays arr1 and arr2, the task is to find the minimum absolute distance between any two elements of arr1 and arr2.

Solution:

1. Brute Force Approach:

Iterate over each element in arr1 and calculate the absolute difference between it and each element in arr2.
Keep track of the minimum absolute difference encountered.
Time complexity: O(n*m), where n is the length of arr1 and m is the length of arr2.

2. Optimized Approach:

Sort both arrays in ascending order.
Initialize two pointers, i and j, to the first element of arr1 and arr2, respectively.
Calculate the absolute difference between arr1[i] and arr2[j].
If the absolute difference is less than the minimum absolute difference found so far, update the minimum.
If arr1[i] is smaller than arr2[j], increment i.
If arr1[i] is greater than arr2[j], increment j.
Continue until both pointers reach the end of their respective arrays.
Time complexity: O(nlog(n) + mlog(m)), which is dominated by sorting.

Python Implementation:

def find_minimum_distance_value(arr1, arr2):
    """Finds the minimum absolute distance between two elements of two sorted arrays.

    Parameters:
    arr1: The first sorted array.
    arr2: The second sorted array.

    Returns:
    The minimum absolute distance.
    """

    # Sort both arrays.
    arr1.sort()
    arr2.sort()

    # Initialize pointers.
    i = 0
    j = 0

    # Minimum absolute difference found so far.
    min_diff = float('inf')

    # Iterate until both pointers reach the end of their respective arrays.
    while i < len(arr1) and j < len(arr2):
        # Calculate the absolute difference.
        diff = abs(arr1[i] - arr2[j])

        # Update the minimum absolute difference if necessary.
        min_diff = min(min_diff, diff)

        # Increment the pointers accordingly.
        if arr1[i] < arr2[j]:
            i += 1
        elif arr1[i] > arr2[j]:
            j += 1

    return min_diff

Real-World Applications:

Finding the closest airport to a given location.
Determining the best route for a delivery vehicle.
Optimizing the placement of objects in a warehouse.

make_the_string_great

Problem Statement:

Given a string s, return the minimum number of deletions needed to make the string a palindrome. A palindrome is a string that reads the same forward and backward.

Example:

Input: s = "aebcbda"
Output: 2
Explanation: The shortest palindrome that can be made by deleting is "aebda".

Solution Overview:

We can use dynamic programming to solve this problem. We will create a 2D table dp, where dp[i][j] represents the minimum number of deletions needed to make the substring s[i:j+1] a palindrome.

To fill in the dp table, we can use the following rules:

If s[i] == s[j], then dp[i][j] = dp[i+1][j-1]. This means that if the first and last characters of the substring are the same, then we don't need to delete any characters to make it a palindrome.
If s[i] != s[j], then we have two options:
- Delete the first character: dp[i][j] = dp[i+1][j] + 1.
- Delete the last character: dp[i][j] = dp[i][j-1] + 1.
We choose the option with the minimum number of deletions.

Once we have filled in the dp table, the minimum number of deletions needed to make the entire string s a palindrome is dp[0][s.length - 1].

Python Implementation:

def min_deletions_to_make_palindrome(s):
    """
    Returns the minimum number of deletions needed to make the given string a palindrome.

    Args:
        s (str): The string to make a palindrome.

    Returns:
        int: The minimum number of deletions needed.
    """

    # Create the dp table.
    dp = [[0 for _ in range(len(s))] for _ in range(len(s))]

    # Fill in the dp table.
    for i in range(len(s) - 1, -1, -1):
        for j in range(len(s)):
            if s[i] == s[j]:
                dp[i][j] = dp[i + 1][j - 1]
            else:
                dp[i][j] = min(dp[i + 1][j], dp[i][j - 1]) + 1

    # Return the minimum number of deletions needed to make the entire string a palindrome.
    return dp[0][len(s) - 1]

Time Complexity:

The time complexity of the above solution is O(n^2), where n is the length of the given string. This is because we are filling in a 2D table of size n x n.

Space Complexity:

The space complexity of the above solution is also O(n^2), as we are storing the dp table.

Real-World Applications:

This problem can be used in a variety of real-world applications, such as:

Spelling correction: When a user types a word into a search engine, the search engine can use this algorithm to suggest possible corrections for the word.
DNA sequencing: Scientists can use this algorithm to identify palindromic sequences in DNA, which can be useful for identifying genes and other important biological features.
Data compression: This algorithm can be used to compress strings by removing redundant characters.

count_good_triplets

Problem Statement

Given an array of integers nums, a "good" triplet is a triplet (i, j, k) that satisfies the following conditions:

0 <= i < j < k < nums.length
nums[i] < nums[j] < nums[k]

Return the number of "good" triplets.

Solution

The brute force solution would be to generate all possible triplets and check if each triplet is good. This would take O(n^3) time, where n is the length of the array.

To improve the time complexity, we can use a two-pointer approach. We can iterate over the array with two pointers, i and j, and keep track of the maximum value of nums[j] encountered so far. For each value of nums[i], we can use binary search to find the number of elements in the range (i, j) that are greater than nums[i] and less than the maximum value. This would take O(n^2 * log n) time.

Here's the Python code for the two-pointer approach:

def count_good_triplets(nums):
  """
  Returns the number of good triplets in the given array.

  Parameters:
    nums: An array of integers.

  Returns:
    The number of good triplets.
  """

  # Initialize the count of good triplets.
  good_triplets = 0

  # Iterate over the array with two pointers.
  for i in range(len(nums)):
    for j in range(i + 1, len(nums)):
      # Find the maximum value of nums[j] encountered so far.
      max_val = nums[j]

      # Use binary search to find the number of elements in the range (i, j) that are greater than nums[i] and less than max_val.
      left, right = j + 1, len(nums) - 1
      while left <= right:
        mid = (left + right) // 2
        if nums[mid] > nums[i] and nums[mid] < max_val:
          good_triplets += 1
          left = mid + 1
        else:
          right = mid - 1

  # Return the count of good triplets.
  return good_triplets

Applications

The problem of counting good triplets has applications in data analysis and statistics. For example, it can be used to find the number of triples of data points that satisfy certain conditions, such as being in ascending order or having a certain difference between them. This information can be useful for identifying trends and patterns in data.

find_common_characters

Problem Statement

Given two strings, find all the characters that are common to both strings.

Example

Input: "hello", "world" Output: ["h", "e", "l", "o", "w"]

Approach

Convert the strings to sets:
- A set is a collection of unique elements.
- We can convert a string to a set using the set() function.
Find the intersection of the sets:
- The intersection of two sets is the set of elements that are common to both sets.
- We can find the intersection of two sets using the & operator.

Code

def find_common_characters(str1, str2):
    """
    Finds all the characters that are common to two strings.

    Args:
        str1 (str): The first string.
        str2 (str): The second string.

    Returns:
        list[str]: A list of the common characters.
    """

    set1 = set(str1)
    set2 = set(str2)

    common_chars = set1 & set2

    return list(common_chars)

Real-World Applications

Natural Language Processing: Find common words or characters in a corpus of text.
Database Queries: Find records that have common fields or values.
Data Analysis: Identify common patterns or trends in data sets.

last_stone_weight

The "last_stone_weight" problem on LeetCode is a puzzle where you are given an array of positive integers representing the weights of stones. The goal is to remove stones from the array one at a time until there is only one stone left. The rules for removing stones are as follows:

If there are two stones of the same weight, you can remove both stones at once.
If there is only one stone left, it is the last stone weight.
Otherwise, you can remove any stone from the array.

The task is to find the last stone weight after all the stones have been removed.

For example, if the input array is [2, 7, 4, 1, 8, 1], the last stone weight is 1 because:

We can remove the two stones with weight 1 to get [2, 7, 4, 8].
We can remove the two stones with weight 2 to get [7, 4, 8].
We can remove the two stones with weight 4 to get [7, 8].
We can remove the two stones with weight 7 to get [8].
Finally, we can remove the stone with weight 8 to get [1].

Therefore, the last stone weight is 1.

Here is a simplified and commented Python solution for this problem:

def last_stone_weight(stones):
  """
  Finds the last stone weight after all the stones have been removed.

  :param stones: An array of positive integers representing the weights of stones.
  :return: The last stone weight.
  """

  # Sort the stones in descending order.
  stones.sort(reverse=True)

  # While there is more than one stone, remove two stones of the same weight.
  while len(stones) > 1:
    # Get the weights of the two heaviest stones.
    stone1 = stones[0]
    stone2 = stones[1]

    # If the two stones have the same weight, remove both stones.
    if stone1 == stone2:
      stones.pop(0)
      stones.pop(0)
    # Otherwise, remove the heavier stone.
    else:
      stones[0] = stone1 - stone2
      stones.pop(1)

  # If there is only one stone left, return its weight.
  return stones[0] if stones else 0

Here is an example of how to use this function:

stones = [2, 7, 4, 1, 8, 1]
last_stone = last_stone_weight(stones)
print(last_stone)  # Output: 1

This problem can be solved in O(n log n) time, where n is the number of stones. Sorting the stones takes O(n log n) time, and removing the two heaviest stones takes O(1) time. This process is repeated until there is only one stone left, so the total time complexity is O(n log n).

This problem can be applied to real-world problems such as finding the maximum weight of a set of objects that can be packed into a container of a certain size. The objects can be represented by stones, and the container size can be represented by the last stone weight. By removing the two heaviest objects at a time, we can find the maximum weight that can be packed into the container.

defanging_an_ip_address

Problem Statement:

Given an IP address, return the defanged version of the IP address.

The defanged version of an IP address is the IP address with all the dots (.) replaced by "[.]".

Example: Input: "1.1.1.1" Output: "1[.]1[.]1[.]1"

Implementation:

Python Solution:

    def defangIPaddr(self, address: str) -> str:
        return address.replace(".", "[.]")

Breakdown:

The replace() method is used to replace all the occurrences of a substring (".") with a new substring ("[.]").
The defangIPaddr() method takes an IP address as input and returns the defanged version of it.

Time complexity: O(n), where n is the length of the IP address. Space complexity: O(n), since a new string is created.

Real-world Application:

Defanging IP addresses is useful when you want to store or display IP addresses in a way that prevents them from being confused with URLs. For example, if you have an IP address like "1.1.1.1", you might want to defang it to "1[.]1[.]1[.]1" to prevent it from being interpreted as a URL.

maximum_69_number

Problem Statement: Given a positive integer num, return the maximum possible integer you can get by replacing exactly one digit by '9'.

Example:

Input: num = 9669
Output: 9969

Approach:

Convert the number to a string: Convert the integer num to a string to make it easy to manipulate individual digits.
Loop through the digits: Iterate through each digit in the string.
Check if the digit is not '9': If the current digit is not '9', then replacing it with '9' will increase the number.
Replace the digit with '9': Create a new string where the current digit is replaced with '9'.
Convert the new string to an integer: Convert the new string back to an integer.
Return the maximum integer: Return the maximum integer obtained from any of the replacements.

Code Implementation:

def maximum69Number (num):
    # Convert the number to a string
    num_str = str(num)

    # Maximum possible number
    max_num = 0

    # Loop through the digits
    for i in range(len(num_str)):
        # Check if the digit is not '9'
        if num_str[i] != '9':
            # Create a new string where the current digit is replaced with '9'
            new_num_str = num_str[:i] + '9' + num_str[i+1:]
            
            # Convert the new string to an integer
            new_num = int(new_num_str)
            
            # Update the maximum number
            max_num = max(max_num, new_num)
            
    # Return the maximum integer
    return max_num

Real World Applications:

Financial analysis: Maximum 69 Number can be used to analyze financial data and identify potential opportunities for growth.
Logistics and supply chain management: It can help optimize routes and schedules to maximize efficiency and minimize costs.
Data analysis: It can be used to identify outliers and patterns in large datasets.

range_sum_of_bst

Problem: Given the root node of a binary search tree (BST) and a target range [low, high], return the sum of values of all nodes in the range [low, high].

Example:

Input:
        10
       /  \
      5    15
     / \   /  \
    3   7 13  18
Target range: [6, 10]
Output: 23
(Note: 6 + 7 + 10 = 23)

Efficient Solution: 1. Recursive Approach:

This solution recursively traverses the BST and adds the values of nodes within the target range to the sum.

Implementation:

def range_sum_of_bst(root, low, high):
    if not root:
        return 0

    # If node value is within range, add it to sum
    sum = root.val if low <= root.val <= high else 0

    # Recursively traverse left and right subtrees
    sum += range_sum_of_bst(root.left, low, high)
    sum += range_sum_of_bst(root.right, low, high)

    return sum

Time Complexity: O(N), where N is the number of nodes in the BST.

Space Complexity: O(H), where H is the height of the BST.

Applications:

Calculating the total value of items in a specific price range in an e-commerce inventory system.
Determining the total weight of people within a certain age group in a database.

Example Usage:

# Create a BST
root = TreeNode(10)
root.left = TreeNode(5)
root.right = TreeNode(15)
root.left.left = TreeNode(3)
root.left.right = TreeNode(7)
root.right.left = TreeNode(13)
root.right.right = TreeNode(18)

# Calculate sum of values within range [6, 10]
sum = range_sum_of_bst(root, 6, 10)
print(sum)  # Output: 23

delete_characters_to_make_fancy_string

Problem Statement:

Given a string containing lowercase letters, delete the minimum number of characters to make the string "fancy". A fancy string is defined as a string where for every group of characters the characters are sorted in ascending order, and the length of each group is at least 3.

Example:

Input: "leeetcode" Output: "leetcode" (Delete 'e')

Breakdown:

What is a Fancy String?
A fancy string is a string that has the following properties:
- It consists of characters in lowercase.
- The characters in each group are sorted in ascending order.
- Each group has at least 3 characters.
Strategy:
We need to delete the minimum number of characters to make the string fancy. To do this, we can follow these steps:
1. Iterate through the string.
2. For each character, check if it forms a group of at least 3 characters in sorted order.
3. If the group has less than 3 characters or the characters are not in sorted order, delete the current character.
Code Implementation:

def makeFancyString(s: str) -> str:
    # Initialize a new string to hold the result.
    result = ""

    # Iterate through the given string.
    for char in s:
        # Check if the result is empty or the current character is different from the last character in the result.
        if not result or result[-1] != char:
            # If so, append the current character to the result.
            result += char
        # Otherwise, check if the current character is the same as the last two characters in the result.
        elif result[-1] == result[-2]:
            # If so, skip the current character.
            continue
        # Otherwise, append the current character to the result.
        else:
            result += char

    # Return the result.
    return result

# Example:
s = "leeetcode"
print(makeFancyString(s))  # Output: "leetcode"

Applications:

The problem of making a string fancy has applications in various fields, including:

Text processing: Cleaning up text data for analysis or display.
Data visualization: Creating visually appealing charts and graphs with sorted and grouped data.
Document layout: Optimizing the appearance of text documents by ensuring that groups of characters are visually distinct.
Database management: Ensuring that data is organized and stored in a consistent manner.

greatest_common_divisor_of_strings

Greatest Common Divisor of Strings

Problem Statement:

Given two strings str1 and str2, your task is to find the "Greatest Common Divisor" (GCD) string between them.

Solution:

The GCD of two strings is the longest string gcd that evenly divides both str1 and str2. In other words, gcd is a substring of both str1 and str2 that repeats an integer number of times to form them.

Naive Approach:

One approach is to try all possible lengths for gcd, from 1 to the length of the shorter string. For each length, check if gcd appears at the beginning of both str1 and str2 by dividing them by gcd. If it does, then gcd is the GCD of the strings.

Efficient Approach (Euclidean Algorithm):

The Euclidean Algorithm can be used to find the GCD of two strings in O(max(m, n)) time, where m and n are the lengths of the strings.

Euclidean Algorithm for Strings:

Let s1 and s2 be the strings with m >= n.
If s2 is empty, the GCD is s1.
Otherwise, perform the following steps until s2 is empty: a. Find the remainder of s1 divided by s2 using the modulus operation, resulting in r. b. Update s1 to s2, and s2 to r.
Return s1.

Python Implementation:

def gcd_of_strings(str1: str, str2: str) -> str:
    """
    Finds the greatest common divisor of two strings using the Euclidean Algorithm.

    Args:
    str1 (str): The first string.
    str2 (str): The second string.

    Returns:
    str: The greatest common divisor of the two strings.
    """

    # Check if one of the strings is empty
    if not str1 or not str2:
        return ""

    # Make sure str1 is always the longer string
    if len(str1) < len(str2):
        str1, str2 = str2, str1

    # Perform the Euclidean Algorithm
    while str2:
        str1, str2 = str2, str1 % str2

    return str1

Example:

str1 = "ABCABC"
str2 = "ABC"

gcd = gcd_of_strings(str1, str2)
print(gcd)  # Output: "ABC"

Real-World Applications:

Cryptography: Finding the GCD of two strings can be used to encrypt and decrypt messages using the GCD algorithm.
String Matching: The GCD of two strings can be used to find the longest common subsequence between them, which is useful for string matching and comparison.
Text Analysis: The GCD of two texts can be used to identify their similarity or difference, which is useful for plagiarism detection and text analysis.

verifying_an_alien_dictionary

Problem:

Given a list of words in a foreign language, verify if it's a valid alien dictionary. In a valid alien dictionary, every word appears lexicographically after the preceding word.

Solution:

Breakdown:

Create an Adjacency List: For each consecutive pair of words, add an edge from the first letter of the first word to the first letter of the second word.
Perform Topological Sort:
- Create a queue queue and a set of visited nodes visited. Initially, add all unique first letters to queue.
- While queue is not empty:
  - Remove the first letter u from queue.
  - For each letter v adjacent to u:
    If v is not in visited:
    Add v to visited.
    Add v to queue.
Check for Validity:
- If the number of visited nodes is equal to the number of unique letters, then it's a valid alien dictionary.

Python Implementation:

from collections import defaultdict, deque

def isAlienSorted(words):
    # Create adjacency list
    adj = defaultdict(list)
    for i in range(1, len(words)):
        w1, w2 = words[i-1], words[i]
        for j in range(min(len(w1), len(w2))):
            if w1[j] != w2[j]:
                adj[w1[j]].append(w2[j])
                break

    # Perform topological sort
    queue = deque(set(c for w in words for c in w))
    visited = set()
    while queue:
        u = queue.popleft()
        visited.add(u)
        for v in adj[u]:
            if v not in visited:
                queue.append(v)

    # Check for validity
    return len(visited) == len({c for w in words for c in w})

Real-World Applications:

Lexicography: Sorting words in a foreign language where the alphabet is unknown.
Data Management: Organizing data where the order of elements is important.
Natural Language Processing: Understanding the relationships between words in a language.

duplicate_zeros

Problem Statement:

Given an integer array, duplicate each occurrence of zeroshifting the remaining elements to the right.

For Example:

Input: [1,0,2,3,0,4,5,0]
Output: [1,0,0,2,3,0,0,4,5,0]

Solution:

Approach:

Iterate the array from start to end.
If the current element is 0, duplicate it by shifting the remaining elements to the right.

Implementation:

def duplicate_zeros(arr):
    # Create a new list to store the modified array
    new_arr = []

    # Iterate over the input array
    for element in arr:
        # If the current element is 0, duplicate it by appending it twice
        if element == 0:
            new_arr.append(0)
            new_arr.append(0)
        # Otherwise, just append the element to the new array
        else:
            new_arr.append(element)

    # Return the new array
    return new_arr

Time Complexity: O(n), where n is the length of the input array.

Space Complexity: O(n), as we need to create a new array to store the modified array.

Applications in Real World:

Data Manipulation: Duplicating zeros can be useful in data manipulation tasks where we need to add filler elements or expand certain sections of an array.
Image Processing: In image processing, duplicate zeros can be used to create blank spaces or borders around images.

count_largest_group

Problem Statement:

Given an integer array containing only positive integers, the task is to find the number of distinct elements in all the non-empty subsets.

Solution Explanation:

The best solution to this problem is to use the bitmasking technique. Bitmasking allows us to represent a subset as a bitmask where each bit represents an element in the array.

Generate all subsets: We can generate all subsets using bitmasking. For n elements, there are a total of 2^n possible subsets.
Count distinct elements: For each subset, we count the number of set bits in the corresponding bitmask. The count of set bits represents the number of distinct elements in the subset.
Store subset counts: We store the count of distinct elements for each subset in an array.
Return distinct element count: Finally, we return the sum of distinct element counts for all non-empty subsets (i.e., all subsets except the empty subset).

Complete Code Implementation:

def count_largest_group(nums):
    """
    :type nums: List[int]
    :rtype: int
    """
    # Initialize bitmask to 0 (represents the empty subset)
    bitmask = 0
    # Initialize count array to store distinct element counts
    counts = [0] * (len(nums) + 1)
    
    for num in nums:
        # Perform bitwise OR to add the bit corresponding to the element
        bitmask |= (1 << num)
        # Increment the count of distinct elements for the current bitmask
        counts[bitmask] += 1
    
    # Find the maximum count of distinct elements
    max_count = max(counts[1:])  # Ignore the empty subset (count = 0)
    
    # Count the number of subsets with the maximum count
    return counts.count(max_count)

Applications in Real World:

Subset counting is used in various applications, such as:
- Combinations and permutations
- Probability and statistics
- Data analysis and mining
- Cryptography

count_odd_numbers_in_an_interval_range

Problem Statement: Given two integer numbers, left and right, count the number of odd numbers in the interval [left, right].

Example:

left = 2, right = 7
Output: 3 (3, 5, 7)

Solution:

We can observe that the odd numbers in the interval [left, right] are:

left + 1, left + 3, ..., right - 2, right - 0

which is an arithmetic progression with the first term left + 1, the common difference 2, and the last term right.

The number of terms in an arithmetic progression is given by the formula:

n = (last - first) / difference + 1

So, the number of odd numbers in the interval [left, right] is:

n = (right - (left + 1)) / 2 + 1

def count_odd_numbers_in_an_interval_range(left: int, right: int) -> int:
    """
    Count the number of odd numbers in the interval [left, right].

    :param left: The lower bound of the interval.
    :param right: The upper bound of the interval.
    :return: The number of odd numbers in the interval [left, right].
    """
    n = (right - (left + 1)) // 2 + 1
    return n

Example Usage:

left = 2
right = 7
result = count_odd_numbers_in_an_interval_range(left, right)
print(result)  # Output: 3

Time Complexity: The time complexity of the solution is O(1) because it takes constant time to compute the number of odd numbers in the interval [left, right].

Space Complexity: The space complexity of the solution is O(1) because it uses a constant amount of memory.

Applications in Real World:

Counting odd numbers in an interval can be useful in various applications, such as:

Counting the number of odd pages in a book
Finding the total number of odd-numbered houses in a street
Generating random odd numbers within a specified range

generate_a_string_with_characters_that_have_odd_counts

Problem Statement:

Given an integer n, generate a string consisting of characters that have an odd count in their ASCII codes.

Python Implementation:

def generate_odd_count_string(n: int) -> str:
    """
    Generates a string consisting of characters that have an odd count in their ASCII codes.

    Args:
    n: The length of the string to generate.

    Returns:
    A string consisting of characters that have an odd count in their ASCII codes.
    """

    # Create a set to store the characters with odd ASCII codes.
    odd_characters = set()

    # Iterate over the ASCII codes from 0 to 127.
    for i in range(128):
        # If the ASCII code has an odd count, add the corresponding character to the set.
        if bin(i).count('1') % 2 == 1:
            odd_characters.add(chr(i))

    # Generate a string by concatenating n random characters from the set.
    return ''.join(random.sample(odd_characters, n))

Explanation:

The function generate_odd_count_string takes an integer n as input and generates a string consisting of n characters that have an odd count in their ASCII codes.

Create a Set to Store Odd Characters: The function first creates a set called odd_characters to store the characters that have an odd count in their ASCII codes.
Iterate Over ASCII Codes: It then iterates over the ASCII codes from 0 to 127.
Check for Odd Count: For each ASCII code, the function checks if it has an odd count of 1's in its binary representation. This is done using the bin(i).count('1') % 2 expression, where i is the ASCII code. If the result is 1, it means the ASCII code has an odd count of 1's, and the corresponding character is added to the set.
Generate Odd String: Finally, the function generates a string by concatenating n random characters from the odd_characters set using the ''.join(random.sample(odd_characters, n)) expression.

Applications in Real World:

Data Obfuscation: This function can be used to obfuscate data by replacing characters with those that have an odd ASCII count.

Encryption: The generated string can be used as a key for encryption algorithms to enhance security.

Puzzle Generation: The odd count string can be used in puzzle games where players need to identify characters based on their unique ASCII code properties.

check_if_a_number_is_majority_element_in_a_sorted_array

Problem:

Determine if a given number is the "majority element" in a sorted array. A majority element appears more than half the size of the array.

Simplified Explanation:

Let's imagine an array as a line of people, with the number we want to check in the middle. We'll count the people on both sides of our number to see if the count on one side is more than half the total number of people.

Implementation:

def is_majority_element(nums, target):
    """
    Checks if a given number is the majority element in a sorted array.

    Args:
        nums: A sorted list of integers.
        target: The number to check.

    Returns:
        True if target is the majority element, False otherwise.
    """

    # Find the index of the target number in the array.
    index = nums.index(target)

    # Count the number of occurrences of target on the left and right sides of the index.
    left_count = index
    right_count = len(nums) - index - 1

    # Check if the count on either side is more than half the total number of elements.
    return left_count > len(nums) // 2 or right_count > len(nums) // 2

Example:

nums = [1, 2, 3, 3, 3, 4, 4, 4, 5]
target = 3

result = is_majority_element(nums, target)

print(result)  # Output: True

Real-World Applications:

This algorithm has applications in various fields, including:

Data Analysis: Identifying the most frequently occurring element in a large dataset.
Election Polling: Estimating the support for a particular candidate by analyzing poll data.
Quality Control: Detecting defects in products by identifying the most common types.
Financial Analysis: Determining the average or median income of a population.
Machine Learning: Identifying patterns and anomalies in large datasets.

find_the_difference_of_two_arrays

Problem Statement: Given two arrays nums1 and nums2, return the difference between the two arrays. The difference between two arrays is defined as the set of elements that are in one array but not in the other.

Example:

Input: nums1 = [1, 2, 3], nums2 = [2, 4]
Output: [1, 3]
Explanation: The difference between the two arrays is [1, 3] because 1 and 3 are in nums1 but not in nums2, and 4 is in nums2 but not in nums1.

Solution: The most efficient way to solve this problem is to use a hash set to store the elements of one array. Then, we can iterate through the other array and check if each element is present in the hash set. If it is not present, then we add it to the difference list.

Here is the Python code for this solution:

def find_the_difference(nums1, nums2):
    """
    :type nums1: List[int]
    :type nums2: List[int]
    :rtype: List[int]
    """
    # Create a hash set to store the elements of nums1
    nums1_set = set(nums1)

    # Create a list to store the difference
    difference = []

    # Iterate through nums2
    for num in nums2:
        # If the number is not in nums1_set, add it to the difference list
        if num not in nums1_set:
            difference.append(num)

    # Return the difference list
    return difference

Time Complexity: The time complexity of this solution is O(n), where n is the length of the longer array.

Space Complexity: The space complexity of this solution is O(n), where n is the length of the shorter array.

Applications: This problem has several applications in real-world scenarios. For example, it can be used to:

Find the difference between two lists of files
Find the difference between two lists of customers
Find the difference between two lists of products

rearrange_spaces_between_words

Leetcode Problem:

Rearrange Spaces Between Words

Problem: Given a string containing words separated by spaces, rearrange the spaces so that there is exactly one space between each word and at least one space in front of the first word and at least one space after the last word.

Example: Input: " this is a sentence " Output: "this is a sentence"

Solution:

Breakdown:

Split into Words: Split the input string into a list of words using the split() function.
Count Spaces: Count the number of spaces in the original string.
Calculate Number of Spaces to Distribute: Determine the number of spaces needed between each word and at the beginning and end of the string. This is calculated as (space_count - word_count + 1) / (word_count + 2).
Add Spaces to Words: Insert the calculated number of spaces between each word and at the beginning and end of the list.
Join Words: Concatenate the modified list of words into a new string to get the final result.

Python Implementation:

def rearrangeSpaces(text):
    words = text.split()
    spaces = text.count(' ')
    space_need = (spaces - len(words) + 1) // (len(words) + 2)
    spaces_between = ' ' * space_need
   
    return spaces_between.join(words)

Real-World Applications:

Text Justification: Aligning text to the left, right, or center.
Whitespace Removal: Cleaning up text to remove extra spaces.
String Manipulation: Transforming text in various ways for data analysis or processing.

check_if_a_word_occurs_as_a_prefix_of_any_word_in_a_sentence

Python Implementation:

def isPrefixOfWord(sentence, target):
    """
    :type sentence: str
    :type target: str
    :rtype: int
    """
    words = sentence.split()

    for i, word in enumerate(words):
        if word.startswith(target):
            return i + 1

    return -1

Breakdown and Explanation:

Breakdown:

Split the Sentence into Words: The split() function separates the sentence into individual words based on whitespace.
Iterate Over the Words: We loop through each word in the sentence.
Check Prefix: For each word, we check if it starts with the target prefix using the startswith() method.
Return Word Index: If a word is found to be a prefix of the target, we return its index in the original sentence (starting from 1).
Return -1: If no word is found to be a prefix of the target, we return -1.

Explanation:

Suppose we have the sentence "hello how world" and the target is "he".

Splitting the sentence gives us words = ["hello", "how", "world"].
Iterating through words, we:
- Check if "hello" starts with "he". It does, so we return 1.

If the target is "wo", it will not match any word in the sentence, so we return -1.

Complete Code Implementation:

# Example 1
sentence = "i love eating burger"
target = "burg"
print(isPrefixOfWord(sentence, target))  # Output: 4

# Example 2
sentence = "hello how world"
target = "he"
print(isPrefixOfWord(sentence, target))  # Output: 1

# Example 3
sentence = "this problem is really tough"
target = "tou"
print(isPrefixOfWord(sentence, target))  # Output: -1

Real-World Applications:

This problem is often encountered in natural language processing and text mining tasks, such as:

Autocompletion: Suggesting words that begin with a user's typed prefix.
Search Optimization: Filtering results based on specified prefixes to improve search efficiency.
Grammar Checking: Detecting and correcting grammatical errors related to prefixes.

lucky_numbers_in_a_matrix

Problem Statement:

Given a matrix of numbers, find the sum of all the "lucky" numbers in the matrix. A lucky number is a number that is the minimum in its row and maximum in its column.

Approach:

Traverse Rows: For each row in the matrix, find the minimum value in that row.
Traverse Columns: For each column in the matrix, find the maximum value in that column.
Check Lucky Numbers: Compare the row minimums with the column maximums. If any row minimum is also the column maximum, it is a lucky number.
Sum Lucky Numbers: Add up all the lucky numbers found in the matrix.

Implementation in Python:

def lucky_numbers_in_a_matrix(matrix):
  """
  Finds the sum of lucky numbers in a matrix.
  
  Args:
    matrix: A list of lists of numbers representing the matrix.
  
  Returns:
    The sum of lucky numbers in the matrix.
  """

  # Create lists to store row minimums and column maximums.
  row_mins = [min(row) for row in matrix]
  col_maxs = [max(col) for col in zip(*matrix)]

  # Initialize the sum of lucky numbers to 0.
  lucky_sum = 0

  # Iterate through the matrix.
  for i in range(len(matrix)):
    for j in range(len(matrix[0])):
      # Check if the current element is a lucky number.
      if matrix[i][j] == row_mins[i] and matrix[i][j] == col_maxs[j]:
        # Add the lucky number to the sum.
        lucky_sum += matrix[i][j]

  # Return the sum of lucky numbers.
  return lucky_sum

Example:

matrix = [[3, 7, 8],
         [9, 11, 13],
         [15, 16, 17]]

result = lucky_numbers_in_a_matrix(matrix)
print(result)  # Output: 17

Explanation:

Row minimums: [3, 9, 15]
Column maximums: [8, 11, 17]
Lucky numbers: [8, 11, 17]
Sum of lucky numbers: 36

Applications in Real World:

Data Analysis: Identifying outliers and finding extreme values in datasets.
Optimization: Finding optimal values in constrained systems.
Finance: Detecting undervalued or overvalued assets.

find_all_the_lonely_nodes

Problem: Given a binary search tree (BST), return the number of lonely nodes in the tree. A node is considered lonely if the node is a leaf (has no children) AND the node is not a child of another node.

Example 1: Input: [1,2,3,null,4] Output: 2

Example 2: Input: [1,2,3,4,5] Output: 0

Explanation:

In the first example, the lonely nodes are 2 and 4. The node 2 is a leaf and it is not a child of another node. The node 4 is also a leaf and it is not a child of another node.
In the second example, there are no lonely nodes.

Approach: To find all the lonely nodes in a BST, we can use a recursive approach. In each recursive call, we will check if the current node is a leaf and if it is not a child of another node. If both of these conditions are met, then the node is a lonely node. We can then increment the count of lonely nodes and continue the recursion.

Code:

def find_all_the_lonely_nodes(root):
    """
    :type root: TreeNode
    :rtype: int
    """
    if not root:
        return 0

    count = 0

    # Check if the current node is a leaf
    if not root.left and not root.right:
        count += 1

    # Check if the current node is not a child of another node
    if (not root.parent or root.parent.left != root) and (not root.parent or root.parent.right != root):
        count += 1

    # Recursively find the lonely nodes in the left and right subtrees
    count += find_all_the_lonely_nodes(root.left)
    count += find_all_the_lonely_nodes(root.right)

    return count

Complexity Analysis:

Time complexity: O(N), where N is the number of nodes in the BST. We visit each node in the BST once.
Space complexity: O(H), where H is the height of the BST. The recursion stack can grow up to a depth of H.

Potential Applications:

Identifying isolated nodes in a network: A lonely node in a BST can represent an isolated node in a network. Finding all the lonely nodes in a BST can help identify isolated nodes in a network, which can be useful for network optimization and resilience.
Finding leaf nodes in a tree: A lonely node in a BST is also a leaf node. Finding all the lonely nodes in a BST can help identify all the leaf nodes in a tree, which can be useful for tree traversal and tree manipulation algorithms.

how_many_numbers_are_smaller_than_the_current_number

How Many Numbers Are Smaller Than the Current Number

Given an array of integers nums, return an array answer such that answer[i] is the number of elements in nums that are strictly smaller than nums[i].

Breakdown of the Problem

The problem can be broken down into the following steps:

Iterate through each element in the nums array.
For each element, count the number of other elements in the array that are smaller than it.
Store the count in the answer array.

Code Implementation

def howManyNumbersAreSmallerThanTheCurrentNumber(nums):
  """
  Counts the number of elements in an array that are smaller than the current element.

  Parameters:
    nums: The array of integers.

  Returns:
    The array of counts.
  """

  # Create an array to store the counts.
  counts = [0] * len(nums)

  # Iterate through each element in the array.
  for i in range(len(nums)):
    # Count the number of other elements in the array that are smaller than it.
    for j in range(len(nums)):
      if i != j and nums[j] < nums[i]:
        counts[i] += 1

  # Return the array of counts.
  return counts

Real-World Applications

This problem has applications in many real-world scenarios, such as:

Ranking: This problem can be used to rank elements in a dataset. For example, in a list of students' test scores, we can use this problem to rank the students by their scores.
Selection: This problem can be used to select the top or bottom elements in a dataset. For example, in a list of job applications, we can use this problem to select the top 10 applications.
Counting: This problem can be used to count the number of elements in a dataset that satisfy a certain condition. For example, in a list of sales records, we can use this problem to count the number of sales that were made in a specific region.

Potential Applications in Real World

Potential applications in real world include:

Data Analysis: This problem can be used to analyze data and identify trends. For example, in a dataset of customer purchases, we can use this problem to identify the products that are most popular with customers.
Decision Making: This problem can be used to make decisions. For example, in a dataset of job applications, we can use this problem to identify the applications that are most likely to be successful.
Resource Allocation: This problem can be used to allocate resources. For example, in a dataset of student grades, we can use this problem to identify the students who need the most help.

unique_email_addresses

Problem Description:

You have a list of email addresses. Each email address consists of a local name and a domain name, separated by the '@' sign.

For example, in the email address "john.smith@example.com", "john.smith" is the local name and "example.com" is the domain name.

The local name can contain letters, numbers, and dots. The domain name can contain letters, numbers, and dots, and it must end with a period ('.').

Some email addresses in the list may be invalid. An email address is considered invalid if it does not follow the above rules.

Your task is to find the number of unique email addresses in the list. Two email addresses are considered unique if they have different local names.

Constraints:

The length of the list is at most 1000.
Each email address in the list is at most 100 characters long.

Example Input:

["john.smith@example.com", "john.smith@example.com", "john.smith@example.com", "mary.johnson@example.com"]

Example Output:

Explanation:

The list contains four email addresses:

"john.smith@example.com"
"john.smith@example.com"
"john.smith@example.com"
"mary.johnson@example.com"

The first three email addresses are all the same, so they are considered duplicates. The fourth email address is different, so it is considered unique. Therefore, the output is 2.

Solution:

The following Python code implements a solution to this problem:

def numUniqueEmails(emails):
  """
  Returns the number of unique email addresses in the list.

  Args:
    emails: A list of email addresses.

  Returns:
    The number of unique email addresses.
  """

  # Create a set to store the unique email addresses.
  unique_emails = set()

  # Loop over the list of email addresses.
  for email in emails:

    # Split the email address into the local name and domain name.
    local_name, domain_name = email.split('@')

    # Remove all the dots from the local name.
    local_name = local_name.replace('.', '')

    # Add the unique email address to the set.
    unique_emails.add(f"{local_name}@{domain_name}")

  # Return the number of unique email addresses.
  return len(unique_emails)

Explanation:

The solution works by first splitting each email address into the local name and the domain name. Then, it removes all the dots from the local name. Finally, it adds the unique email address to the set. The number of unique email addresses is then returned.

Real-World Applications:

This problem has many real-world applications, such as:

Email marketing: Email marketing campaigns often send out emails to a large number of recipients. It is important to ensure that each recipient receives only one email, even if they have multiple email addresses.
Customer relationship management (CRM): CRM systems often store multiple email addresses for each customer. It is important to be able to identify which email addresses belong to the same customer.
Fraud detection: Fraudsters often use multiple email addresses to create fake accounts. It is important to be able to identify these email addresses and prevent them from being used for fraudulent activities.

number_of_days_in_a_month

Problem: Given a month and a year, return the number of days in that month.

Optimal Solution:

def number_of_days_in_a_month(month, year):
    if month in (1, 3, 5, 7, 8, 10, 12):
        return 31
    elif month == 2:
        if is_leap_year(year):
            return 29
        else:
            return 28
    else:
        return 30

def is_leap_year(year):
    return (year % 4 == 0 and year % 100 != 0) or year % 400 == 0

Explanation:

The code first checks if the month is one of the months with 31 days (January, March, May, July, August, October, December). If so, the function returns 31.

If the month is February, the function checks if the year is a leap year. A leap year is a year that is divisible by 4 but not by 100, or a year that is divisible by 400. If the year is a leap year, the function returns 29. Otherwise, it returns 28.

If the month is not one of the months with 31 days or February, the function returns 30.

Real-World Applications:

This function can be used in a variety of applications, such as:

Calculating the number of days until a specific date
Creating a calendar
Determining the length of a month in a specific year
Calculating the number of days in a fiscal year

prime_arrangements

ERROR OCCURED prime_arrangements

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.
```

shift_2d_grid

Problem Statement:

Given a 2D grid of size m x n, filled with either 0s or 1s, shift the grid to the right by k columns. If any rows are shifted out of the grid, they rotate back to the top of the grid.

Implementation:

def shift_2d_grid(grid, k):
    """
    Shifts the given 2D grid to the right by k columns.

    Args:
        grid: The 2D grid to be shifted.
        k: The number of columns to shift the grid by.

    Returns:
        The shifted 2D grid.
    """

    # Get the dimensions of the grid.
    m, n = len(grid), len(grid[0])

    # Shift the grid to the right by k columns.
    for i in range(k):
        # Store the last column in a temporary variable.
        last_column = grid[0][n-1]

        # Shift each column to the right by one position.
        for j in range(n-1, 0, -1):
            grid[0][j] = grid[0][j-1]

        # Shift the last column to the first row.
        for j in range(m-1, 0, -1):
            grid[j][n-1] = grid[j-1][n-1]

        # Store the last row in the temporary variable.
        last_row = grid[m-1][n-1]

        # Shift each row to the right by one position.
        for j in range(m-2, -1, -1):
            grid[m-1][j+1] = grid[m-1][j]

        # Set the last row to the temporary variable.
        grid[m-1][0] = last_row

        # Set the last column to the temporary variable.
        grid[0][n-1] = last_column

    # Return the shifted grid.
    return grid

Example:

grid = [[1,2,3],[4,5,6],[7,8,9]]
k = 1

shifted_grid = shift_2d_grid(grid, k)

print(shifted_grid)
# Output: [[9,1,2],[3,4,5],[6,7,8]]

Explanation:

The code works by iterating over the grid and shifting each column to the right by one position. The last column is stored in a temporary variable and then shifted to the first row. The last row is also stored in a temporary variable and then shifted to the last column. This process is repeated until the grid has been shifted by the specified number of columns.

Applications:

Image processing: Shifting a 2D grid can be used to apply image filters or transformations.
Game development: Shifting a 2D grid can be used to move objects in a game world.
Data analysis: Shifting a 2D grid can be used to align data for comparison or analysis.

can_make_arithmetic_progression_from_sequence

Problem Statement:

Given a sequence of integers, determine if it is possible to rearrange the elements to form an arithmetic progression (AP). An AP is a sequence in which the difference between any two consecutive elements is constant.

Example:

Input: [1, 3, 5, 7] Output: True Explanation: This sequence can be rearranged to [1, 3, 5, 7], which forms an AP with a common difference of 2.

Input: [1, 3, 4, 5] Output: False Explanation: This sequence cannot be rearranged to form an AP because the difference between 3 and 4 is 1, while the difference between 4 and 5 is 2.

Solution:

The key to solving this problem is to first sort the sequence in ascending order. Once the sequence is sorted, we can check if the difference between consecutive elements is the same. If so, then the sequence can be rearranged to form an AP.

Python Implementation:

def can_make_arithmetic_progression_from_sequence(sequence):
  """
  Checks if a sequence of integers can be rearranged to form an arithmetic progression.

  Parameters:
    sequence: A list of integers.

  Returns:
    True if the sequence can be rearranged to form an AP, False otherwise.
  """
  
  # Sort the sequence in ascending order.
  sorted_sequence = sorted(sequence)
  
  # Check if the difference between consecutive elements is the same.
  for i in range(1, len(sorted_sequence)):
    if sorted_sequence[i] - sorted_sequence[i - 1] != sorted_sequence[1] - sorted_sequence[0]:
      return False
  
  return True

Applications in Real World:

Signal processing: In signal processing, APs are used to represent periodic signals.
Time series analysis: In time series analysis, APs are used to detect trends and seasonality in data.
Financial modeling: In financial modeling, APs are used to predict future values of stock prices and other financial data.

maximize_sum_of_array_after_k_negations

Maximize Sum of Array After K Negations

Problem Statement:

Given an integer array 'nums' and an integer 'k', you want to maximize the sum of the array after performing at most 'k' negations on elements in the array. What is the maximum sum you can achieve?

Simplified Breakdown:

What is Negation? Negating a number means changing its sign. For example, -5 is the negation of 5, and -2.5 is the negation of 2.5.
Maximize Sum: Your goal is to make the sum of the numbers in the array as large as possible.
K Negations: You can negate any element in the array up to 'k' times.

Greedy Solution:

The greedy solution involves two steps:

Sort the Array: Sort the array in ascending order.
Negate Negative Numbers: Starting from the beginning of the sorted array, negate as many negative numbers as you can until you reach 'k' negations or the end of the array.

Python Implementation:

def maximize_sum_after_negations(nums, k):
  # Sort the array in ascending order
  nums.sort()
  
  # Negate negative numbers
  for i in range(len(nums)):
    if nums[i] < 0 and k > 0:
      nums[i] = -nums[i]
      k -= 1
  
  # Return the sum of the array
  return sum(nums)

Real-World Applications:

This problem can be applied in various real-world scenarios:

Financial Modeling: A financial advisor may use this technique to optimize investment portfolios by maximizing the overall return on their investments.
Resource Allocation: A project manager may use this approach to distribute resources (e.g., equipment, personnel) to maximize the efficiency and output of their team.
Data Analysis: A data scientist may employ this method to clean and transform data to extract meaningful insights from complex datasets.

three_consecutive_odds

Problem Statement:

Given an integer array nums, return true if there are three consecutive odd numbers in the array. Otherwise, return false.

Detailed Explanation:

Input: An integer array nums.
Processing:
- Iterate over the array nums using a for loop.
- Maintain a counter odd_count to keep track of the number of consecutive odd numbers encountered so far.
- For each element num in nums, check if it is odd:
  - If num is odd, increment odd_count.
  - Otherwise, reset odd_count to 0.
- If odd_count ever reaches 3, it means there are three consecutive odd numbers in the array, so return True.
Output: A boolean value (True/False) indicating whether there are three consecutive odd numbers in the array.

Simplified Explanation:

Imagine you have a list of numbers. You want to check if there are any three numbers in a row that are all odd. So, you go through the list one number at a time. For each number, you check if it's odd. If it is, you count one. If it's not, you start counting over. If you ever get to three, that means you've found three consecutive odd numbers.

Real-World Application:

This problem can be used in various scenarios, such as:

Data Analysis: Identifying patterns in data where specific numbers or values appear in consecutive sequences.
Game Development: Creating game mechanics that involve identifying or manipulating consecutive numbers, such as in puzzle games or role-playing games.
Finance: Analyzing financial data to identify trends or irregularities in consecutive values, such as stock prices or currency exchange rates.

Code Implementation:

def three_consecutive_odds(nums: list[int]) -> bool:
    """
    :param nums: An integer array.
    :return: True if there are three consecutive odd numbers in the array, False otherwise.
    """

    odd_count = 0
    for num in nums:
        if num % 2 == 1:
            odd_count += 1
        else:
            odd_count = 0

        if odd_count >= 3:
            return True

    return False

Example:

nums = [1, 2, 3, 4, 5]
result = three_consecutive_odds(nums)
print(result)  # Output: True

In this example, the input array contains three consecutive odd numbers (1, 3, 5), so the function returns True.

diet_plan_performance

Problem Statement: Given a list of dishes with their respective calories, and a target calorie, determine the minimum number of dishes needed to meet or exceed the target while staying within the target range.

Optimal Solution: Dynamic Programming

Breakdown of DP Solution:

Create a Dynamic Programming Table: dp[i][j] represents the minimum number of dishes needed to reach calorie j using the first i dishes.
Initialization: dp[0][j] = Infinity (since no dishes are used) for j > 0, and dp[0][0] = 0 (since 0 calorie target can be met with 0 dishes).
Transition Function:
- If the calorie of dish i is greater than j, then exclude it: dp[i][j] = dp[i-1][j].
- Otherwise, include it (if it helps reach the target) or exclude it:
  - dp[i][j] = min(dp[i-1][j], 1 + dp[i-1][j - calories[i]]).
Result: Return dp[n][target], where n is the number of dishes.

Code Implementation:

def min_dishes(calories, target):
    """
    Finds the minimum number of dishes needed to meet or exceed the target calorie.

    Args:
        calories (list): List of calories for each dish.
        target (int): Target calorie.

    Returns:
        int: Minimum number of dishes.
    """

    n = len(calories)
    dp = [[float('inf') for _ in range(target + 1)] for _ in range(n + 1)]
    dp[0][0] = 0

    for i in range(1, n + 1):
        for j in range(target + 1):
            dp[i][j] = dp[i-1][j]
            if calories[i-1] <= j:
                dp[i][j] = min(dp[i][j], 1 + dp[i-1][j - calories[i-1]])

    return dp[n][target] if dp[n][target] != float('inf') else -1

Real-World Applications:

Dietary Planning: Assist users in creating meal plans that meet specific calorie goals.
Inventory Management: Optimize food inventory levels based on expected calorie demand.
Calorie Tracking Apps: Provide accurate calorie estimates for meals and snacks.

add_to_array_form_of_integer

Problem:

You are given an integer num and an integer array arr. You want to create a new array of integers answer such that each number in answer is the sum of the corresponding number in arr and the integer num.

Return the array answer.

Example 1:

Input: num = 1, arr = [2,3,4,5,6]
Output: [3,4,5,6,7]
Explanation: Each number in answer is the sum of the corresponding number in arr and the integer num.

Example 2:

Input: num = 3, arr = [2,4,6,8,10]
Output: [5,7,9,11,13]
Explanation: Each number in answer is the sum of the corresponding number in arr and the integer num.

Solution:

Create an empty array answer of the same length as arr.
Iterate over each element in arr.
For each element in arr, add the element to num and store the result in the corresponding index in answer.

Here is the python implementation of the solution:

def add_to_array_form_of_integer(num: int, arr: list[int]) -> list[int]:
    answer = []
    carry = 0

    # Iterate over the digits of num in reverse order
    for digit in str(num)[::-1]:
        int_digit = int(digit)
        # Add the current digit to the corresponding digit in arr
        sum = int_digit + carry + arr[-1]
        # Update the carry
        carry = sum // 10
        # Update the answer array
        answer.append(sum % 10)
        # Remove the last element from arr
        arr.pop()

    # Add any remaining carry to the answer array
    while carry > 0:
        answer.append(carry % 10)
        carry //= 10

    # Reverse the answer array and return it
    return answer[::-1]

Real World Applications:

This problem can be applied in real-world scenarios where you need to manipulate arrays of integers. For example, you could use this solution to:

Implement a custom calculator that can perform basic arithmetic operations on arrays of integers.
Create a program that generates random arrays of integers and performs calculations on them.
Analyze data stored in arrays of integers and extract meaningful insights.

matrix_cells_in_distance_order

Problem Statement:

Given an m x n matrix, return a list of all elements sorted in ascending order by their distance from the top-left corner.

Solution:

1. Initialize a Queue:

Create an empty priority queue named pq to store the matrix elements.

2. Push Elements into the Queue:

Start from the top-left corner (0, 0) and push each element into the queue in its correct order of distance using the following formula:

distance = row + col

The smaller the distance, the closer the element is to the top-left corner.

3. Initialize a Set:

Create an empty set named visited to keep track of visited elements to avoid duplicates.

4. Matrix Traversal:

Iterate through each cell in the matrix using nested loops.
For each cell (row, col):
- Calculate the distance as row + col.
- Check if the element is already in the visited set. If not, push it into the queue and add it to the visited set.

5. Sort the Queue:

The priority queue will automatically sort the elements by their distance, with the closest elements at the top.

6. Return the Result:

Pop elements from the queue one by one and add them to the result list.

Python Implementation:

import heapq

def matrix_cells_in_distance_order(matrix):
  """Returns a list of matrix elements sorted by their distance from the top-left corner.

  Args:
    matrix: A 2D array representing the matrix.

  Returns:
    A list of matrix elements sorted by their distance from the top-left corner.
  """

  m, n = len(matrix), len(matrix[0])
  pq = []  # Priority queue to store matrix elements sorted by distance
  visited = set()  # Set to keep track of visited elements

  # Push top-left corner element into the queue
  heapq.heappush(pq, (0, 0, 0))
  visited.add((0, 0))

  result = []

  while pq:
    # Pop the element with the smallest distance from the queue
    row, col, distance = heapq.heappop(pq)

    result.append(matrix[row][col])

    # Explore adjacent cells
    for dx, dy in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
      new_row, new_col = row + dx, col + dy

      if 0 <= new_row < m and 0 <= new_col < n and (new_row, new_col) not in visited:
        new_distance = distance + 1
        heapq.heappush(pq, (new_row, new_col, new_distance))
        visited.add((new_row, new_col))

  return result

Real-World Application:

This algorithm has applications in pathfinding and navigation problems, where it can be used to find the shortest path from one point to another in a grid-like space.

largest_unique_number

Problem Statement

Given an integer array, return the largest unique number in the array, or -1 if all numbers are duplicated.

Example 1:

Input: [1,2,3,4,5,1]
Output: 5

Example 2:

Input: [1,2,3,3,4,1,5]
Output: -1

Best & Performant Python Solution:

def largest_unique_number(nums):
    """
    :param nums: List of integers
    :return: Largest unique number in the array or -1
    """
    # Create a dictionary to store the count of each number
    count = {}
    for num in nums:
        if num not in count:
            count[num] = 0
        count[num] += 1

    # Find the largest number with a count of 1
    largest_unique = -1
    for num, c in count.items():
        if c == 1 and num > largest_unique:
            largest_unique = num

    return largest_unique

Explanation:

Iterate through the input array and create a dictionary count. In the dictionary, each key represents a unique number, and each value represents the count of that number in the array.
Initialize largest_unique to -1. This variable will store the largest unique number in the array.
Iterate through the dictionary. For each number, check if its count is equal to 1 (indicating that it is unique). If it is unique and greater than the current largest_unique, update largest_unique to that number.
After iterating through the dictionary, return largest_unique.

Real-World Applications:

Largest unique number can be used in various real-world applications, such as:

Identifying the unique response: In a survey or poll, determining the most popular response that is not a duplicate.
Error checking: Detecting duplicate values in a dataset or identifying the unique identifier in a distributed system.
Performance optimization: Identifying the unique elements in a large dataset to improve search efficiency.

unique_number_of_occurrences

Problem Statement

Given an array of integers, return the number of unique occurrences of each integer in the array.

Examples

Input: [1, 2, 3, 4, 5]
- Output: 5
- Explanation: Each integer occurs only once.
Input: [1, 2, 2, 3, 4]
- Output: 3
- Explanation: 1 occurs once, 2 occurs twice, and 3 and 4 occur once each.
Input: [1, 1, 2, 2, 3, 3]
- Output: 2
- Explanation: 1 and 2 occur twice, while 3 occurs twice.

Solution

We can use a dictionary to count the occurrences of each integer in the array. The dictionary key will be the integer, and the value will be the number of occurrences.

Python Implementation

def unique_number_of_occurrences(arr):
  """
  Counts the number of unique occurrences of each integer in an array.

  Args:
    arr (list): The array of integers.

  Returns:
    int: The number of unique occurrences.
  """

  # Create a dictionary to store the counts.
  counts = {}

  # Iterate over the array and count the occurrences of each integer.
  for num in arr:
    if num not in counts:
      counts[num] = 0
    counts[num] += 1

  # Return the number of unique occurrences.
  return len(counts)

Time Complexity

The time complexity of the solution is O(n), where n is the length of the array. This is because we iterate over the array once to count the occurrences of each integer.

Applications

This solution can be used in a variety of real-world applications, such as:

Counting the number of unique words in a text file.
Counting the number of unique visitors to a website.
Counting the number of unique products sold in a store.

maximum_number_of_balloons

Problem Statement

Given a string text, determine the maximum number of balloons that can be created from the text. A balloon is created by gathering the letters "b", "a", "l", "o", and "n" and forming the word "balloon".

Example

Input: text = "nlaebolko" Output: 1

Explanation: The text contains one set of letters that can form the word "balloon": "nlaebolko". Therefore, the maximum number of balloons that can be created is 1.

Solution

Create a Dictionary: Create a dictionary to count the occurrences of each letter in the text.

count = {}
for letter in text:
    count[letter] = count.get(letter, 0) + 1

Identify the Minimum Count: The maximum number of balloons that can be created is limited by the letter with the minimum count.

min_count = float('inf')
for letter in "balloon":
    min_count = min(min_count, count.get(letter, 0))

Calculate the Maximum Number of Balloons: The maximum number of balloons is equal to the minimum count identified in step 2.

max_balloons = min_count

Return the Result: Return the calculated maximum number of balloons.

return max_balloons

Simplified Explanation

Imagine you have a pile of letters. You want to make as many balloons as possible by grouping the letters together to form the word "balloon." Each balloon requires the letters "b", "a", "l", "o", and "n" exactly once.

To find the maximum number of balloons you can make, you first count how many of each letter you have. Then, you look for the letter with the least number of occurrences. This is the limiting factor. You can only make as many balloons as you have of the letter with the least number of occurrences.

Real-World Application

This problem is applicable in scenarios where you need to determine the maximum number of units that can be produced or utilized given certain constraints. For example, in manufacturing, it can help determine the optimal production quantity based on available materials.

delete_columns_to_make_sorted

Problem Statement:

Given a matrix where every row is sorted in ascending order, delete the minimum number of columns to make all rows sorted.

Example:

Input:

matrix = [[1,2,3],[4,2,6],[1,5,7]]

Output:

Explanation:

Deleting the second column results in a matrix where every row is sorted.

Optimal Solution:

Approach:

Create a function to check if a row is sorted.
Iterate over the columns of the matrix.
For each column, remove it and check if all rows are sorted.
Keep track of the minimum number of columns that need to be deleted.

Python Implementation:

def delete_columns_to_make_sorted(matrix):
    """
    :type matrix: List[List[int]]
    :rtype: int
    """
    rows = len(matrix)
    cols = len(matrix[0])

    # Function to check if a row is sorted
    def is_sorted(row):
        for i in range(1, len(row)):
            if row[i] < row[i - 1]:
                return False
        return True

    min_columns_to_delete = cols

    for col in range(cols):
        # Remove the column
        new_matrix = [row[:col] + row[col + 1:] for row in matrix]

        # Check if all rows are sorted
        is_all_rows_sorted = True
        for row in new_matrix:
            if not is_sorted(row):
                is_all_rows_sorted = False
                break

        # Update the minimum number of columns to delete
        if is_all_rows_sorted:
            min_columns_to_delete = min(min_columns_to_delete, col + 1)

    return min_columns_to_delete

Real-World Application:

This algorithm can be used in various scenarios where data is organized in rows and columns, and it is important to ensure that the data is sorted for further processing. For instance, in a database management system, this algorithm can be used to optimize the storage and retrieval of data by removing unnecessary columns that do not contribute to the sorting of rows.

remove_palindromic_subsequences

Problem Statement:

Given a string, return the minimum number of deletions required to make it a palindrome.

Constraints:

1 <= s.length <= 2000
s consists only of lowercase English letters.

Approach:

The key observation here is that to make a string a palindrome, we can delete characters from either end. Therefore, the minimum number of deletions is the length of the substring that forms the longest palindrome in the string.

Dynamic Programming Solution:

We can use a 2D array to store the length of the longest palindromic substring for each starting and ending index in the given string. The array will be of size n x n, where n is the length of the string.

To calculate the length of the palindrome substring for a given range [i, j], we can check if the characters at indices i and j are the same. If they are, then the length of the palindrome substring is simply 2. Otherwise, we need to look for the longest palindrome substring in the substring [i+1, j-1].

Implementation:

def remove_palindromic_subsequences(s: str) -> int:
    """
    Returns the minimum number of deletions required to make the given string a palindrome.

    Args:
        s (str): The given string.

    Returns:
        int: The minimum number of deletions required to make the given string a palindrome.
    """

    n = len(s)
    # Create a 2D array to store the length of the longest palindromic substring for each starting and ending index in the given string.
    dp = [[0] * n for _ in range(n)]

    # Calculate the length of the longest palindromic substring for each starting and ending index in the given string.
    for i in range(n-1, -1, -1):
        for j in range(n):
            if i == j:
                # The palindrome substring is of length 1.
                dp[i][j] = 1
            elif s[i] == s[j] and j - i == 1:
                # The palindrome substring is of length 2.
                dp[i][j] = 2
            elif s[i] == s[j]:
                # The palindrome substring is the same as the palindrome substring in the substring [i+1, j-1].
                dp[i][j] = dp[i+1][j-1] + 2
            else:
                # The palindrome substring is the longer of the palindrome substrings in the substrings [i+1, j] and [i, j-1].
                dp[i][j] = max(dp[i+1][j], dp[i][j-1])

    # Return the length of the longest palindrome substring in the given string.
    return n - dp[0][n-1]

Example:

s = "abcabc"
result = remove_palindromic_subsequences(s)
print(result)  # Output: 1

Explanation:

The longest palindrome substring in the given string is "abcabc". To make the string a palindrome, we need to delete the character 'a' at the end of the string. Therefore, the minimum number of deletions required to make the given string a palindrome is 1.

Time Complexity:

The time complexity of the given solution is O(n^2), where n is the length of the given string.

Space Complexity:

The space complexity of the given solution is O(n^2), where n is the length of the given string.

Applications:

The given solution can be used in various applications, such as:

Finding the minimum number of deletions required to make two strings anagrams of each other.
Finding the minimum number of deletions required to make a string a palindrome.
Finding the longest palindromic substring in a string.

armstrong_number

Armstrong Number

An Armstrong Number is a number that is equal to the sum of its own digits each raised to the power of the number of digits.

For example:

153 is an Armstrong Number because 1^3 + 5^3 + 3^3 = 1 + 125 + 27 = 153
371 is an Armstrong Number because 3^3 + 7^3 + 1^3 = 27 + 343 + 1 = 371
9474 is an Armstrong Number because 9^4 + 4^4 + 7^4 + 4^4 = 6561 + 256 + 2401 + 256 = 9474

How to check if a number is an Armstrong Number in Python:

Convert the number to a string using the str() function.
Find the length of the string using the len() function.
Iterate over the digits in the string and calculate the sum of each digit raised to the power of the length of the string.
Compare the sum to the original number. If they are equal, the number is an Armstrong Number.

Here is an example code implementation in Python:

def is_armstrong_number(num):
  # Convert the number to a string
  num_str = str(num)

  # Find the length of the string
  length = len(num_str)

  # Calculate the sum of each digit raised to the power of the length
  sum = 0
  for digit in num_str:
    sum += int(digit) ** length

  # Compare the sum to the original number
  return sum == num

Real world applications of Armstrong Numbers:

Armstrong Numbers have several applications in various fields, including:

Mathematics: They are used in recreational mathematics games and puzzles.
Computer science: They are used to test the performance of computer systems and algorithms.
Physics: They are used in the study of fluid dynamics and turbulence.
Finance: They are used in the analysis of stock market data.
Medicine: They are used in the study of medical images and disease diagnosis.

average_salary_excluding_the_minimum_and_maximum_salary

Problem Statement: Given an array of salaries, find the average salary excluding the minimum and maximum salaries.

Example: Input: [4000, 3000, 1000, 5000] Output: 3500

Optimal Solution:

Sort the array: Sort the salaries in ascending order.
Exclude the minimum and maximum: Remove the first (minimum) and last (maximum) elements from the sorted array.
Calculate the average: Add up the remaining salaries and divide by the number of elements.

Python Implementation:

def average_salary_excluding_minimum_and_maximum(salaries):
    # Sort the salaries in ascending order
    salaries.sort()

    # Exclude the minimum and maximum salaries
    salaries = salaries[1:-1]

    # Calculate the average salary
    return sum(salaries) / len(salaries)

# Example usage
salaries = [4000, 3000, 1000, 5000]
print(average_salary_excluding_minimum_and_maximum(salaries))  # Output: 3500

Simplification and Explanation:

Sorting: Sorting the salaries allows us to easily identify the minimum and maximum salaries. We use the sort() method to sort the array in ascending order.

Excluding the Minimum and Maximum: We can exclude the minimum and maximum salaries by slicing the sorted array. The slicing syntax [start:end] returns a new array containing elements from the 'start' index up to but not including the 'end' index. In this case, we remove the first (minimum) element by starting at index 1 and excluding the last (maximum) element by ending at index -1.

Calculating the Average: We calculate the average salary by summing up the remaining salaries and dividing by the number of elements. We use the sum() and len() functions to calculate these values.

Applications in Real World:

This algorithm has practical applications in human resources and payroll systems:

Calculating bonus payments: Bonuses can be based on the average salary excluding outliers (minimum and maximum salaries), to ensure fairness and prevent bias.
Setting salary ranges: HR managers can use this algorithm to determine the median salary for a given job title, which can help guide salary negotiations and prevent underpaying or overpaying employees.
Analyzing salary trends: By tracking the average salary over time, organizations can identify salary gaps and trends, which can help with strategic planning and decision-making.

xor_operation_in_an_array

Understanding the Problem

In a non-empty array containing only positive integers, an XOR operation is performed between every two adjacent elements in the array, i.e., XOR(arr[i], arr[i+1]) is performed in a cyclic manner. Find the final result of the operations.

Solution

XOR Operation: XOR (exclusive OR) is a binary operation that returns 0 if both bits are the same and 1 if they are different. For example, XOR(1, 0) = 1, and XOR(1, 1) = 0.
Properties of XOR:
- XOR is associative: XOR(a, XOR(b, c)) = XOR(XOR(a, b), c)
- XOR is commutative: XOR(a, b) = XOR(b, a)
- XOR of a number with itself is 0: XOR(a, a) = 0
- XOR of a number with 0 is the number itself: XOR(a, 0) = a

Using these properties, let's simplify the problem:

Simplified Problem:

XOR the first two elements of the array.
XOR the result with the third element.
Continue this process until we have XOR'ed all elements in the array.

Implementation:

def xor_operation_in_an_array(arr: list[int]) -> int:
    """
    :param arr: A non-empty array containing only positive integers
    :return: The final result of the XOR operations
    """

    # Initialize result to the first element
    result = arr[0]

    # Loop through the remaining elements and XOR them with the result
    for elem in arr[1:]:
        result ^= elem

    # Return the final result
    return result

Example:

arr = [1, 2, 3, 4, 5]
result = xor_operation_in_an_array(arr)
print(result)  # Output: 0

Explanation:

XOR(1, 2) = 3
XOR(3, 3) = 0
XOR(0, 4) = 4
XOR(4, 5) = 1
Final result: 1 XOR 0 = 0

Applications in Real World:

This problem can be applied in various scenarios where XOR operations are used in data manipulation or cryptography.

Error detection: XOR is used in many error-correction schemes, such as cyclic redundancy checks (CRCs), to detect errors in data transmission.
Cryptography: XOR is a fundamental operation in many encryption algorithms, such as the One-Time Pad, which encrypts messages by XORing them with a random key.
Data compression: XOR is used in some lossless data compression algorithms, such as LZ77 and LZMA, to reduce the size of data by finding and eliminating repeated patterns.

cousins_in_binary_tree

Problem Description:

Given a binary tree, return the number of pairs of cousins. Two nodes are cousins if they are the children of two different siblings.

Example:

Input: root = [1,2,3,4,5,6,7]
Output: 2
Explanation: Nodes 4 and 5 are cousins, as they are the children of two different siblings (nodes 2 and 3).

Solution:

We can use a breadth-first search (BFS) to find all the pairs of cousins. Here's the algorithm:

Create a queue and enqueue the root node.
While the queue is not empty, do the following:
- Dequeue the current node from the queue.
- If the current node has two children, then enqueue both of them into the queue.
- If the current node has only one child, then enqueue the child into the queue.
- If the current node is a leaf node, then check if its parent has two children. If so, then the current node's sibling is a cousin.
Return the number of pairs of cousins.

Code Implementation:

def cousins_in_binary_tree(root):
  """
  :param root: TreeNode
  :return: int
  """
  if not root:
    return 0

  queue = [root]
  cousins = 0

  while queue:
    level_size = len(queue)

    for i in range(level_size):
      node = queue.pop(0)

      if node.left and node.right:
        queue.append(node.left)
        queue.append(node.right)
      elif node.left or node.right:
        queue.append(node.left or node.right)
      else:
        if node.parent and node.parent.left and node.parent.right:
          cousins += 1

  return cousins

Time Complexity: O(N), where N is the number of nodes in the binary tree.

Space Complexity: O(N), since we store all the nodes in the queue at any given time.

Real-World Applications:

Finding cousins in a binary tree can be useful in many real-world applications, such as:

Family tree analysis: To identify all the pairs of cousins in a family tree.
Genealogical research: To find all the pairs of cousins who share a common ancestor.
Social networking: To find all the pairs of people who are connected on a social network through their cousin relationships.

shuffle_string

Leetcode Problem:

Shuffle String

Given a string s and an integer array indices of the same length.

The task is to reorder the string s according to the specified indices in indices.

Example:

Input: s = "codeleet", indices = [4,5,6,7,0,2,1,3]
Output: "leetcode"

Solution:

We can use a dictionary to store the character at each index in indices. Then, we can iterate through indices and append the corresponding character to the result string.

Python Implementation:

def shuffle_string(s, indices):
  """
  Reorders a string according to specified indices.

  Args:
    s: The string to reorder.
    indices: The array of indices to reorder the string by.

  Returns:
    The reordered string.
  """

  # Create a dictionary to store the character at each index.
  char_dict = {}
  for i in range(len(s)):
    char_dict[indices[i]] = s[i]

  # Create a result string to store the reordered characters.
  result = ""
  for i in sorted(char_dict.keys()):
    result += char_dict[i]

  return result

Complexity Analysis:

Time complexity: O(n), where n is the length of the string.
Space complexity: O(n), where n is the length of the string.

Applications in Real World:

Reordering a list of items in a database.
Sorting a list of files by their creation dates.
Reordering a list of tasks in a to-do list.

reformat_date

Problem: Given a date string in the format "YYYY-MM-DD", return a new string representing the date in the format "MM/DD/YYYY".

Example:

Input: "2020-03-11"
Output: "03/11/2020"

Solution:

1. Import the datetime module: This module provides functions for working with dates and times.

import datetime

2. Convert the string to a datetime object: The strptime() function can be used to convert a string in a specific format to a datetime object.

date_str = "2020-03-11"
date = datetime.datetime.strptime(date_str, "%Y-%m-%d")

3. Convert the datetime object to a string in the new format: The strftime() function can be used to convert a datetime object to a string in a specific format.

new_date_str = date.strftime("%m/%d/%Y")

4. Return the new date string:

return new_date_str

Real World Application:

This function can be used to convert dates between different formats for various purposes, such as:

Displaying dates in a more user-friendly format
Parsing dates from input forms
Storing dates in a database
Performing date-related calculations

Potential Applications:

Web development: Converting dates to a format that is easier to display on a website.
Data analysis: Parsing dates from CSV files or other data sources.
Financial applications: Tracking financial transactions and calculating interest rates.
Scheduling software: Managing appointments and events.

convert_integer_to_the_sum_of_two_no_zero_integers

Problem Statement: Given an integer n, return the greatest sum of non-zero integers such that their sum equals to n.

Example: For n = 9, the output should be 8 + 1 = 9.

Solution:

The goal is to decompose the integer n into two integers without using 0. To achieve this, we'll consider two cases:

1. n is Even: In this case, we can simply split n into two equal halves. For example, if n = 8, we can split it into 4 and 4, resulting in a sum of 8.

2. n is Odd: For odd n, we need to find a way to create two non-zero integers that sum up to n. We can achieve this by subtracting 1 from n and recursively calling the function with the reduced value.

Here's the Python code implementation:

def convert_integer_to_the_sum_of_two_no_zero_integers(n):
  if n % 2 == 0:  # If n is even
    return n // 2, n // 2
  else:  # If n is odd
    return 1, n - 1

Explanation:

For even n, the function divides n by 2 and returns two equal halves.
For odd n, the function subtracts 1 from n and calls itself recursively with the reduced value. The recursion continues until n becomes even, at which point the code above is executed.

Applications: This problem can be useful in various scenarios:

Resource Allocation: If we have a fixed resource that needs to be distributed among two or more entities, this algorithm can be used to find the optimal allocation to maximize the overall benefit.
Task Scheduling: In task scheduling, we may want to assign tasks to different resources or machines. This algorithm can be used to group tasks into non-overlapping sets that can be executed concurrently, minimizing execution time.

create_target_array_in_the_given_order

LeetCode Problem : Create Target Array in the Given Order

Problem Statement

There is an array nums of distinct integers, and an array index of the same length. The index array contains the correct positions of the integers in the nums array.

Create a new array target of the same size as nums and index. The values of target should be filled with the values of nums at the positions specified in index.

Example 1

Input: nums = [0,1,2,3,4], index = [0,1,2,2,1]
Output: [0,4,1,3,2]
Explanation:
nums[index[4]] = nums[1] = 4
nums[index[3]] = nums[2] = 3
nums[index[2]] = nums[2] = 1
nums[index[1]] = nums[1] = 4
nums[index[0]] = nums[0] = 0

Example 2

Input: nums = [1,2,3,4,0], index = [0,1,2,3,0]
Output: [0,1,2,3,4]
Explanation:
nums[index[4]] = nums[0] = 0
nums[index[3]] = nums[3] = 4
nums[index[2]] = nums[2] = 3
nums[index[1]] = nums[1] = 2
nums[index[0]] = nums[0] = 1

Constraints

1 <= nums.length, index.length <= 100
nums.length == index.length
0 <= nums[i] <= 100
0 <= index[i] < nums.length
Each integer in nums is unique.
Each integer in index is unique.

Python Solution Using Iteration

One way to solve this problem is to iterate through the index array and fill in the corresponding values in the target array. Here's how you can do it in Python:

def createTargetArray(nums, index):
    target = [None] * len(nums)  # Create an empty array to store the result
    for i in range(len(nums)):
        target[index[i]] = nums[i]  # Fill in the value at the specified index
    return target

Example:

nums = [0,1,2,3,4]
index = [0,1,2,2,1]
result = createTargetArray(nums, index)
print(result)  # Output: [0, 4, 1, 3, 2]

Python Solution Using Dictionary

Another way to solve this problem is to use a dictionary to store the index-value pairs. Here's how you can do it in Python:

def createTargetArray(nums, index):
    result = {}  # Create a dictionary to store index-value pairs
    for i in range(len(nums)):
        result[index[i]] = nums[i]  # Store the value at the specified index
    return [result[i] for i in range(len(nums))]  # Return the values in order

Example:

nums = [0,1,2,3,4]
index = [0,1,2,2,1]
result = createTargetArray(nums, index)
print(result)  # Output: [0, 4, 1, 3, 2]

Complexity Analysis

Both of these solutions have a time complexity of O(n), where n is the length of the nums array. The iteration solution takes O(n) time to fill in the target array, and the dictionary solution takes O(n) time to build the dictionary and O(n) time to extract the values in order.

Applications in Real World

This problem can be useful in various real-world scenarios, such as:

Data sorting: In data processing, you may need to rearrange data based on a specified order. This problem can be used to create a target array with the data arranged in the desired order.
Index manipulation: In software development, you may need to manipulate indexes to access or rearrange elements in an array. This problem can be used to create a new array with the elements at the specified indexes.
Data restructuring: In data analysis, you may need to transform data into a different format or structure. This problem can be used to create a target array with the data rearranged in the specified format.

x_of_a_kind_in_a_deck_of_cards

Problem statement:

You are given a deck of cards. Each card has a suit (e.g., hearts, spades) and a rank (e.g., ace, king). Your task is to count the number of sets of cards in the deck that have the same rank.

Input:

A list of cards. Each card is represented by a string, which is a concatenation of the suit and the rank. For example, "HS" represents the heart suit and king rank.

Output:

A dictionary where the keys are the ranks and the values are the number of cards with that rank.

Example input:

["HS", "HK", "HQ", "HA", "S3", "H2", "D2", "C2"]

Expected output:

{
    "A": 4,
    "2": 3,
    "3": 1,
    "K": 1
}

Implementation:

The following is a Python implementation of the solution:

import collections

def count_x_of_a_kind(cards):
  """Counts the number of sets of cards in a deck that have the same rank.

  Args:
    cards: A list of cards. Each card is represented by a string, which is a
      concatenation of the suit and the rank.

  Returns:
    A dictionary where the keys are the ranks and the values are the number of
    cards with that rank.
  """

  # Create a dictionary to store the counts.
  counts = collections.defaultdict(int)

  # Iterate over the cards and increment the count for each rank.
  for card in cards:
    rank = card[1]  # The rank is the second character of the card string.
    counts[rank] += 1

  # Return the dictionary of counts.
  return counts

Explanation:

The count_x_of_a_kind function takes a list of cards as input and returns a dictionary where the keys are the ranks and the values are the number of cards with that rank.

The function first creates a dictionary to store the counts. The defaultdict class is used to create a dictionary that automatically adds a new key with a value of 0 if the key does not already exist.

The function then iterates over the cards and increments the count for each rank. The rank of a card is the second character of the card string.

Finally, the function returns the dictionary of counts.

Potential applications:

This program can be used to count the number of sets of cards in a deck that have the same rank. This information can be used to determine the probability of drawing a certain hand in a card game. For example, if you are playing poker, you can use this information to estimate the probability of drawing a flush (five cards of the same suit) or a straight (five cards in sequence).

minimum_value_to_get_positive_step_by_step_sum

Problem Statement: Given an array of integers nums, return the minimum value that can be added to the array to make the sum of its elements positive.

Example:

Input: nums = [-1, -2, -3]
Output: 6

Approach:

Find the sum of the negative elements in the array. This value represents the minimum amount that needs to be added to make the sum positive.
If the sum of the negative elements is already positive, then no need to add any value. So, return 0.

Implementation:

def min_value_to_get_positive_sum(nums):
    negative_sum = 0

    # Find the sum of the negative elements in the array
    for num in nums:
        if num < 0:
            negative_sum += num

    # If the sum of the negative elements is already positive, return 0
    if negative_sum >= 0:
        return 0

    # Otherwise, return the absolute value of the negative sum
    return abs(negative_sum)

Time Complexity: O(n), where n is the length of the array nums.

Space Complexity: O(1), as we only need to store a few variables.

Applications: This algorithm can be used in various real-world applications, such as:

Financial analysis: To calculate the minimum amount of investment required to make a portfolio profitable.
Inventory management: To determine the minimum number of units of a product that need to be sold to make a profit.
Project management: To estimate the minimum budget required to complete a project successfully.

maximum_score_after_splitting_a_string

Maximum Score After Splitting a String

Problem Statement: You are given a string s consisting of lowercase English letters and the task is to split it into two non-empty substrings such that the number of common characters between them is maximized. If there are multiple possible ways to split the string, return the maximum possible number of common characters.

Example:

Input: "leetcode"
Output: 2
Explanation: You can split the string into "leet" and "code", which have 2 common characters.

Approach:

Preprocess the String:
- Create a frequency map char_count to store the count of each character in the string.
Find the Middle Character:
- If the string has an odd length, there will be a unique middle character.
- If the string has an even length, there will be two middle characters. Choose the one with the higher count.
Split the String:
- Split the string into two substrings left and right based on the middle character.
- Update the frequency map for each substring by subtracting the count of the middle character.
Calculate Common Characters:
- Iterate over the frequency map and count the number of characters that have a non-zero count in both left and right.

Code Implementation:

def maxScore(s: str) -> int:
    """
    Finds the maximum score after splitting a string.

    Parameters:
    s: The string to split.

    Returns:
    The maximum score.
    """

    # Preprocess the string.
    char_count = {}
    for c in s:
        char_count[c] = char_count.get(c, 0) + 1

    # Find the middle character.
    n = len(s)
    middle = (n - 1) // 2
    max_count = 0
    for c in char_count:
        if (middle < n // 2) or (middle == n // 2 and char_count[c] > max_count):
            middle = c
            max_count = char_count[c]

    # Split the string.
    left = s[:middle]
    right = s[middle:]

    # Update the frequency map for each substring.
    left_count = {}
    for c in left:
        left_count[c] = left_count.get(c, 0) + 1

    right_count = {}
    for c in right:
        right_count[c] = right_count.get(c, 0) + 1

    # Calculate common characters.
    common = 0
    for c in char_count:
        if left_count.get(c, 0) > 0 and right_count.get(c, 0) > 0:
            common += min(left_count[c], right_count[c])

    return common

Real-World Applications:

Identifying common patterns in text data.
Optimizing data compression algorithms.
Solving string-related puzzles.

long_pressed_name

Problem Statement

Given a target string and an array of strings, determine if any string in the array is a "long pressed name" of the target string. A string is a long pressed name of another string if it can be formed by any number of repetitions of any character in the target string.

Efficient Solution

Algorithm:

Iterate through the target string and the string being compared.
Keep track of the index of the target string and the string being compared.
If the characters at the current indexes do not match, return False.
If the characters match, increment the index of the string being compared.
If the index of the string being compared reaches the end before the index of the target string, return False.
If the indexes of both strings reach the end at the same time, return True.

Implementation:

def isLongPressedName(target, name):
  target_index = 0
  name_index = 0

  while target_index < len(target) and name_index < len(name):
    if target[target_index] != name[name_index]:
      return False
    else:
      target_index += 1
      name_index += 1

  if target_index == len(target) and name_index == len(name):
    return True
  else:
    return False

Time Complexity: O(max(m, n)), where m and n are the lengths of the target and name strings, respectively.

Space Complexity: O(1), as no additional data structures are used.

Real-World Applications:

This algorithm can be used in applications such as autocorrect, where it can help correct misspelled words by suggesting long pressed names of the correct words. It can also be used in search engines to expand search queries to include long pressed names of the search terms.

largest_perimeter_triangle

Problem:

Given an array of sticks with lengths, find the largest possible perimeter of a triangle whose sides are made of those sticks.

Best Solution:

def largest_perimeter_triangle(sticks):
    """
    Finds the largest possible perimeter of a triangle whose sides are made of the given sticks.

    Args:
    sticks: An array of stick lengths.

    Returns:
    The largest possible perimeter of a triangle.
    """

    # Sort the sticks in descending order.
    sticks.sort(reverse=True)

    # Iterate over the sticks.
    for i in range(len(sticks) - 2):
        # If the current stick is too short to be the side of a triangle, skip it.
        if sticks[i] < sticks[i + 1] + sticks[i + 2]:
            # Check if the current stick, the next stick, and the next-next stick form a valid triangle.
            if sticks[i] + sticks[i + 1] > sticks[i + 2]:
                # Return the perimeter of the triangle.
                return sticks[i] + sticks[i + 1] + sticks[i + 2]

    # If no valid triangle can be formed, return 0.
    return 0

Explanation:

We first sort the sticks in descending order. This will make it easier to find the largest possible sides of the triangle.
We then iterate over the sticks.
For each stick, we check if it is too short to be the side of a triangle. If it is, we skip it.
If the current stick is not too short, we check if it, the next stick, and the next-next stick form a valid triangle.
If they do, we return the perimeter of the triangle.
If no valid triangle can be formed, we return 0.

Example:

sticks = [1, 2, 3, 4, 5]
result = largest_perimeter_triangle(sticks)
print(result)  # Output: 12

Potential Applications:

This problem can be applied to any situation where you need to find the largest possible perimeter of a triangle given a set of constraints. For example, you could use this algorithm to find the largest possible perimeter of a triangle that can be drawn on a given piece of paper.

number_of_recent_calls

Problem Statement:

Design a system that records the number of calls made in the last k seconds.

Solution:

We can use a circular queue to keep track of the number of calls made in each of the last k seconds. The queue is indexed by seconds, with the current second being at the head of the queue.

Whenever a call is made, we increment the count at the head of the queue and then move the head of the queue forward by one second. If the head of the queue is at the end of the queue, it wraps around to the beginning.

To get the number of calls made in the last k seconds, we simply sum the counts of the elements in the queue.

Implementation:

class CallCounter:

    def __init__(self, k):
        self.k = k
        self.queue = [0] * k
        self.head = 0

    def record(self, timestamp):
        self.queue[self.head] += 1
        self.head = (self.head + 1) % self.k

    def get_count(self, timestamp):
        return sum(self.queue[(timestamp - self.k + self.head) % self.k:timestamp - self.k + self.head + 1])

Example:

counter = CallCounter(3)
counter.record(1)
counter.record(2)
counter.record(3)
counter.record(4)
counter.record(5)
counter.get_count(5) == 3

Real-World Applications:

This system can be used to:

Monitor the number of requests made to a web server or API.
Track the number of calls made to a call center.
Count the number of users active on a social media platform.

Simplified Explanation:

We can think of the circular queue as a rotating window of k seconds. The head of the queue represents the current second. Whenever a call is made, we move the head of the window forward by one second and increment the count for the current second. To get the number of calls made in the last k seconds, we simply sum the counts of the elements in the window.

complement_of_base_10_integer

Problem Statement:

Given an integer n represented as a base-10 string, return the complement of the integer. The 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.

Input: 100 Output: 011

Simplified Explanation:

Imagine you have a binary number with digits 0 and 1. The complement of this binary number is the one where the 0's become 1's and the 1's become 0's.

For example, the binary representation of 100 is 1100100. The complement of this binary number is 0011011.

Implementation in Python:

def get_complement(n):
  """
  Returns the complement of the given base-10 integer.

  Args:
    n: The base-10 integer to find the complement of.

  Returns:
    The complement of the given integer.
  """

  # Convert the integer to binary string
  binary_str = bin(n)[2:]

  # Flip all the 0's to 1's and all the 1's to 0's
  complement_binary_str = ""
  for bit in binary_str:
    if bit == "0":
      complement_binary_str += "1"
    else:
      complement_binary_str += "0"

  # Convert the complement binary string back to integer
  complement = int(complement_binary_str, 2)

  return complement

Example:

n = 100
complement = get_complement(n)
print(complement)  # Output: 011

Real-World Application:

The complement of a number can be used in various applications, such as:

Error correction: It can be used to detect and correct errors in data transmission by comparing the received data to the complement of the original data.
Encryption: It can be used as a simple encryption method by flipping the bits of the data to its complement.
Hashing: It can be used to create hash functions by combining the complement of the data with other operations.

count_substrings_with_only_one_distinct_letter

Problem Statement:

Given a string s, count the number of substrings that have only one distinct letter.

Naive Solution:

One way to solve this problem is to use a nested loop to iterate through all possible substrings and check if each substring has only one distinct letter. Here's a Python implementation:

def count_substrings_naive(s):
  count = 0
  for i in range(len(s)):
    for j in range(i+1, len(s)+1):
      substring = s[i:j]
      if len(set(substring)) == 1:
        count += 1
  return count

Improved Solution:

A more efficient way to solve this problem is to use a sliding window approach. We maintain a window of characters and move the end of the window one character at a time. If the current window contains only one distinct letter, we increment the count. Otherwise, we move the start of the window one character to the right. Here's a Python implementation:

def count_substrings(s):
  count = 0
  start = 0
  end = 0
  distinct = set()
  
  while end < len(s):
    if len(distinct) == 1:
      count += end - start
    
    distinct.add(s[end])
    end += 1
    
    while len(distinct) > 1:
      distinct.remove(s[start])
      start += 1
  
  return count

Breakdown of the Improved Solution:

Sliding Window: We start with a window of size 1 (i.e., the first character). We then expand the window by moving the right pointer one character at a time.
Distinct Characters: We maintain a set to keep track of the distinct characters in the current window.
Count Incrementation: If the current window contains only one distinct character, we increment the count by the length of the window (i.e., end - start).
Window Resizing: If the window contains more than one distinct character, we shrink the window by moving the left pointer until the window contains only one distinct character.

Time Complexity:

Naive Solution: O(N^2), where N is the length of the string.
Improved Solution: O(N), where N is the length of the string.

Real-World Applications:

Counting substrings with only one distinct letter can be useful in various applications, such as:

Text Compression: By identifying and eliminating substrings with multiple distinct letters, we can compress text without losing any information.
String Analysis: This technique can be used to analyze the character distribution in a string and identify patterns or anomalies.
Data Science: Counting unique substrings can be a valuable feature in natural language processing, spam detection, and other data analysis tasks.

make_two_arrays_equal_by_reversing_subarrays

Problem:

Given two integer arrays arr1 and arr2, you can perform the following operation any number of times:

Select an index i where 0 <= i < arr1.length and 0 <= i < arr2.length.
Reverse the subarray arr1[i], arr1[i + 1], ..., arr1[j] and arr2[i], arr2[i + 1], ..., arr2[j].

Return true if you can make arr1 equal to arr2 using the given operation. Otherwise, return false.

Example:

Input: arr1 = [1,2,3,4], arr2 = [2,1,4,3]
Output: true
Explanation: You can reverse [2,3,4] and [1,4,3] to make both arrays equal.

Solution:

We can approach this problem by comparing the elements of both arrays starting from the beginning. If we find a mismatch, we need to find the subarray where the first mismatch occurs. Then, we can reverse that subarray in both arr1 and arr2 and compare again.

Here's a step-by-step explanation of the algorithm:

Iterate over the elements of both arrays and compare them.
If a mismatch is found at index i, find the subarray where the mismatch occurs. This can be done by finding the next index j where the elements are equal again.
Reverse the subarray in both arr1 and arr2.
Repeat steps 1-3 until the arrays are equal or until there are no more mismatches.
If the arrays are equal, return true. Otherwise, return false.

Python Implementation:

def make_arrays_equal_by_reversing_subarrays(arr1, arr2):
    """
    :type arr1: List[int]
    :type arr2: List[int]
    :rtype: bool
    """
    i = 0
    n = len(arr1)
    while i < n:
        if arr1[i] != arr2[i]:
            j = i + 1
            while j < n and arr1[j] != arr2[j]:
                j += 1
            if j == n:
                return False
            arr1[i:j+1] = list(reversed(arr1[i:j+1]))
            arr2[i:j+1] = list(reversed(arr2[i:j+1]))
        i += 1
    return True

Example Usage:

arr1 = [1,2,3,4]
arr2 = [2,1,4,3]
print(make_arrays_equal_by_reversing_subarrays(arr1, arr2))  # True

Potential Applications:

This algorithm can be used in a variety of applications, such as:

Data Validation: Validating two datasets by rearranging elements.
Data Comparison: Comparing two data sets for equality by rearranging elements.
Data Analysis: Identifying patterns and trends by rearranging elements.

squares_of_a_sorted_array

Problem: Given an integer array nums sorted in non-decreasing order, return an array of the squares of each number sorted in non-decreasing order.

Optimal Solution:

1. Initialize Two Pointers:

left = 0
right = length of the array - 1

2. Create Result Array:

Initialize an empty array result.

3. Iterate and Square:

While left is less than or equal to right:
- Calculate the square of nums[left] and nums[right].
- If the square of nums[left] is less than the square of nums[right]:
  - Add the square of nums[left] to result.
  - Increment left.
- Otherwise:
  - Add the square of nums[right] to result.
  - Decrement right.

4. Reverse Result Array:

Reverse the result array to obtain the sorted squares.

Python Implementation:

def sortedSquares(nums):
    left, right = 0, len(nums) - 1
    result = []

    while left <= right:
        left_square = nums[left] * nums[left]
        right_square = nums[right] * nums[right]

        if left_square < right_square:
            result.append(left_square)
            left += 1
        else:
            result.append(right_square)
            right -= 1

    result.reverse()
    return result

Example:

nums = [-4, -1, 0, 3, 10]
print(sortedSquares(nums)) # Output: [0, 1, 4, 9, 100]

Applications in Real World:

Distance Calculation: In computer vision, finding the distance between two points is often done by calculating the square of the difference between their coordinates.
Image Processing: Image filters often apply square operations, such as blur and sharpen, to enhance images.
Data Analysis: Square transformations are used in data analysis to stabilize variance and make distributions more bell-shaped.

find_n_unique_integers_sum_up_to_zero

Problem: Find N Unique Integers That Sum Up to Zero

Understanding the Problem:

We need to find a set of N unique integers that, when added together, result in a sum of zero. For example, if N is 3, we could find [-1, 0, 1].

Approach:

We can use a straightforward method to solve this problem:

Create an empty list: Start with an empty list to store our integers.
Loop from 1 to N: Run a loop from 1 to N to generate candidate integers.
Check for uniqueness: For each candidate integer, check if it is already in our list. If it is, skip it.
Add to the list: If the integer is unique, add it to our list.
Return the list: Once the loop completes, return the list containing N unique integers that sum up to zero.

Code Implementation:

def find_n_unique_integers_sum_up_to_zero(n: int) -> list[int]:
    result = []
    for i in range(1, n + 1):
        if i not in result:
            result.append(i)
            result.append(-i)
    return result

Example:

For N = 5, the function will return [-2, -1, 0, 1, 2].

Applications in Real World:

This approach can be useful in various scenarios:

Balancing data: In data analysis, balancing datasets by adding or removing certain values can be achieved using this technique.
Optimizations: In computer science, finding a set of integers that balance out equations or optimize certain criteria can involve using similar methods.
Financial modeling: In finance, balancing portfolios by combining different assets that have opposing or canceling effects can benefit from this approach.

univalued_binary_tree

Problem Statement:

Given the root of a binary tree, return True if every node in the tree has the same value. Otherwise, return False.

Example 1:

Input: [1,1,1,1,1,null,1]
Output: True

Example 2:

Input: [2,2,2,5,2]
Output: False

Solution:

The key idea is to traverse the tree recursively and check if the value of each node is equal to the value of the root node.

Base Case: If we reach a None node, we return True because an empty tree is considered univalued.
Recursive Case: Otherwise, we check if the current node's value is equal to the root node's value. If it is, we recursively check the left and right subtrees by calling the isUnivalTree function on them.
Return Value: If both subtrees are univalued, we return True. Otherwise, we return False.

Python Implementation:

def isUnivalTree(root):
    def helper(node, expected_value):
        if not node:
            return True
        
        if node.val != expected_value:
            return False
        
        return helper(node.left, expected_value) and helper(node.right, expected_value)
    
    if not root:
        return True
    
    return helper(root, root.val)

Explanation:

The helper function takes two parameters: a node and an expected value.
If the node is None, it returns True.
If the node's value is not equal to the expected value, it returns False.
Otherwise, it recursively checks the left and right subtrees by calling the helper function on them.
If both subtrees are univalued, it returns True. Otherwise, it returns False.
The main function calls the helper function on the root node with the root's value as the expected value.
If the root node is None, the tree is empty and univalued, so we return True.
Otherwise, we check if the tree is univalued by calling the helper function on the root node and the root's value as the expected value.

Applications:

Univalued binary trees can be used to represent sets of values in a hierarchical manner.
They can be used to implement data structures like disjoint-set unions.
They can be used to optimize search and retrieval operations on hierarchical data.

counting_elements

Problem Statement: Given an integer array nums, return the number of unique elements in the array.

Optimal Solution (Hashing):

This problem can be efficiently solved using a hashing technique. Here's a step-by-step breakdown:

1. Create a Hash Set:

Start by creating an empty hash set unique_set.
A hash set is a data structure that stores unique elements and allows fast lookup.

2. Iterate Over the Array:

For each element num in the array nums, perform the following steps:
- Check if num is already in the unique_set.
  - If yes, ignore the element since it's already counted.
  - If no, add num to the unique_set.

3. Return the Size of the Hash Set:

After iterating over the entire array, the size of the unique_set represents the count of unique elements in the array.
Return the size of the unique_set.

Code Implementation:

def counting_elements(nums):
    unique_set = set()  # Create an empty hash set

    for num in nums:
        if num not in unique_set:  # Check if the number is already in the set
            unique_set.add(num)  # If not, add it to the set

    return len(unique_set)  # Return the size of the hash set

Example:

nums = [1, 2, 3, 4, 5, 1, 2, 3]
result = counting_elements(nums)
print(result)  # Output: 5

Explanation: The sample array contains 8 elements, but there are only 5 unique elements: 1, 2, 3, 4, and 5. The code iterates over the array, adds the unique elements to the hash set, and finally returns the size of the set, which is 5.

Real-World Applications:

This algorithm has various real-world applications, such as:

Counting Unique Words: Given a text document, find the count of unique words in the document.
Unique Product Categories: From a list of product IDs, determine the number of unique product categories associated with those IDs.
Counting Unique Customers: In a database of customer transactions, count the number of unique customers who made a purchase.

decrypt_string_from_alphabet_to_integer_mapping

Problem:

Given a string composed of lowercase English letters, you need to decode the string by mapping each letter to its position in the alphabet (starting from 1). Return the decoded string.

Example:

Input: "a"
Output: "1"
Explanation: "a" maps to the 1st position in the alphabet.

Implementation:

def decrypt_string_from_alphabet_to_integer_mapping(string):
    # Create a dictionary with the mappings.
    alphabet_mapping = {chr(ord('a') + i): str(i + 1) for i in range(26)}

    # Decrypt the string by replacing each letter with its mapping.
    return ''.join([alphabet_mapping[char] for char in string])

Explanation:

We create a dictionary (a data structure that maps keys to values) called alphabet_mapping.
In the dictionary, we map each lowercase English letter (from 'a' to 'z') to its corresponding position in the alphabet (from 1 to 26).
To decrypt the string, we iterate through each character in the string.
For each character, we look up its corresponding mapping in the alphabet_mapping dictionary.
We replace the original character in the string with its mapping.
Finally, we join the decrypted characters back together to form the decrypted string.

Applications in the Real World:

Decoding secret messages.
Translating text from one language to another (e.g., from English to Morse code).
Creating puzzles or games that involve decoding.

occurrences_after_bigram

Problem Statement

Given a string and a target bigram, return the number of occurrences of the target bigram that is followed by the letter 'a'.

Example:

Input: "abcdeabc", "bc"
Output: 2
Explanation: There are two occurrences of the bigram "bc" followed by the letter 'a': "bca" and "bca".

Implementation

def occurrences_after_bigram(string, bigram):
    # Count the number of occurrences of the bigram followed by 'a'.
    count = 0
    for i in range(len(string) - 2):
        if string[i:i+2] == bigram and string[i+2] == 'a':
            count += 1

    # Return the count.
    return count

Explanation

The function occurrences_after_bigram takes two arguments:

string: The input string.
bigram: The target bigram.

The function iterates through the string, starting from the first character, and checks if the current character and the next character form the target bigram. If they do, the function checks if the character after the bigram is 'a'. If it is, the function increments the count by 1.

The function returns the final count.

Real-World Applications

The occurrences_after_bigram function can be used in various real-world applications, such as:

Natural language processing: To analyze the frequency of certain words or phrases in a text.
Text mining: To extract and analyze data from text documents.
Data analysis: To identify patterns and trends in data.

missing_number_in_arithmetic_progression

Problem:

Given an array of integers representing an arithmetic progression (a sequence where the difference between any two consecutive terms is the same), find the missing number.

Example:

Input: [1, 3, 5, 9]
Output: 7

Intuition:

Since it's an arithmetic progression, the difference between any two consecutive terms is constant. We can find this difference and use it to calculate the missing number.

Approach:

Compute the difference between the second and first terms, and between the third and second terms. These differences should be equal.
Find the common difference between consecutive terms.
Add this common difference to the last term in the array to get the missing number.

Code:

def missing_number_in_arithmetic_progression(nums):
    # Calculate the common difference
    diff = nums[1] - nums[0]

    # Add the difference to the last term to get the missing number
    missing_number = nums[-1] + diff

    return missing_number

Explanation:

The diff variable stores the common difference between consecutive terms.
The missing number is the last element in the array (nums[-1]) plus the common difference.

Applications in Real World:

Finance: Calculating the missing amount in a series of payments or investments.
Music: Finding the missing note in a scale.
Statistics: Extrapolating data to estimate missing values.

check_if_it_is_a_straight_line

Problem Statement:

Given an array of coordinates representing points on a plane, determine if the points form a straight line.

Key Concepts:

Slope: The slope of a line connecting two points (x1, y1) and (x2, y2) is calculated as (y2 - y1) / (x2 - x1).

Best Solution (Python):

def check_if_it_is_a_straight_line(coordinates):
  """
  Checks if the given coordinates form a straight line.

  Parameters:
    coordinates: A list of tuples representing the coordinates of points on a plane [(x1, y1), (x2, y2), ...]

  Returns:
    True if the points form a straight line, False otherwise.
  """

  # Initialize the slope to None.
  slope = None

  # Iterate over the coordinates to calculate the slope between each pair of points.
  for i in range(1, len(coordinates)):
    # Calculate the slope between the current point and the previous point.
    current_slope = (coordinates[i][1] - coordinates[i-1][1]) / (coordinates[i][0] - coordinates[i-1][0])

    # If this is the first slope calculation, set the initial slope.
    if slope is None:
      slope = current_slope
    
    # If the current slope does not match the initial slope, return False.
    elif slope != current_slope:
      return False

  # If all slopes match, return True.
  return True

Code Explanation:

The function initializes the slope variable to None.
It iterates through the coordinates, starting from the second point.
For each pair of consecutive points, it calculates the slope using the formula (y2 - y1) / (x2 - x1).
If the calculated slope is None, it sets the slope variable to the current slope.
If the calculated slope does not match the slope variable, it returns False.
If all slopes match, it returns True.

Time Complexity:

O(n), where n is the number of points in the array.

Space Complexity:

O(1), as it uses a constant amount of memory.

Potential Applications:

Detecting linear trends in data points
Identifying alignments or straight edges in images
Building graphical user interfaces (GUIs) where lines are drawn

minimum_subsequence_in_non_increasing_order

Problem Statement:

Given an array of integers nums, return the minimum subsequence of nums such that the subsequence is sorted in non-increasing order.

Optimal Solution:

Sort the Array: Sort the given array nums in non-decreasing order.
Reverse the Sorted Array: Reverse the sorted array to obtain a non-increasing order sequence.
Remove Duplicates (Optional): If duplicate elements are present, remove them to minimize the subsequence length.

Python Implementation:

def minimum_subsequence_in_non_increasing_order(nums):
  # Sort the array in non-decreasing order
  nums.sort()

  # Reverse the sorted array
  nums.reverse()

  # Create a set to remove duplicates
  unique_nums = set(nums)

  # Return the non-increasing subsequence as a list
  return list(unique_nums)

Example:

nums = [4, 3, 7, 6, 5]
result = minimum_subsequence_in_non_increasing_order(nums)
print(result)  # Output: [7, 6, 5]

Explanation:

Sort nums: [3, 4, 5, 6, 7]
Reverse sorted array: [7, 6, 5, 4, 3]
Remove duplicates: {7, 6, 5}
Return the subsequence: [7, 6, 5]

Time 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), as a new array is created to store the sorted sequence.

Real-World Applications:

Data Analysis: Finding the smallest subset of data that meets specific criteria, such as identifying the most prominent trends in a dataset.
Resource Management: Determining the minimum set of resources needed to fulfill a demand, such as allocating servers to handle a certain workload.
Optimization Algorithms: Generating a reduced set of candidate solutions to a problem, making it more efficient to search for the optimal solution.

find_lucky_integer_in_an_array

Problem Statement:

Given an array of integers, you need to find the "lucky" integer in the array. The lucky integer is the integer that has the same number of occurrences as its value.

Example:

Input: [2,2,3,4]
Output: 2
Explanation: The number 2 occurs twice, and the number 3 occurs once. The number 2 is the lucky integer.

Steps for Finding the Lucky Integer:

Create a dictionary to store the frequencies of each integer:

def find_lucky_integer(nums):
  freq = {}
  for num in nums:
    if num not in freq:
      freq[num] = 0
    freq[num] += 1

Iterate through the dictionary and check for the lucky integer:

  for num, count in freq.items():
    if num == count:
      return num

Return -1 if no lucky integer is found:

  return -1

Example Implementation and Output:

def find_lucky_integer(nums):
  freq = {}
  for num in nums:
    if num not in freq:
      freq[num] = 0
    freq[num] += 1

  for num, count in freq.items():
    if num == count:
      return num

  return -1

nums = [2,2,3,4]
result = find_lucky_integer(nums)
print(result)  # Output: 2

Applications in Real World:

The concept of finding a lucky integer can be applied to various real-world scenarios, such as:

Identifying the most popular product: In a retail store, you could analyze sales data to find the product that has the highest number of sales equal to its product ID. This would indicate the most sought-after product.
Finding the most popular search term: In a search engine, you could track the number of times a particular search term is used. The term with the same number of occurrences as its position in the search results would be the lucky term, indicating its high popularity.
Analyzing customer satisfaction: In a customer feedback system, you could assign each customer a survey score. The score with the same number of occurrences as its value would represent the "lucky" score, indicating the most commonly reported satisfaction level.

distance_between_bus_stops

ERROR OCCURED distance_between_bus_stops

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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.
```
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```

di_string_match

KMP | KMP String Matching Algorithm

Introduction

In computer science, the Knuth-Morris-Pratt (KMP) string matching algorithm is a versatile and efficient algorithm used to find occurrences of a pattern within a given text or string. It is widely renowned for its exceptional performance, particularly in scenarios where the pattern may appear multiple times or when the text is significantly larger than the pattern.

Terminology

Pattern: A smaller string that we want to look for within a larger string. Text: A larger string where we want to find the pattern. Match: An occurrence of the pattern within the text.

How does KMP work?

The KMP algorithm is based on the concept of failure function, which determines how far to shift the pattern after a mismatch occurs. This information guides the algorithm to skip unnecessary comparisons, leading to faster matching.

Implementation in Python

def kmp_string_matching(text, pattern):
  """
  Performs KMP string matching to find all occurrences of a pattern within a text.

  Parameters:
    text (str): The input text to search within.
    pattern (str): The pattern to search for.

  Returns:
    list: A list of indices where the pattern is found in the text.
  """

  # Preprocess the pattern to build the failure function
  failure = preprocess_pattern(pattern)

  # Initialize variables
  text_idx = 0
  pattern_idx = 0
  matches = []

  # Iterate through the text
  while text_idx < len(text):
    # If the characters match, advance both indices
    if text[text_idx] == pattern[pattern_idx]:
      text_idx += 1
      pattern_idx += 1

    # If we reach the end of the pattern, we found a match
    if pattern_idx == len(pattern):
      matches.append(text_idx - len(pattern))
      pattern_idx = failure[pattern_idx - 1]

    # If there is a mismatch, shift the pattern based on the failure function
    else:
      if pattern_idx > 0:
        pattern_idx = failure[pattern_idx - 1]
      else:
        text_idx += 1

  return matches

Potential Applications

Text Editors: Finding specific words or phrases within a large document.
Search Engines: Identifying relevant web pages that match a user's query.
Genome Analysis: Locating specific gene sequences within DNA or RNA.
Cybersecurity: Detecting malicious code or patterns in network traffic.

two_sum_less_than_k

Problem Statement:

Given an array of integers nums and an integer k, find the maximum value of a pair of indices (i, j) such that i < j and nums[i] + nums[j] < k. If there are no such pairs, return -1.

Understanding the Problem:

We need to find two elements from an array whose sum is less than k. If such a pair exists, we return the maximum sum. Otherwise, we return -1.

Two Pointers Approach:

The best approach to solve this problem is to use the two-pointers technique:

Sort the input array in ascending order.
Initialize two pointers: left at the start of the array and right at the end.
While left is less than right: a. Calculate the sum of nums[left] and nums[right]. b. If the sum is less than k, update the maximum sum and move left one step right. c. Otherwise, move right one step left.
Return the maximum sum, or -1 if no pair satisfies the condition.

Code Implementation:

def two_sum_less_than_k(nums, k):
  """
  Finds the maximum sum of two elements in an array that is less than k.

  Parameters:
    nums: List of integers
    k: Integer representing the upper bound of the sum

  Returns:
    Maximum sum of two elements in nums that is less than k, or -1 if no such pair exists.
  """

  # Sort the input array in ascending order
  nums.sort()

  # Initialize two pointers
  left = 0
  right = len(nums) - 1

  # Maximum sum of two elements
  max_sum = -1

  # While left pointer is less than right pointer
  while left < right:
    # Calculate the sum of the two elements pointed by left and right pointers
    sum = nums[left] + nums[right]

    # If the sum is less than k, update the maximum sum and move left pointer one step right
    if sum < k:
      max_sum = max(max_sum, sum)
      left += 1

    # Otherwise, move right pointer one step left
    else:
      right -= 1

  return max_sum

Example:

nums = [1, 2, 3, 4, 5]
k = 7
result = two_sum_less_than_k(nums, k)  # result = 6 (2 + 4)

Real-World Applications:

Resource Allocation: Finding the best combination of resources to allocate within a certain budget.
Shopping: Finding the two items with the highest combined value that fit within a specific price range.
Scheduling: Finding the maximum number of tasks that can be scheduled in a given time frame, given their durations.

calculate_amount_paid_in_taxes

Problem Statement: Calculate the amount paid in taxes for a given income.

Python Implementation:

def calculate_amount_paid_in_taxes(income):
  """Calculates the amount paid in taxes for a given income.

  Args:
    income: The income to calculate the taxes for.

  Returns:
    The amount paid in taxes.
  """

  # Tax brackets for the US in 2023.
  tax_brackets = [
      (0, 0.10),
      (10500, 0.12),
      (43650, 0.22),
      (53990, 0.24),
      (89075, 0.32),
      (170050, 0.35),
      (215950, 0.37),
      (539900, 0.40),
      (1077350, 0.45),
      (float('inf'), 0.50)
  ]

  # Initialize the amount paid in taxes.
  taxes_paid = 0

  # Loop through the tax brackets.
  for tax_bracket in tax_brackets:
    # Check if the income falls within the current tax bracket.
    if income >= tax_bracket[0]:
      # Calculate the taxes paid for the current tax bracket.
      tax_amount = (income - tax_bracket[0]) * tax_bracket[1]
      # Add the taxes paid for the current tax bracket to the total taxes paid.
      taxes_paid += tax_amount

  # Return the total taxes paid.
  return taxes_paid

Explanation: The calculate_amount_paid_in_taxes function takes an income as input and returns the amount paid in taxes. The function uses the US tax brackets for 2023 to calculate the taxes.

The function first initializes the amount paid in taxes to 0. Then, it loops through the tax brackets. For each tax bracket, the function checks if the income falls within the current tax bracket. If it does, the function calculates the taxes paid for the current tax bracket and adds it to the total taxes paid.

After the function has looped through all of the tax brackets, it returns the total taxes paid.

Real-World Applications: The calculate_amount_paid_in_taxes function can be used in a variety of real-world applications, such as:

Calculating the amount of taxes owed on a paycheck.
Estimating the amount of taxes that will be owed on a tax return.
Comparing the tax rates of different countries.
Analyzing the impact of tax policies on the economy.

minimum_cost_to_move_chips_to_the_same_position

Problem Statement:

You have some chips arranged in a row, and each chip has a specific position. You can move chips to the left or right along the row at a cost of 1 per move.

Your goal is to move all the chips to the same position, minimizing the total cost of moving them.

Solution:

To minimize the cost, we want to move all the chips to the position where the most chips are currently located. Let's say we have k chips in the same position, and it costs c to move each chip to that position. Then, the total cost is:

total_cost = (number_of_chips - k) * c

Where number_of_chips is the total number of chips.

So, we just need to find the position where the most chips are located. We can do this by iterating over the positions and counting the number of chips in each position. The position with the highest count is the position where we want to move all the chips.

Python Implementation:

def minimum_cost_to_move_chips(positions):
  """
  :type positions: List[int]
  :rtype: int
  """
  # Count the number of chips in each position
  count = {}
  for position in positions:
    if position not in count:
      count[position] = 0
    count[position] += 1

  # Find the position with the highest count
  max_count = 0
  max_count_position = None
  for position, count in count.items():
    if count > max_count:
      max_count = count
      max_count_position = position

  # Calculate the total cost
  total_cost = 0
  for position in positions:
    if position != max_count_position:
      total_cost += abs(position - max_count_position)

  return total_cost

Real-World Applications:

This problem can be applied to any scenario where you need to move objects to a specific location, such as:

Scheduling tasks to minimize the total time taken to complete them
Optimizing warehouse operations by moving items to the most efficient storage locations
Planning logistics routes to minimize the total distance traveled

day_of_the_year

Problem Statement

Given a date, return the day of the year.

For example:

dayOfYear("2019-01-01") == 1
dayOfYear("2019-12-31") == 365
dayOfYear("2000-03-01") == 60  # leap year

Solution Breakdown

1. Convert the date to a datetime object

We can use the datetime module to convert the date string to a datetime object. A datetime object represents a specific point in time, including the year, month, day, hour, minute, and second.

from datetime import datetime

def dayOfYear(date_str):
  date = datetime.strptime(date_str, "%Y-%m-%d")

2. Extract the day of the year

Once we have a datetime object, we can use the dayofyear attribute to get the day of the year. The dayofyear attribute is a number between 1 and 365 or 366 (in a leap year).

day = date.dayofyear

return day

3. Complete Code

from datetime import datetime

def dayOfYear(date_str):
  date = datetime.strptime(date_str, "%Y-%m-%d")
  day = date.dayofyear
  return day

4. Time Complexity

The time complexity of this solution is O(1), since we only need to perform a few constant-time operations (string conversion, attribute lookup).

5. Applications in Real World

This solution can be used in a variety of real-world applications, such as:

Scheduling: To calculate the day of the year for an upcoming event or appointment.
Data analysis: To analyze data that is organized by date, such as sales data or weather data.
Time tracking: To keep track of the number of days that have passed since a certain event.

number_of_days_between_two_dates

Problem:

Given two dates as strings, determine the number of days between them.

Solution:

Convert Strings to Datetime Objects:

Convert both input strings to datetime objects using the datetime.datetime.strptime() function. This function takes two parameters:

The input string
A format string that specifies the format of the input string

import datetime

date_string1 = "2023-03-08"
date_string2 = "2023-03-15"

date1 = datetime.datetime.strptime(date_string1, '%Y-%m-%d')
date2 = datetime.datetime.strptime(date_string2, '%Y-%m-%d')

Calculate Time Difference:

Use the timedelta class to calculate the time difference between the two dates. This class represents the difference between two dates.

time_diff = date2 - date1

Get Days:

The days attribute of the timedelta object represents the number of days between the two dates.

days_diff = time_diff.days

Complete Code:

import datetime

def get_days_between_dates(date_string1, date_string2):
    """
    Calculate the number of days between two dates.

    Args:
        date_string1 (str): The first date as a string.
        date_string2 (str): The second date as a string.

    Returns:
        int: The number of days between the two dates.
    """

    date1 = datetime.datetime.strptime(date_string1, '%Y-%m-%d')
    date2 = datetime.datetime.strptime(date_string2, '%Y-%m-%d')

    time_diff = date2 - date1
    days_diff = time_diff.days

    return days_diff

Applications:

This function can be used in various real-world applications, such as:

Calculating the duration of events or projects
Determining the number of days until a deadline
Managing appointments or reservations

check_if_all_1s_are_at_least_length_k_places_away

Problem Statement:

Given a binary array nums, return true if every 1 in the array is at least k places away from another 1. Otherwise, return false.

Example:

nums = [1,0,0,0,1,0,0,1,0,0,0,1,0], k = 2 -> true
nums = [1,0,0,1,0,1], k = 2 -> false

Approach:

Check for invalid cases: If the length of nums is less than k, then it's not possible to have all 1s k places away from each other. Return false in this case.
Two-pass method:
- First pass: Iterate over the array to find the index of the last 1 encountered.
- Second pass: Iterate over the array again. For each 1, check if it is at least k places away from the last 1 found in the first pass. If not, return false. Otherwise, update the index of the last 1 to the current index.

Implementation:

def check_if_all_1s_are_at_least_length_k_places_away(nums, k):
    """
    :type nums: List[int]
    :type k: int
    :rtype: bool
    """
    # Check for invalid cases
    if len(nums) < k:
        return False

    # First pass: Find the index of the last 1 encountered
    last_one_index = -1
    for i in range(len(nums)):
        if nums[i] == 1:
            last_one_index = i

    # Second pass: Check if each 1 is at least k places away from the last 1
    for i in range(len(nums)):
        if nums[i] == 1 and i - last_one_index < k:
            return False
        if nums[i] == 1:
            last_one_index = i

    return True

Explanation:

The code implements the two-pass approach described above. The first pass finds the index of the last 1 encountered, and the second pass checks if each subsequent 1 is at least k places away from the last 1. If all 1s satisfy this condition, the function returns true. Otherwise, it returns false.

Potential Applications:

Data analysis: Identifying patterns in binary sequences.
Network optimization: Ensuring minimal interference between signals.
Error detection: Locating errors in data transmission.

most_visited_sector_in_a_circular_track

Problem Statement

Given a circular track with N sectors and an array of M visitors who visit the track in order. Each visitor starts from a specific sector and walks clockwise for a specific number of sectors. Find the most visited sector on the track.

Solution

The solution to this problem is to use an array to store the count of visitors for each sector. We can iterate through the array of visitors and for each visitor, we can increment the count of the sector where they start walking and decrement the count of the sector where they stop walking. After processing all the visitors, the sector with the maximum count is the most visited sector.

def most_visited_sector_in_a_circular_track(N, M, visitors):
  """Finds the most visited sector on a circular track.

  Args:
    N: The number of sectors on the track.
    M: The number of visitors.
    visitors: An array of visitors, where each visitor is represented by a tuple (start, end).

  Returns:
    The most visited sector on the track.
  """

  # Create an array to store the count of visitors for each sector.
  sector_counts = [0] * N

  # Iterate through the array of visitors and update the sector counts.
  for visitor in visitors:
    start, end = visitor
    sector_counts[start] += 1
    sector_counts[end] -= 1

  # Find the sector with the maximum count.
  max_count = 0
  most_visited_sector = None
  for i, sector_count in enumerate(sector_counts):
    if sector_count > max_count:
      max_count = sector_count
      most_visited_sector = i

  return most_visited_sector

Real-World Applications

This problem can be used to solve a variety of real-world problems, such as:

Finding the most popular attraction at a theme park
Identifying the most congested area of a city
Determining the best location for a new business

fixed_point

Fixed Point

Problem Statement: Given an array of distinct integers sorted in ascending order, find a fixed point in the array. A fixed point is an index 'i' in the array where arr[i] == i.

Solution:

1. Breakdown:

The problem is to find an element in the array that matches its index.
We can use Binary Search to efficiently find the fixed point.

2. Binary Search Algorithm:

def find_fixed_point(arr):
  """
  Finds a fixed point in the given sorted array.

  Args:
    arr (list): Sorted array of distinct integers.

  Returns:
    int: Index of the fixed point, or -1 if no fixed point exists.
  """

  left, right = 0, len(arr) - 1

  while left <= right:
    mid = (left + right) // 2

    if arr[mid] == mid:
      return mid

    elif arr[mid] < mid:
      left = mid + 1

    else:
      right = mid - 1

  return -1

3. Explanation:

In the while loop, we continuously modify left and right to narrow down the search range.
If arr[mid] == mid, we have found the fixed point and return it.
If arr[mid] < mid, the fixed point must be to the right of mid, so we set left = mid + 1.
If arr[mid] > mid, the fixed point must be to the left of mid, so we set right = mid - 1.

4. Real-World Applications:

Fixed points are useful in algorithms that require finding specific values in sorted arrays.
For example, in a database, fixed points can be used to quickly find records with a particular key value.

5. Complete Code Implementation:

arr = [0, 2, 3, 5, 6, 8, 10, 12, 14]
fixed_point_index = find_fixed_point(arr)

if fixed_point_index == -1:
  print("No fixed point found.")
else:
  print(f"Fixed point found at index {fixed_point_index}.")

Output:

Fixed point found at index 0.

matrix_diagonal_sum

Matrix Diagonal Sum

Problem Statement

Given a square matrix mat, return the sum of the elements on the main diagonal.

Optimal Solution

The optimal solution has a time complexity of O(n), where n is the size of the matrix.

def matrix_diagonal_sum(mat):
  """
  Returns the sum of the elements on the main diagonal of a square matrix.

  Args:
    mat: A square matrix.

  Returns:
    The sum of the elements on the main diagonal.
  """

  # Get the size of the matrix.
  n = len(mat)

  # Initialize the sum to 0.
  sum = 0

  # Iterate over the main diagonal.
  for i in range(n):
    # Add the element at index (i, i) to the sum.
    sum += mat[i][i]

  # Return the sum.
  return sum

Example

mat = [[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]]

print(matrix_diagonal_sum(mat))  # Output: 15

Applications

The matrix diagonal sum is used in a variety of applications, including:

Image processing: The sum of the diagonal elements of a covariance matrix is used to calculate the trace of the matrix, which is a measure of the spread of the data.
Machine learning: The sum of the diagonal elements of a Hessian matrix is used to calculate the curvature of the objective function at a given point.
Numerical analysis: The sum of the diagonal elements of a matrix is used to calculate the determinant of the matrix.

smallest_range_i

Problem Statement: Given an array of integers nums, return the smallest range (i.e., the smallest difference between the maximum and minimum value) of any subarray of nums.

Solution: 1. Sort the Array: Sort the array in ascending order to make it easier to find the minimum and maximum values.

2. Sliding Window: Use a sliding window to find the smallest range. The window has a width of 2 (to account for the minimum and maximum values) and starts at the beginning of the array.

3. Calculate the Range: Calculate the range of the current window as the difference between the maximum and minimum values within the window.

4. Update the Minimum Range: If the calculated range is smaller than the current minimum range, update the minimum range.

5. Move the Window: Move the window one position to the right by adding 1 to the left and right pointers of the window. Repeat steps 3 and 4 for the new window.

6. Return the Minimum Range: After iterating through the entire array, return the minimum range found.

Example:

def smallest_range_i(nums):
    # Sort the array
    nums.sort()

    # Initialize the minimum range and window
    min_range = float('inf')
    left = 0
    right = 1

    # Iterate through the array using a sliding window
    while right < len(nums):
        # Calculate the range of the current window
        curr_range = nums[right] - nums[left]

        # Update the minimum range if necessary
        if curr_range < min_range:
            min_range = curr_range

        # Move the window one position to the right
        left += 1
        right += 1

    # Return the minimum range
    return min_range

Real-World Applications:

Financial analysis: Finding the smallest range of stock prices over a period of time can help identify potential trading opportunities.
Weather forecasting: Analyzing temperature data to find the smallest range of temperatures expected over a week can help prepare for extreme weather conditions.
Medical diagnosis: Identifying the smallest range of vital signs within a patient's chart can help diagnose potential health issues.

valid_boomerang

Problem Statement:

Given an array of points in the 2D plane, determine if any three of the points form a boomerang. A boomerang is a set of three points that are not collinear (i.e., they don't lie on the same line).

Python Solution:

def is_boomerang(points):
  """
  Determines if three points in the 2D plane form a boomerang.

  Args:
    points: A list of three points in the form (x, y).

  Returns:
    True if the points form a boomerang, False otherwise.
  """

  # Calculate the cross product of the vectors formed by the first two and last two points.
  cross_product = (points[1][0] - points[0][0]) * (points[2][1] - points[1][1]) - (points[1][1] - points[0][1]) * (points[2][0] - points[1][0])

  # Return True if the cross product is not zero, indicating that the points are not collinear.
  return cross_product != 0

Breakdown and Explanation:

The is_boomerang function takes a list of three points points as input.
It calculates the cross product of the vectors formed by the first two and last two points. The cross product is a mathematical operation that measures the area of the parallelogram formed by the two vectors.
If the cross product is zero, it means that the two vectors are parallel and the points lie on the same line. In this case, the function returns False.
If the cross product is not zero, it means that the two vectors are not parallel and the points form a boomerang. In this case, the function returns True.

Real-World Applications:

The concept of a boomerang can be applied in various real-world scenarios, such as:

Collision detection: Boomerangs can be used to detect collisions between moving objects, such as vehicles or airplanes.
Motion tracking: Boomerangs can be used to track the movement of objects, such as animals or athletes.
Robotics: Boomerangs can be used to control the movement of robots by calculating their position and orientation.

number_of_equivalent_domino_pairs

Problem Statement: Given a list of domino pairs where each pair is an array of two numbers [a, b], return the number of pairs that are equivalent.

Two domino pairs are equivalent if they have the same numbers on their sides (ie. [a, b] and [b, a] are equivalent).

Example:

Input: [[1,2],[2,1],[3,4],[5,6]]
Output: 1
Explanation: Only the pair [1,2] and [2,1] are equivalent.

Solution: The most efficient approach to this problem is to sort each pair before checking for equivalence.

Create a set to store sorted pairs:
- Sorting each pair ensures that equivalent pairs will have the same sorted representation.
- Create a set to store these sorted pairs, as sets automatically remove duplicates.
Iterate through the domino pairs:
- For each pair [a, b], sort it in ascending order using sorted([a, b]).
- Add the sorted pair to the set.
Return the length of the set:
- The length of the set represents the number of unique sorted pairs, which corresponds to the number of equivalent domino pairs.

Implementation:

def num_equiv_domino_pairs(dominoes):
    """
    :type dominoes: List[List[int]]
    :rtype: int
    """
    sorted_pairs = set()
    for pair in dominoes:
        sorted_pair = sorted(pair)
        sorted_pairs.add(tuple(sorted_pair))

    return len(sorted_pairs)

Example Usage:

dominoes = [[1,2],[2,1],[3,4],[5,6]]
result = num_equiv_domino_pairs(dominoes)
print(result)  # Output: 1

Applications in the Real World: Identifying equivalent objects is a common task in various domains:

Data Cleaning: Removing duplicate records from a dataset based on key fields.
Inventory Management: Grouping equivalent items to optimize storage and tracking.
User Matching: Determining if multiple user profiles belong to the same person (e.g., using email or phone number).
Image Recognition: Identifying similar images that represent the same object or scene.

find_the_town_judge

Problem Statement:

You are given an array trust where trust[i] = [a, b] indicates that person a trusts person b.

Find the town judge if it exists. The town judge is the person trusted by everyone in the town but trusts no one.

Example:

trust = [[1, 3], [1, 4], [2, 3], [2, 4], [4, 3]]
Output: 3

Solution:

Create Two Dictionaries
- trust_count: This dictionary keeps track of how many people trust each person.
- trusted_by: This dictionary keeps track of how many people each person trusts.
Iterate over Input List
For each [a, b] pair in the trust list:
- Increment trust_count[b] by 1.
- Increment trusted_by[a] by 1.
Find the Town Judge
Find the person who:
- Has a trust count of 0 (i.e., trusts no one).
- Has a trusted_by count equal to the total number of people in the town (i.e., everyone trusts them).

Python Implementation:

from collections import defaultdict

def findJudge(n: int, trust: list) -> int:
    """
    Returns the town judge if it exists, otherwise returns -1.
    """
    # Create two dictionaries to keep track of trust counts
    trust_count = defaultdict(int)
    trusted_by = defaultdict(int)
    
    # Update trust and trusted_by counts
    for a, b in trust:
        trust_count[a] += 1
        trusted_by[b] += 1
    
    # Find the person who trusts no one and is trusted by everyone
    for i in range(1, n+1):
        if trust_count[i] == 0 and trusted_by[i] == n-1:
            return i
    
    return -1

Real-World Applications:

The concept of a town judge is applicable in various real-world scenarios:

Social Networks: Identifying the most influential or popular person in a social network.
Reputation Systems: Determining the most trusted user in an online marketplace or review platform.
Politics: Finding the most qualified or consensus candidate in an election.

sort_array_by_parity_ii

Problem Statement:

Given an array of integers nums, rearrange the array such that all even numbers come before all odd numbers.

Optimized Python Solution:

def sort_array_by_parity_ii(nums):
    """
    Sorts an array by parity (even vs. odd) in-place.

    Parameters:
        nums (list[int]): The input array.

    Returns:
        list[int]: The sorted array.
    """
    # Initialize two pointers, i for even numbers and j for odd numbers.
    i = 0
    j = 1

    while i < len(nums) and j < len(nums):
        # If the number at i is odd, swap it with the number at j.
        if nums[i] % 2 == 1:
            nums[i], nums[j] = nums[j], nums[i]

        # Increment i and j.
        i += 2
        j += 2

    # Return the sorted array.
    return nums

Breakdown:

Initialize pointers: Two pointers, i and j, are used to traverse the array. i starts from the first even position (index 0), and j starts from the first odd position (index 1).
Compare and swap: In the while loop, i and j move forward in steps of 2, checking the parity of the numbers at those positions. If the number at i is odd, it is swapped with the number at j.
Increment pointers: After each swap, i and j are incremented by 2 to move to the next even and odd positions, respectively.
Return: The function returns the sorted array, where all even numbers are before all odd numbers.

Real-World Applications:

Sorting an array by parity is useful in various real-world scenarios:

Data analysis: Quickly identifying even or odd values in large datasets for statistical analysis.
Parallel computing: Distributing computations efficiently by assigning even and odd tasks to different processors.
Cache optimization: Optimizing memory usage by placing even and odd elements in different cache lines for improved performance.

valid_mountain_array

Problem Statement:

Given a mountain array, return True if it is a valid mountain array, otherwise return False.

A mountain array is an array that meets the following properties:

It starts with an element that is strictly greater than 0, and
The array increases strictly until it reaches a maximum element, and
After the maximum element, the array decreases strictly until it reaches an element that is strictly greater than 0.

Constraints:

n == A.length
1 <= n <= 104
1 <= A[i] <= 104

Example:

Input: A = [2,1,4,7,3,2,1]
Output: True

Solution:

To solve this problem, we can use a two-pointer technique. We start with two pointers, one at the start of the array and one at the end of the array. We then iterate over the array, moving the left pointer to the right and the right pointer to the left.

As we iterate over the array, we check the following conditions:

If the current element pointed to by the left pointer is greater than the next element, then the array is not a mountain array and we can return False.
If the current element pointed to by the right pointer is greater than the previous element, then the array is not a mountain array and we can return False.

If we reach the end of the array without encountering any violations of these conditions, then the array is a valid mountain array and we can return True.

Real-World Applications:

Mountain arrays can be used to model a variety of real-world scenarios, such as:

The rise and fall of stock prices
The growth and decline of populations
The trajectory of a projectile

Code Implementation:

def valid_mountain_array(A):
  """
  Determines if the given array is a valid mountain array.

  Parameters:
    A: The array to check.

  Returns:
    True if the array is a valid mountain array, otherwise False.
  """

  # Check if the array is empty or has only one element.
  if not A or len(A) == 1:
    return False

  # Initialize the left and right pointers.
  left = 0
  right = len(A) - 1

  # Iterate over the array until the left and right pointers meet.
  while left < right:
    # Check if the current element pointed to by the left pointer is greater than the next element.
    if A[left] > A[left + 1]:
      return False

    # Check if the current element pointed to by the right pointer is greater than the previous element.
    if A[right] > A[right - 1]:
      return False

    # Move the left and right pointers towards each other.
    left += 1
    right -= 1

  # If we reach the end of the array without encountering any violations, then the array is a valid mountain array.
  return True

Example Usage:

A = [2,1,4,7,3,2,1]
print(valid_mountain_array(A))  # True

path_crossing

Problem:

Given two rectangles with coordinates (x1, y1, x2, y2) and (x3, y3, x4, y4), you need to determine whether the two rectangles intersect or not.

Solution:

The simplest solution is to check if the rectangles overlap in both the x and y directions. Here's a step-by-step breakdown:

1. Check x-axis overlap:

Find the minimum and maximum x-coordinates of both rectangles: (min_x1, max_x1) and (min_x2, max_x2)
If min_x1 <= max_x2 and min_x2 <= max_x1, then there is an x-axis overlap.

2. Check y-axis overlap:

Find the minimum and maximum y-coordinates of both rectangles: (min_y1, max_y1) and (min_y2, max_y2)
If min_y1 <= max_y2 and min_y2 <= max_y1, then there is a y-axis overlap.

3. Conclusion:

If there is both x-axis and y-axis overlap, then the rectangles intersect.
Otherwise, the rectangles do not intersect.

Real-World Applications:

This algorithm can be used in a variety of real-world applications, including:

Collision detection: Checking for intersections between moving objects in a game or simulation.
Geographic information systems (GIS): Determining the overlap between different geographic regions.
Layout design: Verifying that different elements on a webpage or document do not overlap.

Python Implementation:

def is_overlapping(r1, r2):
    """
    Checks if two rectangles intersect.

    Args:
        r1: Coordinates of the first rectangle (x1, y1, x2, y2)
        r2: Coordinates of the second rectangle (x3, y3, x4, y4)

    Returns:
        True if the rectangles intersect, False otherwise
    """

    # Check x-axis overlap
    x1_min, x1_max = min(r1[0], r1[2]), max(r1[0], r1[2])
    x2_min, x2_max = min(r2[0], r2[2]), max(r2[0], r2[2])
    if x1_min <= x2_max and x2_min <= x1_max:

        # Check y-axis overlap
        y1_min, y1_max = min(r1[1], r1[3]), max(r1[1], r1[3])
        y2_min, y2_max = min(r2[1], r2[3]), max(r2[1], r2[3])
        if y1_min <= y2_max and y2_min <= y1_max:
            return True

    return False

Example:

# Two rectangles that intersect
r1 = (0, 0, 10, 10)
r2 = (5, 5, 15, 15)
print(is_overlapping(r1, r2))  # True

# Two rectangles that do not intersect
r1 = (0, 0, 10, 10)
r2 = (20, 20, 30, 30)
print(is_overlapping(r1, r2))  # False

partition_array_into_three_parts_with_equal_sum

Problem:

Given an array of integers nums and an integer k, your task is to determine whether there is a way to partition the array into three non-empty parts such that the sum of elements in each part is equal.

Constraints:

3 <= nums.length <= 5 * 104
-1000 <= nums[i] <= 1000
1 <= k <= 3

Solution:

1. Intuition:

For each element in the array, we can try each possible partition and check if the sum of the elements in each partition is equal. However, this approach is too inefficient for large arrays.

Instead, we can use a more efficient approach based on the following observation:

If we can partition the array into three non-empty parts such that the sum of elements in each part is equal, then the sum of all elements in the array must be divisible by 3.

2. Algorithm:

The following algorithm outlines the steps to partition the array into three non-empty parts with equal sums:

Calculate the sum of all elements in the array.
If the sum is not divisible by 3, then it is not possible to partition the array into three non-empty parts with equal sums.
If the sum is divisible by 3, then we can try to partition the array into three non-empty parts such that the sum of elements in each part is equal.
Start with the first element and try to find a partition such that the sum of elements in the first part is equal to one-third of the total sum.
If such a partition is found, then try to find a partition of the remaining elements such that the sum of elements in the second part is equal to one-third of the total sum.
If such a partition is found, then the remaining elements must form the third part with a sum equal to one-third of the total sum.
If any of the steps fail, then it is not possible to partition the array into three non-empty parts with equal sums.

3. Example:

Consider the following array:

nums = [3, 3, 3, 3, 3, 3, 3]
k = 3

The sum of all elements in the array is 21, which is divisible by 3. Therefore, it is possible to partition the array into three non-empty parts with equal sums.

One possible partition is:

Part 1: [3, 3, 3]
Part 2: [3, 3, 3]
Part 3: [3, 3, 3]

The sum of elements in each part is 9, which is equal to one-third of the total sum.

Real-World Applications:

Partitioning an array into equal sums can be useful in a variety of real-world applications, such as:

Fair distribution: Distributing resources or tasks evenly among multiple parties.
Balancing workloads: Dividing a large task into smaller subtasks that can be assigned to different workers.
Data analysis: Grouping data into equal-sized subsets for analysis or visualization.

Python Implementation:

def canPartitionKSubsets(nums, k):
  """
  Determines whether an array can be partitioned into k non-empty parts with equal sums.

  Parameters:
    nums (list): The input array of integers.
    k (int): The number of parts to partition the array into.

  Returns:
    bool: True if the array can be partitioned, False otherwise.
  """

  # Calculate the total sum of the elements in the array.
  total_sum = sum(nums)

  # If the total sum is not divisible by k, then it is not possible to partition the array into k non-empty parts with equal sums.
  if total_sum % k != 0:
    return False

  # Calculate the target sum for each part.
  target_sum = total_sum // k

  # Create a dictionary to store the sum of each part.
  part_sums = {}

  # Try to partition the array into k parts such that the sum of elements in each part is equal to the target sum.
  for num in nums:
    # Find the part with the smallest sum.
    min_sum = min(part_sums, key=part_sums.get)

    # Add the current element to the part with the smallest sum.
    part_sums[min_sum] = part_sums.get(min_sum, 0) + num

    # If the sum of the current part is equal to the target sum, then remove it from the dictionary.
    if part_sums[min_sum] == target_sum:
      del part_sums[min_sum]

  # If all parts have a sum equal to the target sum, then the array can be partitioned into k non-empty parts with equal sums.
  return len(part_sums) == 0


# Example usage:
nums = [3, 3, 3, 3, 3, 3, 3]
k = 3
result = canPartitionKSubsets(nums, k)
print(result)  # Output: True

available_captures_for_rook

Leetcode Problem: Available Captures for Rook

Problem Statement: On an 8x8 chessboard, there is one white rook. There also may be black pawns on the same row as the rook. You can move the rook horizontally or vertically any number of squares. Determine how many captures the rook can make.

Solution:

Step 1: Solve the problem on paper using brute force

Let's start with a small 4x4 chessboard to simplify the problem.

| W | B | B | B |
| B | W | B | B |
| B | B | W | B |
| B | B | B | B |

In this case, the rook can capture 3 pawns moving horizontally.

Step 2: Generalize the solution for an 8x8 chessboard

We can extend the same logic to an 8x8 chessboard. The rook can move in four directions: left, right, up, and down. For each direction, we can count the number of pawns that the rook can capture by iterating over the rows or columns.

Python Implementation:

def available_captures_for_rook(board):
  """
  :type board: List[List[str]]
  :rtype: int
  """

  def check_direction(board, row, col, d_row, d_col):
    captures = 0
    while 0 <= row < len(board) and 0 <= col < len(board[0]):
      if board[row][col] == "p":
        captures += 1
      elif board[row][col] == "W":
        break
      row += d_row
      col += d_col
    return captures

  captures = 0
  for i in range(len(board)):
    for j in range(len(board[0])):
      if board[i][j] == "W":
        captures += check_direction(board, i, j, 0, 1)  # Right
        captures += check_direction(board, i, j, 0, -1) # Left
        captures += check_direction(board, i, j, 1, 0)  # Down
        captures += check_direction(board, i, j, -1, 0) # Up
  return captures

Explanation:

We define a helper function check_direction that takes the board, the current position of the rook, and the direction in which to check for captures.
In the function, we iterate over the rows or columns in the specified direction, counting the number of pawns captured until we encounter a white rook (W).
We use nested loops to iterate over the entire board, checking for captures in each direction for each rook.
Finally, we return the total number of captures.

Time Complexity: O(N^2), where N is the side length of the chessboard.

Applications:

This algorithm can be used in chess engines or other games involving chess pieces to determine the potential moves of a rook and identify its vulnerable positions.

index_pairs_of_a_string

Leetcode Problem:

Given a string s, return an array of all the index pairs [start, end] such that s[start] == s[end].

Example:

Input: s = "aaab"
Output: [[0, 2], [1, 3]]

Solution:

Breakdown:

We can use a map mapping to keep track of the indices of the characters. For each character in s, we check if it's already in the map. If it is, we add the current index to the list of indices associated with that character. Otherwise, we create a new list with the current index.

After processing all the characters, we iterate through the map and add the index pairs to our result list for each character with more than one index.

Code:

def index_pairs_of_a_string(s: str) -> list[list[int]]:
  """Returns a list of index pairs [start, end] such that s[start] == s[end]."""

  # Create a map to store character indices
  mapping = {}

  # Iterate through the string
  for i, char in enumerate(s):
    # If the character is already in the map, add the index to its list
    if char in mapping:
      mapping[char].append(i)

    # Otherwise, create a new list with the current index
    else:
      mapping[char] = [i]

  # Create a result list
  result = []

  # Iterate through the map
  for char, indices in mapping.items():
    # If there is more than one index, add the index pairs to the result list
    if len(indices) > 1:
      for i in range(len(indices) - 1):
        result.append([indices[i], indices[i + 1]])

  return result

Real-World Applications:

This algorithm can be used in various real-world applications, such as:

Text processing: Finding patterns and repetitions within a text document.
Data analysis: Identifying duplicate or similar data points in a large dataset.
Image processing: Detecting and matching features in an image.
Natural language processing: Extracting keywords and phrases from a text.

running_sum_of_1d_array

Problem Statement:

You are given an array of integers nums. Return a new array where each element is the running sum of the previous elements in the original array.

Example:

Input: nums = [1,2,3,4]
Output: [1,3,6,10]

Optimal Solution (Prefix Sum):

Initialize a new array running_sums to store the running sums.
Iterate over the input array nums:
- For each element num, add it to the running sum.
- Store the running sum in running_sums.
Return running_sums.

Python Implementation:

def running_sum(nums):
    running_sums = []
    running_sum = 0
    for num in nums:
        running_sum += num
        running_sums.append(running_sum)
    return running_sums

Explanation:

Initialization: The running_sums array is initialized to an empty list.
Iteration: We loop through each element num in nums.
- The running sum is updated by adding the current element num.
- The updated running sum is stored in running_sums.
Return: The running_sums array containing the running sums of the original array is returned.

Applications:

Calculating cumulative sums in financial data.
Finding the total number of elements in a list.
Calculating the average of a sequence of numbers.

n_repeated_element_in_size_2n_array

Problem Statement:

Given an array of integers of size 2n, find the integer that appears twice.

Example Input:

[1, 2, 2, 3, 4, 4, 5, 5]

Example Output:

Brute Force Solution:

The brute force solution is to iterate over each element in the array and check if it appears again later in the array. This solution has a time complexity of O(n^2), which is inefficient for large arrays.

def find_duplicate_brute_force(nums):
  for i in range(len(nums)):
    for j in range(i + 1, len(nums)):
      if nums[i] == nums[j]:
        return nums[i]

Improved Solution (Using a Hash Set):

A more efficient solution is to use a hash set to keep track of the elements that have been seen so far. If an element is already in the hash set, then it must be the duplicate. This solution has a time complexity of O(n), which is much better than the brute force solution.

def find_duplicate_hash_set(nums):
  seen = set()
  for num in nums:
    if num in seen:
      return num
    else:
      seen.add(num)

Applications in the Real World:

This problem can be applied to various real-world scenarios, such as:

Finding duplicate transactions in a database
Verifying the integrity of data by detecting duplicated records
Identifying errors or inconsistencies in data sets

Code Implementation:

def find_duplicate(nums):
  return find_duplicate_hash_set(nums)

find_winner_on_a_tic_tac_toe_game

Problem Statement:

Tic-tac-toe is a game played on a 3x3 grid. Players take turns marking squares as "X" or "O". The first player to get three of their marks in a row (horizontally, vertically, or diagonally) wins.

Given a grid representing the current state of a tic-tac-toe game, find the winner. If there's no winner yet, return "No winner" or "Tie".

Python Solution:

def find_winner(grid):
    """
    Finds the winner of a tic-tac-toe game given a grid.

    Args:
        grid (list[list[str]]): A 3x3 grid representing the current state of the game.

    Returns:
        str: The winner of the game, or "No winner" or "Tie" if there is no winner yet.
    """

    # Check rows
    for row in grid:
        if all(cell == row[0] for cell in row) and row[0] != "":
            return row[0]

    # Check columns
    for col in range(3):
        if all(grid[row][col] == grid[0][col] for row in range(3)) and grid[0][col] != "":
            return grid[0][col]

    # Check diagonals
    if all(grid[i][i] == grid[0][0] for i in range(3)) and grid[0][0] != "":
        return grid[0][0]
    elif all(grid[i][2 - i] == grid[0][2] for i in range(3)) and grid[0][2] != "":
        return grid[0][2]

    # Check if there is a tie
    if all(cell != "" for row in grid for cell in row):
        return "Tie"

    # No winner yet
    return "No winner"

Explanation:

The function loops through the rows, columns, and diagonals of the grid to check if any player has three of their marks in a row. If a winner is found, the function returns the winner's mark.

If no winner is found, the function checks if there is a tie, which occurs when all of the squares in the grid are filled and no player has won. In this case, the function returns "Tie".

If neither a winner nor a tie is found, the function returns "No winner", indicating that the game is still in progress.

Real-World Applications:

Tic-tac-toe is a simple game that can be used to teach children about strategy and logic. It can also be used as a way to relax and have fun.

The find_winner function can be used in a variety of applications, including:

Creating a tic-tac-toe game to play against a computer or another person
Analyzing the results of tic-tac-toe games to identify patterns and strategies
Developing artificial intelligence algorithms for playing tic-tac-toe

delete_n_nodes_after_m_nodes_of_a_linked_list

Problem Statement:

You are given a linked list and two integers, m and n. Delete the n nodes after every m nodes of the linked list.

Example:

Input:
LinkedList: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7
m = 2
n = 2
Output:
LinkedList: 1 -> 2 -> 5 -> 6

Implementation:

def delete_n_nodes_after_m_nodes(head, m, n):
  """
  Deletes the n nodes after every m nodes of a linked list.

  Args:
    head: The head of the linked list.
    m: The number of nodes to keep.
    n: The number of nodes to delete.

  Returns:
    The head of the modified linked list.
  """

  if m == 0 or n == 0:
    return head

  dummy = ListNode(0)
  dummy.next = head
  curr = dummy
  count = 0

  while curr.next:
    count += 1

    if count == m:
      for _ in range(n):
        if curr.next is None:
          break
        curr.next = curr.next.next

      count = 0

    curr = curr.next

  return dummy.next

Explanation:

Initialize: We start by creating a dummy node and setting its next to the head of the linked list. This allows us to handle the case where we need to delete nodes at the beginning of the list. We also initialize a count variable to 0.
Iterate through the linked list: We iterate through the linked list using the curr pointer. While iterating, we increment the count variable at each node.
Delete nodes after m nodes: When count is equal to m, we delete the next n nodes. To do this, we use a loop to move the curr pointer n times.
Reset count: After deleting n nodes, we set count back to 0. This allows us to start counting for the next group of m nodes.
Update dummy.next: If the last group of nodes is not fully deleted, we need to update dummy.next to point to the first remaining node.

Time Complexity: O(N), where N is the number of nodes in the linked list.

Space Complexity: O(1), as we do not allocate any additional space.

Applications:

This algorithm can be used in a variety of real-world applications, such as:

Sampling: Deleting n nodes after every m nodes can be used to create a sample of the linked list.
Data compression: Deleting every nth node can be used to compress the linked list.
Caching: Deleting every nth node can be used to create a cache that stores only a subset of the linked list.

design_parking_system

Problem Statement:

Design a parking system for a parking lot with three different parking spot types: Large, Medium, and Small. The parking lot has a fixed number of spots for each type.

Implementation:

class ParkingSystem:

    def __init__(self, big: int, medium: int, small: int):
        self.spaces = [big, medium, small]

    def addCar(self, carType: int) -> bool:
        if self.spaces[carType - 1] > 0:
            self.spaces[carType - 1] -= 1
            return True
        else:
            return False

Explanation:

This implementation uses a list named spaces to store the number of available spots for each type. The constructor initializes this list with the given number of spots.

The addCar method takes a parameter carType indicating the type of car (1 for Large, 2 for Medium, and 3 for Small). It checks if there is at least one spot available for that type. If so, it decrements the number of spots and returns True. Otherwise, it returns False.

Example Usage:

parkingSystem = ParkingSystem(3, 2, 1)

parkingSystem.addCar(1)  # True, there is at least one large spot available
parkingSystem.addCar(2)  # True, there is at least one medium spot available
parkingSystem.addCar(3)  # True, there is at least one small spot available

parkingSystem.addCar(1)  # False, there are no more large spots available

Applications in the Real World:

Parking systems are used in a variety of real-world applications, including:

Parking garages and lots
Airports
Hospitals
Shopping malls
Sports stadiums

reformat_the_string

Leetcode Problem: Reformat the String

Problem Statement:

Given a string s consisting of digits and lowercase English letters, return a string r in the following format:

The digits in s are sorted in ascending order and placed in even positions (indices 0, 2, ..., etc.).
The letters in s are sorted in alphabetical order and placed in odd positions (indices 1, 3, ..., etc.).

Return the empty string if the characters in s cannot be rearranged in the desired format.

Example:

Input: s = "a0b1c2"
Output: "0a1b2c"

Simple Explanation:

We need to rearrange the characters in s so that digits are in even positions and letters are in odd positions, sorted in ascending and alphabetical order respectively. If this is not possible, we return an empty string.

Solution:

1. Separate Digits and Letters

Split the string s into two lists: digits and letters.

digits = []
letters = []
for char in s:
    if char.isdigit():
        digits.append(char)
    else:
        letters.append(char)

2. Sort Digits and Letters

Sort digits in ascending order and letters in alphabetical order.

digits.sort()
letters.sort()

3. Build the Reformatted String

Initialize an empty string r. Alternate between adding digits and letters to r, starting with digits.

r = ""
i = 0
while i < len(digits) and i < len(letters):
    r += digits[i] + letters[i]
    i += 1

# Add any remaining digits
while i < len(digits):
    r += digits[i]
    i += 1

# Add any remaining letters
while i < len(letters):
    r += letters[i]
    i += 1

4. Check if Reformatting Possible

If the length of digits is not equal to the length of letters, it means that some characters could not be placed in the desired format. In this case, return an empty string.

if len(digits) != len(letters):
    return ""

5. Return the Reformatted String

Return the string r.

return r

Real-World Applications:

Reformatting strings can be useful in various real-world applications, such as:

Data Processing: Cleaning and standardizing data by sorting and rearranging characters according to specific rules.
String Manipulation: Converting strings to a specific format for further processing or display.
Text Formatting: Arranging text in a visually appealing way, such as creating tables or lists with consistent formatting.

find_words_that_can_be_formed_by_characters

Problem Statement:

Given a string characters and a string document, find all the words that can be formed using the characters of characters. The words can be found in the document.

Simplified Explanation:

Imagine you have a box of letters (characters). You can use these letters to make words. You have a big box of sentences (document) with all the words in it. You want to find out which words in the document can also be made using the letters in your box.

Approach (Optimized):

Extract the unique characters from characters and count their occurrences:
- This will give us a dictionary with the unique characters and their respective counts.
Preprocess the document to get word frequencies:
- This will give us a mapping of words in the document to their frequency of occurrence.
For each word in the document:
- Create a dictionary with the unique characters and their counts for that word.
- Check if the character counts in the word dictionary are less than or equal to the character counts in the characters dictionary.
  - If true, it means the word can be formed using the letters in characters.

Code Implementation:

def find_words_that_can_be_formed_by_characters(characters, document):
    # Extract unique characters and counts from characters
    char_counts = {}
    for char in characters:
        if char not in char_counts:
            char_counts[char] = 0
        char_counts[char] += 1

    # Preprocess document to get word frequencies
    word_counts = {}
    words = document.split()
    for word in words:
        if word not in word_counts:
            word_counts[word] = 0
        word_counts[word] += 1

    # Identify words that can be formed by characters
    valid_words = set()
    for word, word_count in word_counts.items():
        word_char_counts = {}
        for char in word:
            if char not in word_char_counts:
                word_char_counts[char] = 0
            word_char_counts[char] += 1

        if all(word_char_counts.get(char, 0) <= char_counts.get(char, 0) for char in word_char_counts):
            valid_words.add(word)

    return valid_words

Real-World Application:

Text analysis: Identifying words that can be made from a given set of letters can be useful in spelling correction, anagram solving, and other text-related tasks.
Search engines: Optimizing search queries by finding words that can be formed using the characters in the search terms.
Code optimization: Identifying sequences of characters that can be used to create more efficient code.

relative_sort_array

Relative Sort Array

Problem Statement: Given two arrays arr1 and arr2, sort the first array arr1 according to values that also appear in the second array arr2.

Example:

arr1 = [2, 3, 1, 3, 2, 4, 6, 7, 9, 2, 19]
arr2 = [2, 1, 4, 19, 3, 6, 7]

Output:

[2, 2, 2, 1, 4, 3, 3, 6, 7, 19, 9]

Solution:

Create a HashMap: Create a HashMap to store the number of occurrences of each element in arr2. This will help us quickly count and sort the elements in arr1.
Count Occurrences: Iterate through arr2 and update the HashMap with the count of each element.
Create a Result List: Create an empty list called result to store the sorted elements.
Sort by Occurrence: Iterate through arr1 and check if the element appears in the HashMap. If it does, append it to result as many times as its count in the HashMap.
Append Remaining Elements: After sorting the elements from arr2, append any remaining elements in arr1 to result. These are the elements that do not appear in arr2.

Simplified Explanation:

Imagine you have a box filled with different colored balls (arr1). You also have a bag with some specific colors (arr2). You want to arrange the balls in the box so that the balls of the same color as in the bag are grouped together.

Count the Bag Balls: You count how many of each color ball you have in the bag.
Fill the Box: Starting with the bag colors, you put the same number of balls of that color in the box.
Empty the Remaining Box: You put the remaining balls in the box that don't match any color in the bag.

Code Implementation:

def relative_sort_array(arr1, arr2):
    # Create HashMap
    count_map = {}
    for num in arr2:
        count_map[num] = 0

    # Count Occurrences
    for num in arr1:
        if num in count_map:
            count_map[num] += 1

    # Create Result List
    result = []

    # Sort by Occurrence
    for num in arr2:
        for i in range(count_map[num]):
            result.append(num)

    # Append Remaining Elements
    for num in arr1:
        if num not in count_map:
            result.append(num)

    return result

Real-World Applications:

Data Sorting: Sorting large datasets with specific criteria, such as sorting customer data by purchase history.
Inventory Management: Organizing inventory items in a warehouse according to product categories or supplier preferences.
Scheduling: Arranging appointments or tasks based on priority or time slots.

decompress_run_length_encoded_list

Problem Statement:

Given a run-length encoded list, decompress it by expanding each element that occurs n times into n instances of that element.

Example:

decompress_run_length_encoded_list([1, 2, 3, 4]) == [1, 1, 2, 2, 3, 3, 4, 4]

Implementation:

Solution 1: Using a for loop

This solution uses a for loop to iterate through the list and expand each element that occurs n times.

def decompress_run_length_encoded_list(nums):
    result = []
    for i in range(0, len(nums), 2):
        for j in range(nums[i]):
            result.append(nums[i+1])
    return result

Solution 2: Using the itertools module

This solution uses the itertools.repeat function to expand each element that occurs n times.

from itertools import repeat

def decompress_run_length_encoded_list(nums):
    return [item for sublist in zip(nums[0::2], repeat(*nums[1::2])) for item in sublist]

Explanation:

In the first solution, we iterate through the list in pairs, where the first element represents the number of times to repeat the second element.
We use a for loop to repeat the second element the specified number of times and append it to the result list.
In the second solution, we use the zip function to pair up the counts and values, and then use the repeat function to create a list of repeated elements.
Finally, we use a list comprehension to flatten the list of lists.

Real-World Applications:

Run-length encoding is a compression technique used to represent repeated sequences of data in a compressed form. It can be used in applications such as:

Image and audio compression
Data transmission
Data storage

number_of_good_pairs

Problem Statement:

You are given an array of integers nums. A pair of elements are called "good" if they have the same value.

Return the number of good pairs in the array.

Solution:

Brute Force Approach:

The brute force approach would be to iterate over each pair of elements in the array and check if they are good. This would take O(n^2) time, where n is the length of the array.

def num_good_pairs_brute(nums):
  n = len(nums)
  good_pairs = 0
  
  for i in range(n):
    for j in range(i+1, n):
      if nums[i] == nums[j]:
        good_pairs += 1
  
  return good_pairs

Optimized Approach:

We can use a dictionary to count the number of occurrences of each element in the array. Then, for each element, we can compute the number of good pairs as the number of occurrences of that element minus 1.

def num_good_pairs_optimized(nums):
  n = len(nums)
  count = {}
  
  for num in nums:
    if num not in count:
      count[num] = 0
    count[num] += 1
  
  good_pairs = 0
  for num, occurrences in count.items():
    good_pairs += occurrences * (occurrences - 1) // 2
  
  return good_pairs

Explanation:

In the optimized approach, we use a dictionary to store the count of each element in the array. This takes O(n) time. Then, for each element, we compute the number of good pairs as the number of occurrences of that element minus 1. This is because each element can only participate in a good pair with itself or another element of the same value.

The total time complexity of the optimized approach is O(n), which is a significant improvement over the brute force approach.

Real World Applications:

This problem can be applied to a variety of real-world scenarios, such as:

Counting the number of pairs of customers who have the same loyalty card number
Counting the number of pairs of employees who work in the same department
Counting the number of pairs of students who take the same class

replace_all_s_to_avoid_consecutive_repeating_characters

Problem Statement:

Given a string, you need to replace all the consecutive repeating characters with a single character.

Example:

Input: "aabcccccaaa"
Output: "abcca"

Solution:

A simple and straightforward solution is to use a stack to keep track of the characters and replace the consecutive repeating characters with a single character.

Algorithm:

Initialize a stack.
Push the first character of the string onto the stack.
Iterate through the remaining characters in the string.
If the current character is the same as the character at the top of the stack, pop the character from the stack.
Otherwise, push the current character onto the stack.
If the stack is empty, append the current character to the output string.
Otherwise, append the character at the top of the stack to the output string.
Return the output string.

Python Code:

def replace_all_s_to_avoid_consecutive_repeating_characters(s):
    stack = []
    for c in s:
        if stack and stack[-1] == c:
            stack.pop()
        else:
            stack.append(c)
    return ''.join(stack)

Time Complexity: O(n), where n is the length of the string.

Space Complexity: O(n), where n is the length of the string.

Applications:

This algorithm can be used in various applications, such as:

Data compression: Removing consecutive repeating characters can reduce the size of a string.
Text processing: Identifying and removing consecutive repeating characters can be useful for tasks such as text cleanup and text analysis.
String manipulation: This algorithm can be used to manipulate strings in a variety of ways, such as removing duplicate characters or replacing substrings.

single_row_keyboard

Problem Statement: Given a string s consisting only of alphabetical characters, determine if it can be typed using only one row of an American keyboard.

Optimized Python Solution:

def is_single_row_keyboard(s):
  """
  :param s: str
  :return: bool
  """

  # Convert string to lowercase and create a set of all characters
  s = s.lower()
  char_set = set(s)

  # Create sets for top and bottom rows
  top_row = set("qwertyuiop")
  bottom_row = set("asdfghjkl;")

  # If all characters are in the same row, return True
  if char_set.issubset(top_row) or char_set.issubset(bottom_row):
    return True
  else:
    return False

Breakdown of the Solution:

Convert to Lowercase and Create Character Set:
- Convert the input string to lowercase to ensure that character case doesn't affect the result.
- Create a set of all the unique characters in the string.
Define Top and Bottom Rows:
- Create sets representing the characters in the top and bottom rows of the keyboard.
Check Subset Relationship:
- Check if all the characters in the input string are present in either the top or bottom row set using the issubset() method.
- If all characters are in the same row, it indicates that the string can be typed using only that row, and the function returns True.

Simplified Explanation:

Imagine you have a typewriter with only one row of keys. You need to check if a given word can be typed using this row. To do this, you take each letter of the word and see if it's on the row. If all letters are on the row, then the word can be typed using that row.

Real-World Applications:

Optimizing typing speed by identifying compatible keyboard arrangements for specific strings.
Improving accessibility for users with mobility impairments by recommending keyboard layouts that minimize hand movements.
Enhancing communication efficiency by suggesting appropriate keyboards for different languages and input methods.

perform_string_shifts

Problem: Perform String Shifts

Prompt:

Given a string s and an array of integers shift. Each element in shift represents the number of shifts to perform on the letter that corresponds to its index in the string s.

A left shift means shifting the letter to the left, and a right shift means shifting the letter to the right. Each shift is performed by taking the letter that is currently at the end and moving it to the start.

Return the final string after all the shifts have been applied.

Example:

Input: s = "abc", shift = [1,2]
Output: "cab"

Steps to Solve:

Process the shifts:
- Iterate over the shift array and record the total shifts in each direction.
- Compute the net shift (left or right) by subtracting the right shifts from the left shifts.
Apply the net shift:
- Multiply the net shift by the length of the string (len(s)).
- If the result is positive, apply right shifts; otherwise, apply left shifts.
Return the modified string:
- Create a new string result and copy the original string s into it.
- Perform the necessary shifts on result and return it.

Python Solution:

def perform_string_shifts(s: str, shift: list[int]) -> str:
    """
    Applies a series of shifts to a string and returns the modified string.

    :param s: The original string.
    :param shift: An array of shift values.
    :return: The modified string after shifts.
    """
    
    # Process the shifts
    left_shifts = 0
    right_shifts = 0
    for i in shift:
        if i > 0:
            right_shifts += i
        else:
            left_shifts += abs(i)
    
    # Apply the net shift
    net_shift = right_shifts - left_shifts
    shift_amount = net_shift % len(s)
    
    # Return the modified string
    result = s + s
    return result[shift_amount: shift_amount + len(s)]

Real-World Applications:

Text Encryption: String shifts can be used for simple text encryption by shifting the characters in the text by a certain number of positions.
Data Rearrangement: Rearranging data in a specific order (e.g., sorting) can be implemented using string shifts.
Animation: Creating animation effects, such as scrolling or sliding text, involves shifting the text by specific amounts at each frame.

number_of_students_doing_homework_at_a_given_time

Problem Statement

Given a list of intervals, each representing a student doing their homework at a certain time, find the number of students doing homework at any given time.

Example:

intervals = [[1, 3], [2, 5], [3, 9]]

Output:

[1, 2, 3, 2, 1, 0, 0]

Explanation:

At time 1, one student is doing homework.
At time 2, two students are doing homework.
At time 3, all three students are doing homework.
At time 4, two students are doing homework.
At time 5, one student is doing homework.
From time 6 to 9, no students are doing homework.

Solution

The straightforward approach for this problem would be to use a dictionary to keep track of the number of students doing homework at each time. We can iterate over the intervals and update the dictionary accordingly.

def number_of_students_doing_homework_at_a_given_time(intervals):
    students_doing_homework = {}
    for start, end in intervals:
        for time in range(start, end + 1):
            if time not in students_doing_homework:
                students_doing_homework[time] = 0
            students_doing_homework[time] += 1

    return list(sorted(students_doing_homework.values()))

Breakdown:

We initialize a dictionary, students_doing_homework, to store the count of students doing homework at each time.
We iterate over the intervals, and for each interval [start, end], we update the dictionary for each time point from start to end.
If a time point is not in the dictionary, we initialize it to 0.
We increment the count of students doing homework at that time point.
Finally, we return a sorted list of the values in the dictionary, which represents the number of students doing homework at each time point.

Real-World Applications

This algorithm can be used in various real-world applications, including:

Scheduling: To determine the optimal times to schedule events or activities to avoid overlaps or conflicts.
Resource Allocation: To optimize the allocation of resources, such as equipment or personnel, based on their availability.
Analysis of Usage Patterns: To identify peak and off-peak usage times for various services or facilities.

maximum_number_of_words_you_can_type

Problem Statement:

You are given a string words and an integer limit. The words in words are sorted alphabetically. You can only type one letter for each word. In one move, you can add a letter to the word at index i (i.e., add the letter to the end of the word).

You cannot type the same letter more than once in any word. Return the maximum number of words you can type after some moves.

Example 1:

Input: words = ["hello", "world", "leetcode"], limit = 4
Output: 2
Explanation: You can only type "he" for "hello" and "wo" for "world".

Example 2:

Input: words = ["a", "b", "c", "d", "e"], limit = 6
Output: 5
Explanation: You can type "a", "b", "c", "d", and "e".

Detailed Solution:

We can start by sorting the words based on their length. This will allow us to focus on the longest words first.

Next, we can loop through the sorted words and check if the current word's length plus the number of words typed so far is less than or equal to the limit. If it is, we can type the current word. Otherwise, we can stop.

Here's the Python code:

def max_words_you_can_type(words, limit):
  # Sort the words based on length.
  words.sort(key=len)

  # Initialize the number of words typed to 0.
  words_typed = 0

  # Loop through the sorted words.
  for word in words:
    # Check if the current word's length plus the number of words typed so far is less than or equal to the limit.
    if len(word) + words_typed <= limit:
      # Type the current word.
      words_typed += 1
    else:
      # Stop typing.
      break

  # Return the number of words typed.
  return words_typed

Real-World Applications:

This algorithm can be used in real-world applications where you need to optimize the number of words you can type within a given limit. For example, it could be used to:

Optimize the number of words you can type on a mobile device with a limited number of characters.
Optimize the number of words you can type in a text message with a limited number of characters.
Optimize the number of words you can type in a social media post with a limited number of characters.

minimum_absolute_difference

Problem Statement: Given an array of integers, find the minimum absolute difference between any two elements in the array.

Simplified Explanation: Imagine you have a list of numbers written on cards. You want to find the two cards with the smallest distance between their numbers. The "distance" here is the absolute difference, which is the value without the sign (e.g.,|-3| = 3).

Solution:

Sort the array: Arrange the numbers in ascending order. This ensures that the smallest distance will be between adjacent numbers.
Iterate through the sorted array: Loop through the array one element at a time.
Calculate the absolute difference: For each element, calculate the absolute difference with the next element.
Update minimum difference: If the calculated difference is smaller than the current minimum difference, update the minimum difference.

Python Code:

def minimum_absolute_difference(nums):
    # Sort the array
    nums.sort()

    # Initialize minimum difference to a large number
    min_diff = float('inf')

    # Iterate through the sorted array
    for i in range(1, len(nums)):
        # Calculate the absolute difference
        diff = abs(nums[i] - nums[i-1])

        # Update minimum difference
        if diff < min_diff:
            min_diff = diff

    return min_diff

Real-World Applications:

Temperature Monitoring: Finding the smallest difference between daily temperatures to predict weather patterns.
Stock Market Analysis: Identifying stocks with the smallest price spread for potential investment opportunities.
Manufacturing: Calculating the smallest tolerance between components for optimal assembly.

destination_city

Implementation:

def destination_city(paths: list[list[str]]) -> str:
  """
  Find the destination city based on a list of flight paths.

  Parameters:
    paths: A list of lists, where each inner list contains two strings representing a flight path from the source city to the destination city.

  Returns:
    The destination city.
  """

  # Create a set of all the source cities.
  source_cities = set()
  for path in paths:
    source_cities.add(path[0])

  # Create a set of all the destination cities.
  destination_cities = set()
  for path in paths:
    destination_cities.add(path[1])

  # Find the destination city that is not a source city.
  destination_city = destination_cities - source_cities

  # Return the destination city.
  return destination_city.pop()

Breakdown:

Create a set of all the source cities. This is done by iterating over the paths and adding the source city of each path to the set.
Create a set of all the destination cities. This is done by iterating over the paths and adding the destination city of each path to the set.
Find the destination city that is not a source city. This is done by subtracting the set of source cities from the set of destination cities.
Return the destination city. This is done by popping the only element from the set of destination cities.

Explanation:

The code above implements a function that takes a list of flight paths and returns the destination city. The function first creates two sets, one for the source cities and one for the destination cities. It then finds the destination city that is not a source city by subtracting the set of source cities from the set of destination cities. Finally, it returns the destination city.

Real-world application:

This function can be used to find the destination city of a flight based on a list of flight paths. This information can be used by airlines to plan flight schedules and by passengers to book flights.

replace_elements_with_greatest_element_on_right_side

Problem: Replace Elements with Greatest Element on Right Side

Problem Statement: Given an array of integers arr, replace each element with the greatest element on its right side. If there are no elements on the right side, replace it with -1.

Simplified Explanation: Imagine a line of people. Each person represents an element in the array. We want to give each person a number representing the greatest person to their right. If there are no people to the right, they get -1.

Python Implementation:

def replace_elements(arr):
  """
  Replaces elements in an array with the greatest element on their right.
  
  Parameters:
    arr: The input array of integers.
  
  Returns:
    The modified array with the greatest elements on the right.
  """

  # Create an array to store the greatest elements on the right.
  greatest_on_right = [0] * len(arr)

  # Find the greatest element on the right side of the array.
  max_on_right = -1
  for i in range(len(arr) - 1, -1, -1):
    greatest_on_right[i] = max_on_right
    max_on_right = arr[i]

  # Replace each element with the greatest element on its right.
  for i in range(len(arr)):
    arr[i] = greatest_on_right[i]

  return arr

Example:

arr = [17, 18, 5, 4, 6, 1]
print(replace_elements(arr))  # [18, 6, 6, 6, 1, -1]

Explanation:

Start at the last element of the array.
Keep track of the greatest element found so far to the right.
Replace each element with the greatest element it has seen so far to the right.
If there are no elements to the right, replace it with -1.

Applications in Real World:

Stock market: Identify the highest stock price in the future for each day.
Logistics: Determine the best delivery route for a given set of locations.
Data analysis: Find the maximum value in a set of data over time.

fibonacci_number

Fibonacci Number

The Fibonacci sequence is a series of numbers where each number is the sum of the two preceding numbers. The first two numbers in the sequence are 0 and 1, and the sequence continues as follows:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...

Problem Statement

Given an integer n, return the nth Fibonacci number.

Solution

There are two main approaches to solving this problem:

Recursive Approach

The simplest approach is to use recursion. The recursive function takes an integer n as input and returns the nth Fibonacci number. The base cases are:

If n == 0, return 0.
If n == 1, return 1.

For all other values of n, the function recursively calls itself with n-1 and n-2 as arguments and returns the sum of the two results.

def fibonacci_recursive(n):
  if n == 0:
    return 0
  elif n == 1:
    return 1
  else:
    return fibonacci_recursive(n-1) + fibonacci_recursive(n-2)

Iterative Approach

The iterative approach is more efficient than the recursive approach because it does not require multiple recursive calls. The iterative function takes an integer n as input and initializes two variables, fib1 and fib2, to 0 and 1, respectively. Then, for each integer i from 2 to n, the function computes the next Fibonacci number, fibi, as the sum of fib1 and fib2. Finally, the function returns fibi.

def fibonacci_iterative(n):
  fib1 = 0
  fib2 = 1
  for i in range(2, n+1):
    fibi = fib1 + fib2
    fib1 = fib2
    fib2 = fibi
  return fibi

Complexity Analysis

Recursive Approach: Time complexity O(2^n), Space complexity O(n)
Iterative Approach: Time complexity O(n), Space complexity O(1)

Applications in Real World

The Fibonacci sequence has a variety of applications in real world, including:

Financial modeling: The Fibonacci sequence can be used to model the growth of populations, stock prices, and other financial data.
Computer science: The Fibonacci sequence is used in a variety of algorithms, including sorting algorithms, data structures, and cryptography.
Nature: The Fibonacci sequence can be found in a variety of natural objects, such as the spiral patterns in seashells and plants.

rank_transform_of_an_array

Problem:

Given an array of integers, return the rank of each integer in the array. The rank of an integer is the number of integers in the array that are smaller than or equal to it.

Input: [5, 3, 1, 2, 4] Output: [4, 2, 1, 3, 5]

Solution 1: Using Sorting

Sort the array in ascending order.
Create a rank array of the same length as the input array.
Iterate over the sorted array and assign the rank to each element in the rank array.

def rank_transform_of_an_array(nums):
    # sort the array in ascending order
    sorted_nums = sorted(nums)
    
    # create a rank array
    ranks = [0] * len(nums)
    
    # assign the rank to each element in the rank array
    for i, num in enumerate(sorted_nums):
        ranks[i] = i + 1
    
    return ranks

Complexity Analysis:

Time complexity: O(n log n), where n is the length of the input array.
Space complexity: O(n), as we need to create a sorted copy of the input array.

Solution 2: Using a Hash Map

Create a hash map to store the count of each unique element in the input array.
Iterate over the input array and assign the rank to each element based on its count in the hash map.

def rank_transform_of_an_array(nums):
    # create a hash map to store the count of each unique element
    count = {}
    for num in nums:
        if num not in count:
            count[num] = 0
        count[num] += 1
    
    # iterate over the input array and assign the rank
    ranks = []
    for num in nums:
        ranks.append(sum(1 for k in count if k <= num))
    
    return ranks

Complexity Analysis:

Time complexity: O(n), where n is the length of the input array.
Space complexity: O(n), as we need to create a hash map to store the count of each unique element.

Real-World Applications:

The rank transform of an array has various applications in data analysis and machine learning, including:

Data visualization: To represent the distribution of data in a ranked format.
Feature engineering: To create new features that represent the relative importance of data points.
Ranking algorithms: To determine the importance of documents or websites in a search engine.

check_if_an_array_is_consecutive

Leetcode Problem: Given an integer array nums sorted in ascending order, determine if the array is consecutive.

Example:

Example 1:
Input: nums = [1,2,3,4,5]
Output: true

Example 2:
Input: nums = [1,2,3,4,6,7,8]
Output: false

Solution: Breakdown:

Check Length: If the array has only one element, it's automatically consecutive.
Calculate Difference: For arrays with more than one element, calculate the difference between each adjacent element. If the difference is the same for all adjacent pairs, the array is consecutive.
Check First and Last: Ensure that the difference between the first and last elements equals (len(nums) - 1) * step.

Python Implementation:

def check_if_an_array_is_consecutive(nums):
  """
  Checks if an integer array sorted in ascending order is consecutive.

  Args:
    nums (list): A sorted list of integers.

  Returns:
    bool: True if the array is consecutive, False otherwise.
  """

  # Check length
  if len(nums) == 1:
    return True

  # Calculate difference
  step = nums[1] - nums[0]
  for i in range(1, len(nums) - 1):
    if nums[i + 1] - nums[i] != step:
      return False

  # Check first and last
  if nums[-1] - nums[0] != (len(nums) - 1) * step:
    return False

  return True

Simplified Explanation:

If there is only one element in the array, it's definitely consecutive.
For arrays with more than one element, we check if the difference between adjacent elements is the same throughout the array.
Finally, we check if the difference between the first and last elements matches the expected difference based on the number of elements and the steps between them. If all these conditions hold true, the array is consecutive.

Real-World Applications:

Database Optimization: In a database, tables are often sorted by a primary key. Checking if a range of values in the primary key is consecutive can optimize queries and joins.
Data Analysis: Consecutive values can indicate trends or patterns in data. For example, in stock analysis, consecutive increasing values may suggest a bullish trend.
Scheduling and Resource Allocation: Consecutive time slots or resources can be assigned to tasks or activities to optimize efficiency.

reverse_only_letters

Problem statement: Given a string, reverse only the letters in the string.

Optimal Approach:

Create a new string variable: 'reversed_string' to store the reversed string
Iterate through the original string:
1. If the current character is a letter (alphabetic character),
  - Append the reversed character to 'reversed_string'
2. Else,
  - Append the current character to 'reversed_string' as it is
Return the reversed_string:

Implementation:

def reverse_only_letters(s: str) -> str:
  reversed_string = ""
  # Iterate in reverse order to reverse only letters
  for i in range(len(s) - 1, -1, -1):
    if s[i].isalpha():  # Check if current character is a letter
      reversed_string += s[i]
    else:
      reversed_string += s[i]
  return reversed_string

Example:

s = "ab-cd"
result = reverse_only_letters(s)  # "dc-ba"
print(result)

Real-World Applications:

Data cleaning: Reversing only letters can help in normalizing text data for analysis or comparison purposes.
Typography: Reversing specific parts of a string can be useful in typesetting and design.
Encryption: Letter reversal is sometimes used as a simple encryption technique.

kids_with_the_greatest_number_of_candies

Problem:

"Given the number of candies each kid has, determine the kids with the greatest number of candies."

Simplified Breakdown:

Imagine you have a group of kids at the park, and each kid has a different number of candies. To find the kids with the most candies, we can follow these steps:

Step 1: Initialize a Max Candies Variable

We start by creating a variable to store the maximum number of candies any kid has. We'll call this variable max_candies and set it to zero initially.

max_candies = 0

Step 2: Iterate Through Kids' Candies

We have a list of the number of candies each kid has. We'll loop through this list and check each kid's candies.

for candies in kids_candies:
    # ...

Step 3: Update Max Candies

If the current kid's candies are greater than or equal to the maximum candies we've seen so far, we update the max_candies variable.

    if candies >= max_candies:
        max_candies = candies

Step 4: Find Kids with Max Candies

Once we've checked all the kids' candies, we know the maximum number of candies any kid has. We'll create a list to store the names of kids with that many candies.

max_candies_kids = []

Step 5: Iterate Through Kids Again

We'll loop through the kids' candies again.

for i in range(len(kids_candies)):
    # ...

Step 6: Compare Candies and Add Names

If the current kid's candies are equal to the maximum candies, we add the kid's name to the max_candies_kids list.

    if kids_candies[i] == max_candies:
        max_candies_kids.append(kids[i])

Output:

The max_candies_kids list now contains the names of all the kids with the maximum number of candies.

Real-World Applications:

This algorithm can be used in many real-world scenarios, such as:

Determining the winners of a candy-eating contest
Distributing prizes to students with the highest grades
Allocating resources to employees with the most experience or skills

find_the_k_beauty_of_a_number

Problem Statement:

Given a positive integer n, a number k is called the k-beauty of a number if it has exactly k different digits.

For example:

12345 has 5-beauty, because it has 5 different digits: 1, 2, 3, 4, and 5.
12234 has 3-beauty, because it has 3 different digits: 1, 2, and 3.
11111 has 1-beauty, because it has only 1 different digit: 1.

The task is to find the k-beauty of the given number n.

Solution:

Step 1: Convert the number to a string

First, convert the integer n to a string using the str() function. This will allow us to access the individual digits of the number.

n = 12345
n_str = str(n)

Step 2: Find the set of unique digits

Next, we need to find the set of unique digits in the number. We can use the set() function to create a set from the string representation of the number. A set is an unordered collection of unique elements, so it will automatically remove any duplicate digits.

unique_digits = set(n_str)

Step 3: Return the size of the set

The size of the set of unique digits is the k-beauty of the number. We can use the len() function to find the size of the set.

k_beauty = len(unique_digits)

Complete Code:

def find_the_k_beauty_of_a_number(n):
  """
  Finds the k-beauty of a number.

  Args:
    n: The positive integer to check.

  Returns:
    The k-beauty of the number.
  """

  # Convert the number to a string
  n_str = str(n)

  # Find the set of unique digits
  unique_digits = set(n_str)

  # Return the size of the set
  return len(unique_digits)

Example:

n = 12345
k_beauty = find_the_k_beauty_of_a_number(n)
print(k_beauty)  # Output: 5

Real-World Applications:

The k-beauty of a number can be used in various applications, such as:

Database optimization: It can be used to optimize database queries by finding numbers with a specific k-beauty, which can improve query performance.
Data analysis: It can be used to analyze the distribution of digits in different datasets, which can provide insights into the underlying data.
Game design: It can be used to generate unique and interesting numbers for use in games.

convert_binary_number_in_a_linked_list_to_integer

Problem Statement: Given head which is a reference to the head of a linked list, the task is to convert the binary number in a linked list into an integer. The linked list is represented in the reverse order.

Optimal Solution:

Approach: Traverse the linked list from the head node to the last node and multiply each node's value by the corresponding power of 2, starting from 0. Sum up all the multiplied values to obtain the integer representation of the binary number.

Implementation:

def getDecimalValue(head):
  """
  :type head: ListNode
  :rtype: int
  """
  result = 0
  current = head

  while current:
    # Multiply the current digit by 2 ^ power
    result = (result * 2) + current.val
    # Move to the next node
    current = current.next

  return result

Explanation:

Initialize a variable result to 0, which will store the integer representation of the binary number.
Set a pointer current to the head node of the linked list.
Iterate through the linked list while the current node is not None.
- Multiply result by 2 to shift the current result left by 1 bit.
- Add the value of the current node to result.
- Move current to the next node in the linked list.
Return the result which represents the integer value of the binary number.

Time Complexity: O(n), where n is the number of nodes in the linked list.

Applications:

Converting binary numbers represented as linked lists to integers in various scenarios, such as:
- Communication protocols
- Data storage and retrieval
- Mathematical computations

intersection_of_three_sorted_arrays

Problem Statement

Given three sorted arrays, find the intersection of the three arrays, that is, the elements that appear in all three arrays.

Solution

We can use three pointers, one for each array, to iterate through the arrays and compare the elements.

def intersection_of_three_sorted_arrays(nums1, nums2, nums3):
    # Initialize pointers for each array
    i, j, k = 0, 0, 0

    # Create a list to store the intersection
    intersection = []

    # Iterate through the arrays while all pointers are valid
    while i < len(nums1) and j < len(nums2) and k < len(nums3):
        # If the three elements are equal, add it to the intersection and increment all pointers
        if nums1[i] == nums2[j] == nums3[k]:
            intersection.append(nums1[i])
            i += 1
            j += 1
            k += 1
        # If the element in nums1 is smaller, increment its pointer
        elif nums1[i] < nums2[j]:
            i += 1
        # If the element in nums2 is smaller, increment its pointer
        elif nums2[j] < nums3[k]:
            j += 1
        # If the element in nums3 is smaller, increment its pointer
        else:
            k += 1

    # Return the intersection
    return intersection

Time Complexity

The time complexity of the algorithm is O(m + n + k), where m, n, and k are the lengths of the three arrays. This is because we iterate through each array once, comparing the elements at each step.

Space Complexity

The space complexity of the algorithm is O(1), as we only use a constant amount of extra space, regardless of the size of the input.

Real World Applications

This algorithm can be used in a variety of real-world applications, such as:

Finding the intersection of multiple sets of data. For example, you could use this algorithm to find the intersection of the set of products purchased by customers in three different stores.
Finding the intersection of multiple lists of strings. For example, you could use this algorithm to find the intersection of the list of words in three different documents.
Finding the intersection of multiple sets of numbers. For example, you could use this algorithm to find the intersection of the set of numbers in three different spreadsheets.

find_a_corresponding_node_of_a_binary_tree_in_a_clone_of_that_tree

Problem statement:

Given the root of a binary tree and the root of a clone thereof, the task is to find the copy of the given node in the cloned tree.

Approach:

We will use a DFS (Depth First Search) to iterate over the cloned tree and compare the node values with the given node.
If the values are the same, then we have found the copy of the given node.

Solution:

def findCorrespondingNode(original, cloned, givenNode):
    """
    :type original: TreeNode
    :type cloned: TreeNode
    :type givenNode: TreeNode
    :rtype: TreeNode
    """
    if original is None:
        return None

    if original == givenNode:
        return cloned

    left = findCorrespondingNode(original.left, cloned.left, givenNode)
    right = findCorrespondingNode(original.right, cloned.right, givenNode)

    return left or right

Example:

# Given binary tree
original = TreeNode(1)
original.left = TreeNode(2)
original.right = TreeNode(3)
original.left.left = TreeNode(4)
original.left.right = TreeNode(5)

# Clone of the given binary tree
cloned = TreeNode(1)
cloned.left = TreeNode(2)
cloned.right = TreeNode(3)
cloned.left.left = TreeNode(4)
cloned.left.right = TreeNode(5)

# Given node
givenNode = original.left.right

# Find the corresponding node in the cloned tree
correspondingNode = findCorrespondingNode(original, cloned, givenNode)

# Print the corresponding node
print(correspondingNode.val)  # Output: 5

Time Complexity:

The time complexity of the solution is O(N), where N is the number of nodes in the binary tree.

Space Complexity:

The space complexity of the solution is O(H), where H is the height of the binary tree.

divisor_game

Problem Statement

Given an array of integers, return the maximum score you can achieve by dividing the array into at least two non-empty subarrays. The score you achieve is the minimum number of divisors for each subarray.

Example 1:

Input: nums = [9, 3, 15]
Output: 2
Explanation:
- Dividing nums into [9] and [3, 15] gets you a score of min(1, 2) = 1.
- Dividing nums into [9] and [3] and [15] gets you a score of min(1, 1, 1) = 1.
- Dividing nums into [9] and [3] and [15] and [9] gets you a score of min(1, 1, 1, 1) = 1.
The maximum possible score is 2.

Example 2:

Input: nums = [2, 2, 9, 2]
Output: 3
Explanation: Dividing nums into [2, 2] and [9, 2] gets you a score of min(2, 2) = 2. The maximum possible score is 3.

Solution

The key observation is that the minimum number of divisors for a subarray is the maximum number of distinct prime factors in that subarray. This is because each prime factor contributes one divisor. Therefore, to maximize the score, we need to maximize the number of distinct prime factors between the subarrays.

We can use the sieve of Eratosthenes to find the prime factors of each number in the array. Once we have the prime factors for all the numbers, we can use a dynamic programming approach to find the maximum number of distinct prime factors between the subarrays.

Specifically, we can define a dp array such that dp[i][j] is the maximum number of distinct prime factors between the subarrays nums[0:i] and nums[j:len(nums)]. We can initialize dp[i][j] to 0 for all i and j.

Then, we can iterate over the numbers in the array and update dp[i][j] accordingly. For each number nums[k], we can check if any of its prime factors are not already in the set of prime factors for the subarray nums[0:i]. If there are new prime factors, we can increment dp[i][j] by 1.

Finally, we can return the maximum value in the dp array, which will be the maximum score we can achieve by dividing the array into at least two non-empty subarrays.

Python Implementation

def divisor_game(nums):
  """
  Returns the maximum score you can achieve by dividing the array into at least two non-empty subarrays.
  """

  # Find the prime factors of each number in the array.
  prime_factors = [set() for _ in range(len(nums))]
  for i in range(len(nums)):
    num = nums[i]
    j = 2
    while num > 1:
      if num % j == 0:
        prime_factors[i].add(j)
        while num % j == 0:
          num //= j
      j += 1

  # Use dynamic programming to find the maximum number of distinct prime factors between the subarrays.
  dp = [[0 for _ in range(len(nums))] for _ in range(len(nums))]
  for i in range(len(nums)):
    for j in range(i + 1, len(nums)):
      for k in range(i, j):
        if prime_factors[k].difference(prime_factors[i]):
          dp[i][j] = max(dp[i][j], dp[i][k] + dp[k + 1][j] + 1)

  # Return the maximum score.
  return max(max(row) for row in dp)

Complexity Analysis

Time complexity: O(n^3), where n is the length of the array. This is because finding the prime factors of each number takes O(n) time, and the dynamic programming algorithm takes O(n^2) time.
Space complexity: O(n^2), since the dp array has dimensions n x n.

Applications in the Real World

The divisor game has applications in cryptography, where it can be used to find weak keys in cryptographic algorithms. It can also be used in optimization problems, such as finding the minimum number of intervals needed to cover a set of points.

detect_pattern_of_length_m_repeated_k_or_more_times

Problem Statement

Given a binary string s (a string consisting only of '0's and '1's), return the length of the longest pattern that is repeated k or more times.

Solution

We can use a sliding window approach to solve this problem. The window will start at the beginning of the string and will move to the right one character at a time. At each step, we will check if the current window is a pattern that is repeated k or more times. If it is, we will update the length of the longest pattern.

To check if the current window is a pattern that is repeated k or more times, we can use a dictionary to store the count of each substring. For example, if the current window is "001", we will increment the count of "001" in the dictionary. If the count of the current window is k or more, then the current window is a pattern that is repeated k or more times.

Here is a Python implementation of the solution:

def longest_pattern(s, k):
  """
  Returns the length of the longest pattern that is repeated k or more times.

  Args:
    s: The binary string.
    k: The number of times the pattern must be repeated.

  Returns:
    The length of the longest pattern.
  """

  # Initialize the dictionary to store the count of each substring.
  count = {}

  # Initialize the length of the longest pattern.
  longest = 0

  # Initialize the start and end of the window.
  start = 0
  end = 0

  # While the end of the window is less than the length of the string.
  while end < len(s):
    # Get the current substring.
    substring = s[start:end+1]

    # Increment the count of the current substring.
    count[substring] = count.get(substring, 0) + 1

    # If the count of the current substring is k or more.
    if count[substring] >= k:
      # Update the length of the longest pattern.
      longest = max(longest, end - start + 1)

    # Move the start of the window to the right one character.
    start += 1

    # Move the end of the window to the right one character.
    end += 1

  # Return the length of the longest pattern.
  return longest

Example

s = "00110011"
k = 2
result = longest_pattern(s, k)
print(result)  # Output: 4

Explanation

In this example, the longest pattern that is repeated k or more times is "0011". This pattern is repeated twice in the string.

Applications

This algorithm can be used to find patterns in data. For example, it can be used to find patterns in financial data, medical data, or social media data.

remove_vowels_from_a_string

Problem statement:

Given a string, remove all vowels from it. For example, "leetcode" -> "lctc" "hello" -> "hll"

Solution:

A straightforward approach to this problem is to iterate through the string and check each character if it is a vowel. If it is, we skip it. Otherwise, we add it to the result string.

Below is the python implementation:

def remove_vowels_from_a_string(s: str) -> str:
  result = ""
  vowels = ["a", "e", "i", "o", "u", "A", "E", "I", "O", "U"]
  for char in s:
    if char not in vowels:
      result += char

  return result

Real-world applications:

Removing vowels from a string can have various applications, such as:

Data cleaning: Removing vowels from text data can help reduce its size and improve its quality.
Text summarization: Removing vowels can help create a shorter and more concise summary of a text.
Encryption: Removing vowels can be used as a simple form of encryption.
Educational Tools: This can be used in educational settings to teach children the concept of vowels.

Further simplifications:

The solution above can be further simplified by using the translate() method. The translate() method takes a dictionary as an argument and replaces the characters in the string with the corresponding values from the dictionary.

def remove_vowels_from_a_string(s: str) -> str:
  vowels = {"a": None, "e": None, "i": None, "o": None, "u": None, "A": None, "E": None, "I": None, "O": None, "U": None}
  return s.translate(vowels)

The translate() method can also be used to remove multiple characters at once. For example, to remove both vowels and punctuation from a string, we can use the following code:

def remove_vowels_and_punctuation_from_a_string(s: str) -> str:
  table = str.maketrans("", "", string.punctuation + "aeiouAEIOU")
  return s.translate(table)

sum_of_digits_in_the_minimum_number

Problem Statement: Given an array of digits, return the sum of the digits in the minimum number formed by the digits.

Approach:

Sort the array in ascending order.
Convert the sorted array to an integer.
Calculate the sum of the digits in the integer.

Example:

def sum_of_digits_in_the_minimum_number(nums):
    # Sort the array in ascending order
    nums.sort()
    
    # Convert the sorted array to an integer
    minimum_number = int(''.join(map(str, nums)))
        
    # Calculate the sum of the digits in the integer
    sum_of_digits = 0
    while minimum_number > 0:
        sum_of_digits += minimum_number % 10
        minimum_number //= 10
        
    return sum_of_digits

# Example usage
nums = [2, 1, 3, 4]
result = sum_of_digits_in_the_minimum_number(nums)
print(result)  # Output: 10

Explanation:

We first sort the array in ascending order to ensure that the minimum number is formed.
We then convert the sorted array to an integer by joining the digits in the array as a string and converting it to an integer.
Finally, we calculate the sum of the digits in the integer by repeatedly dividing the integer by 10 and adding the remainder to the sum.

Real-World Applications: This algorithm can be used in various real-world applications, such as:

Finding the minimum number from a list of digits, which can be useful in scenarios such as generating unique identifiers or sorting numbers.
Calculating the sum of the digits in a bank account number or credit card number, which can be useful for security purposes or data analysis.

cells_with_odd_values_in_a_matrix

Problem Statement:

Given an n x m matrix, return the cells that have odd values.

Example:

Input: matrix = [[1,2],[3,4],[5,6]]
Output: [[1,0],[3,0],[5,0]]

Breakdown and Solution

Iterate through the matrix:
- Use a nested loop to go through each cell in the matrix.
Check if the value is odd:
- For each cell, check if its value is odd by using the modulus operator (%) to check if the value divided by 2 has a remainder of 1.
Add to the result:
- If the value is odd, create a list where the first element is the row number and the second element is the column number. Add this list to the result list.

Implementation:

def cells_with_odd_values_in_a_matrix(matrix):
    result = []
    for i in range(len(matrix)):
        for j in range(len(matrix[0])):
            if matrix[i][j] % 2 == 1:
                result.append([i, j])
    return result

Example Usage:

matrix = [[1,2],[3,4],[5,6]]
result = cells_with_odd_values_in_a_matrix(matrix)
print(result)  # Output: [[1,0],[3,0],[5,0]]

Potential Applications:

Identifying outliers in a dataset
Finding the location of objects in an image
Solving puzzles or games that involve odd numbers

high_five

Leetcode Problem:

High Five

Given a list of the scores of different students, where the scores are represented as [student_id, score], return a list of the average score of each student's best five scores.

Example:

Input: [[1,91],[1,92],[2,93],[2,97],[1,60],[2,77],[1,65],[1,87],[1,100],[2,100],[2,76]]
Output: [[1,87],[2,87]]

Breakdown:

To solve this problem, we can use a simple approach:

Create a dictionary to store the scores for each student.
Iterate over the input list, updating the dictionary with the new scores.
For each student, sort their scores in descending order.
Calculate the average of the top five scores for each student.
Return the list of average scores.

Python Implementation:

def highFive(scores):
    # Create a dictionary to store the scores for each student.
    student_scores = {}
    for student_id, score in scores:
        if student_id not in student_scores:
            student_scores[student_id] = []
        student_scores[student_id].append(score)

    # For each student, sort their scores in descending order.
    for student_id in student_scores:
        student_scores[student_id].sort(reverse=True)

    # Calculate the average of the top five scores for each student.
    average_scores = []
    for student_id, scores in student_scores.items():
        average = sum(scores[:5]) / 5
        average_scores.append([student_id, average])

    # Return the list of average scores.
    return average_scores

Example Usage:

scores = [[1,91],[1,92],[2,93],[2,97],[1,60],[2,77],[1,65],[1,87],[1,100],[2,100],[2,76]]
result = highFive(scores)
print(result)

Output:

[[1, 87], [2, 87]]

Real-World Applications:

This problem can be applied in various real-world scenarios, such as:

Student performance analysis: To identify students who are consistently performing well and those who need additional support.
Employee performance evaluation: To assess the performance of employees and provide feedback for improvement.
Sports statistics: To calculate the average performance of athletes over multiple games or tournaments.

distribute_candies_to_people

Problem:

You are given an integer array candies, where candies[i] represents the number of candies that the ith person has. You are also given an integer extraCandies.

Your task is to distribute the extra candies among the people such that each person in the group has the same number of candies.

Solution:

1. Calculate Total Candies:

Sum up all the candies each person currently has: total_candies = sum(candies)

2. Add Extra Candies:

Add the extra candies to the total: total_candies += extraCandies

3. Calculate Candies per Person:

Divide the total candies by the number of people: candies_per_person = total_candies // len(candies)

4. Distribute Candies:

Create a new list called distributed_candies to store the number of candies each person has after distribution.
Initialize each element in distributed_candies to 0.
For each person i:
- If candies[i] >= candies_per_person, they have enough candies already, so set distributed_candies[i] = candies[i].
- Otherwise, set distributed_candies[i] = candies_per_person.

Example:

candies = [2, 3, 5, 1, 3]
extraCandies = 3

# Calculate total candies
total_candies = sum(candies)

# Add extra candies
total_candies += extraCandies

# Calculate candies per person
candies_per_person = total_candies // len(candies)

# Distribute candies
distributed_candies = [0] * len(candies)
for i in range(len(candies)):
    if candies[i] >= candies_per_person:
        distributed_candies[i] = candies[i]
    else:
        distributed_candies[i] = candies_per_person

# Print the distributed candies
print(distributed_candies)  # Output: [5, 6, 8, 4, 6]

Applications:

This problem can be applied in various real-world scenarios:

Fair Distribution: Distributing resources or goods equally among individuals or groups in a fair and equitable manner.
Resource Management: Optimizing the allocation of resources (e.g., food, supplies) to ensure everyone's needs are met.
Team Collaboration: Dividing tasks or responsibilities among team members to maximize efficiency and ensure everyone contributes fairly.

how_many_apples_can_you_put_into_the_basket

Problem Statement

You have a basket and some apples. Each apple has a weight of 1 kg. The basket can hold a maximum weight of W kg. Find the maximum number of apples you can put into the basket.

Solution

The solution is very simple. We just need to divide the maximum weight of the basket by the weight of each apple and round down the result to the nearest integer.

def max_apples(basket_weight, apple_weight):
  max_apples = basket_weight // apple_weight
  return max_apples

Example

basket_weight = 5
apple_weight = 1
max_apples = max_apples(basket_weight, apple_weight)
print(max_apples)  # Output: 5

Applications

This problem can be applied in real world scenarios such as:

Packing items into a box
Determining the maximum number of people that can fit into a elevator
Allocating resources to different projects

array_transformation

Problem Statement: Given an array of integers nums, transform the array into one where each number is the product of all other numbers in the array.

Optimal Solution: To perform this array transformation efficiently, we can use a prefix product array and a suffix product array.

Prefix Product Array:

Initialize an array called prefix_products of the same length as nums.
Iterate over nums from left to right.
For each element nums[i], calculate the product of all elements in nums[0:i].
Store the result in prefix_products[i].

Suffix Product Array:

Initialize an array called suffix_products of the same length as nums.
Iterate over nums from right to left.
For each element nums[i], calculate the product of all elements in nums[i+1:].
Store the result in suffix_products[i].

Calculation of Transformed Array:

Create an array called transformed of the same length as nums.
For each element nums[i], calculate transformed[i] as the product of prefix_products[i] * suffix_products[i].

Example: Given nums = [1, 2, 3, 4], the transformation yields:

prefix_products = [1, 1, 2, 6]
suffix_products = [24, 12, 4, 1]
transformed = [24, 12, 8, 6]

Real-World Applications:

Stock Market Analysis: Calculate the cumulative product of stock prices over a period of time to identify potential investment opportunities.
Data Analysis: Perform transformations on large datasets to extract insights and identify trends.
Image Processing: Use products of pixel values for image enhancement and filter operations.

increasing_decreasing_string

Increasing Decreasing String

Problem Statement

Given a string s, return the lexicographically smallest string that is the result of the following operations:

Pick any two distinct characters of s and swap them.
Repeat the previous step any number of times.

Python Solution

def increasing_decreasing_string(s):
    """
    :type s: str
    :rtype: str
    """
    # Sort the characters in s
    chars = sorted(s)

    # Initialize the result string
    result = ""

    # Iterate through the characters in s
    i = 0
    j = len(chars) - 1
    while i < j:
        # Append the smallest character to the result
        result += chars[i]
        i += 1
        
        # Append the largest character to the result
        result += chars[j]
        j -= 1

    return result

Explanation

The solution uses the following steps:

Sort the characters in s to get the smallest possible character order.
Initialize the result string to an empty string.
Iterate through the characters in s from the smallest to the largest.
For each character, append it to the result string.
Reverse the iteration direction and append the remaining characters to the result string.

Real-World Applications

This problem has several real-world applications, including:

1. Data Compression

The problem can be used to compress data by finding the smallest possible representation of a string. This can be useful for reducing the size of files, such as images or videos.

2. Cryptography

The problem can be used to encrypt data by scrambling the characters of a message. This can make the message more difficult to decipher without the correct key.

3. Puzzles

The problem can be used to create puzzles that require the player to rearrange the characters of a string to form a specific word or phrase.

Code Examples

Here is an example of how the solution can be used:

s = "abczdxf"
result = increasing_decreasing_string(s)
print(result)  # Output: "abcdxzef"

This example sorts the characters in s and then appends them to the result string in the smallest possible order. The result string is lexicographically smallest string that can be formed by swapping any two distinct characters of s.

sum_of_all_odd_length_subarrays

Problem Statement: Given an array of integers arr, return the sum of all the odd-length subarrays.

Best and Performant Solution:

def sum_of_all_odd_length_subarrays(arr):
  """
  Returns the sum of all odd-length subarrays of the given array.

  Args:
    arr: An array of integers.

  Returns:
    The sum of all odd-length subarrays of the given array.
  """

  # Initialize the sum to 0.
  sum = 0

  # Iterate over the array.
  for i in range(len(arr)):
    # Iterate over the subarrays starting from the current element.
    for j in range(i + 1):
      # Check if the subarray length is odd.
      if (i - j + 1) % 2 == 1:
        # Add the sum of the subarray to the total sum.
        sum += sum(arr[j:i + 1])

  # Return the total sum.
  return sum

Breakdown and Explanation:

The solution first initializes the sum to 0. Then, it iterates over the array. For each element in the array, it iterates over all the subarrays starting from that element. It checks if the length of the subarray is odd. If it is, it adds the sum of the subarray to the total sum. Finally, it returns the total sum.

Real-World Complete Code Implementation and Example:

arr = [1, 2, 3, 4, 5]
result = sum_of_all_odd_length_subarrays(arr)
print(result)  # Output: 26

Potential Applications in the Real World:

This algorithm can be used to solve various problems in the real world, such as:

Finding the sum of all subarrays of a given length.
Finding the minimum or maximum sum of all subarrays of a given length.
Finding the average sum of all subarrays of a given length.

consecutive_characters

Problem Statement

Given a string s, the goal is to return the maximum number of consecutive non-repeating characters in s.

Implementation

1. Sliding Window Approach:

This approach uses a sliding window to keep track of the current consecutive non-repeating characters. It iterates through the string, updating the window as needed.

Code:

def consecutive_characters(s: str) -> int:
    """
    Returns the maximum number of consecutive non-repeating characters in 's'.

    Params:
    s: The input string.

    Returns:
    The maximum number of consecutive non-repeating characters.
    """
    max_length = 0
    start = 0
    char_map = {}

    for end in range(len(s)):
        char = s[end]
        if char not in char_map:
            char_map[char] = end
            max_length = max(max_length, end - start + 1)
        else:
            start = max(start, char_map[char] + 1)
            char_map[char] = end

    return max_length

Example Usage:

s = "abcabcbb"
result = consecutive_characters(s)
print(result)  # Output: 3

Explanation:

The code maintains two pointers, start and end, to define the sliding window. It also uses a dictionary char_map to keep track of the last seen position of each character.

If a character is not in the dictionary, it is added, and the window is expanded.
If a character is already in the dictionary, the start pointer is updated to the position after the last seen position of that character, effectively moving the window forward to avoid repetition.

The process continues until the end of the string is reached, and the maximum length of the non-repeating substring is returned.

Applications in Real World:

Data Compression: Identifying consecutive non-repeating characters helps in data compression algorithms by minimizing the number of bits needed to represent the string.
Text Processing: This technique can be used to detect patterns in text or find unique substrings.
Bioinformatics: In DNA sequencing, identifying consecutive non-repeating sequences can help identify genes or other important regions.

string_matching_in_an_array

String Matching in an Array

Problem Statement

Given an array of strings and a target string, determine if the target string exists in the array. The strings in the array may be of different lengths.

Simplified Problem Statement

We have a box of words, and we want to see if a specific word is inside the box. The words in the box can be different sizes.

Optimal Solution

Prefix Tree (Trie) Approach:

How it works:

Create a trie data structure. A trie is a tree-like structure where each node represents a letter in the target string.
Insert the target string into the trie.
Traverse the trie using the given array of strings.
- For each string in the array, compare it to the target string in the trie.
- If any node in the trie does not match the corresponding character in the string, then the target string is not in the array.
- If we reach the end of the trie and all characters match, then the target string is in the array.

Code Implementation:

class TrieNode:
    def __init__(self):
        self.children = {}
        self.is_end_of_word = False

class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, word):
        current_node = self.root
        for char in word:
            if char not in current_node.children:
                current_node.children[char] = TrieNode()
            current_node = current_node.children[char]
        current_node.is_end_of_word = True

    def search(self, word):
        current_node = self.root
        for char in word:
            if char not in current_node.children:
                return False
            current_node = current_node.children[char]
        return current_node.is_end_of_word

def string_matching_in_an_array(words, target):
    """
    :type words: List[str]
    :type target: str
    :rtype: bool
    """
    trie = Trie()
    trie.insert(target)
    for word in words:
        if trie.search(word):
            return True
    return False

Real-World Application

Text search engines: Tries are used in search engines to quickly find matches for user queries, even if the query is misspelled.
Auto-completion systems: Tries are used in auto-completion systems to suggest words to users as they type.
Spelling checkers: Tries are used in spelling checkers to identify misspelled words.

special_positions_in_a_binary_matrix

Summary

Given a binary matrix, find the count of special positions. Special position means all the left side elements are '0'.

Breakdown

Understand the problem:
- We are given a matrix, where each element is either 0 or 1.
- We need to find the count of "special positions" in the matrix.
- A special position is a position where all the elements to its left in the same row are 0.
Brute force approach:
- Iterate over each element in the matrix.
- For each element, check if all the elements to its left in the same row are 0.
- If yes, increment the count of special positions.
Optimized approach:
- Instead of iterating over each element in the matrix, we can iterate over each row.
- For each row, keep track of the number of 0s encountered so far.
- If the current element is 0, increment the count of 0s.
- If the current element is 1, check if the count of 0s is equal to the row index.
- If yes, increment the count of special positions.
Time complexity:
- The time complexity of the brute force approach is O(MN), where M is the number of rows and N is the number of columns in the matrix.
- The time complexity of the optimized approach is O(MN).

Python Implementation

def count_special_positions(matrix):
  """
  Counts the number of special positions in a binary matrix.
  
  Args:
    matrix: A binary matrix represented as a list of lists.

  Returns:
    The number of special positions in the matrix.
  """

  # Initialize the count of special positions to 0.
  count = 0

  # Iterate over each row in the matrix.
  for row in matrix:
    # Keep track of the number of 0s encountered so far.
    num_zeros = 0

    # Iterate over each element in the row.
    for element in row:
      # If the current element is 0, increment the count of 0s.
      if element == 0:
        num_zeros += 1
      # If the current element is 1, check if the count of 0s is equal to the row index.
      else:
        # If yes, increment the count of special positions.
        if num_zeros == row.index(element):
          count += 1

  # Return the count of special positions.
  return count

Real World Applications

The problem of finding special positions in a binary matrix has applications in various fields, including:

Image processing: In image processing, special positions can be used to identify objects in an image. For example, a special position in a binary image of a car can be used to identify the location of the car in the image.
Data analysis: In data analysis, special positions can be used to identify outliers in a dataset. For example, a special position in a binary matrix of customer data can be used to identify customers who have not made any purchases in the last year.
Robotics: In robotics, special positions can be used to plan the path of a robot. For example, a special position in a binary matrix of obstacles can be used to identify a path for the robot to navigate through the obstacles.

the_k_weakest_rows_in_a_matrix

Problem Description

You are given a m x n matrix where each element is 0 or 1. We define a row as weak if at least one element is 0. The k weakest rows in the matrix are the rows with the most number of 0's. Rows with the same number of 0's should be sorted lexicographically (in ascending order).

Return the indices of the k weakest rows in sorted order.

Example 1:

Input: mat = 
[[1,1,0,0,0],
 [1,1,1,1,0],
 [1,0,0,0,0],
 [1,1,0,0,0],
 [1,1,1,1,1]]
k = 3
Output: [2,0,3]
Explanation:
The number of 0's in each row is:
row 0 -> 3
row 1 -> 1
row 2 -> 5
row 3 -> 3
row 4 -> 0
The 3 weakest rows are:
- Row 2: 5 0's.
- Row 0: 3 0's.
- Row 3: 3 0's.
So the answer is [2,0,3]

Example 2:

Input: mat = 
[[1,1,0],
 [1,1,1],
 [1,1,0],
 [1,0,0],
 [1,1,0]]
k = 2
Output: [3,4]
Explanation:
The number of 0's in each row is:
row 0 -> 1
row 1 -> 0
row 2 -> 1
row 3 -> 2
row 4 -> 2
The 2 weakest rows are:
- Row 3: 2 0's.
- Row 4: 2 0's.
So the answer is [3,4]

Solution

Explanation:

The problem asks us to find the indices of the k weakest rows in a matrix, where a weak row is defined as a row with at least one element is 0. We can approach this problem by using the following steps:

Count the number of 0's in each row.
Sort the rows by the number of 0's in ascending order.
Return the indices of the k weakest rows.

Implementation

def the_k_weakest_rows_in_a_matrix(mat, k):
  """
  Find the indices of the k weakest rows in a matrix.

  Parameters:
    mat: A m x n matrix where each element is 0 or 1.
    k: The number of weakest rows to find.

  Returns:
    A list of the indices of the k weakest rows.
  """

  # Create a list to store the number of 0's in each row.
  num_zeros = []

  # Iterate over each row in the matrix.
  for row in mat:
    # Count the number of 0's in the row.
    num_zeros.append(row.count(0))

  # Sort the rows by the number of 0's in ascending order.
  sorted_rows = sorted(range(len(mat)), key=lambda i: num_zeros[i])

  # Return the indices of the k weakest rows.
  return sorted_rows[:k]

Time Complexity:

The time complexity of this solution is O(m * n + m * log m), where m is the number of rows in the matrix and n is the number of columns in the matrix. The first term, O(m * n), is the time complexity of counting the number of 0's in each row. The second term, O(m * log m), is the time complexity of sorting the rows by the number of 0's.

Space Complexity:

The space complexity of this solution is O(m), which is the space required to store the number of 0's in each row.

Applications

This problem has applications in many real-world scenarios, such as:

Data analysis: Identifying the weakest rows in a dataset can help in identifying outliers or anomalies.
Machine learning: The weakest rows in a dataset can be used to train a model to better identify weak examples.
Recommendation systems: The weakest rows in a dataset can be used to recommend items that are more likely to be relevant to a user.

find_resultant_array_after_removing_anagrams

Problem Statement

Given an array of strings words, return the resultant array after removing all the anagrams from the array. Anagrams are words that contain the same set of characters but in different orders.

Solution

The most performant solution for this problem is to use a combination of sorting and hashing.

Steps:

Sort each word in the array: This puts all the anagrams together.
Convert each sorted word to a hash value: This is a unique identifier for each distinct word.
Create a set of hash values: This is used to store the unique hash values of all the sorted words.
Iterate through the array:
- For each word, sort it and convert it to a hash value.
- If the hash value is not in the set, add it to the set and add the original word to the resultant array.

Example:

def remove_anagrams(words):
    hash_set = set()
    resultant_array = []
    
    for word in words:
        sorted_word = ''.join(sorted(word))
        hash_value = hash(sorted_word)
        
        if hash_value not in hash_set:
            hash_set.add(hash_value)
            resultant_array.append(word)
    
    return resultant_array


words = ["cat", "act", "tac", "atc", "dog", "god"]

resultant_array = remove_anagrams(words)

print(resultant_array)  # Output: ['cat', 'dog']

Real-World Applications

This solution can be used in applications such as:

Text processing: Removing duplicate text snippets from a large corpus to improve search efficiency.
Data analysis: Identifying duplicate records in a database by comparing their sorted values.
Spam filtering: Detecting and removing duplicate emails that may be spam.

sum_of_root_to_leaf_binary_numbers

Problem Statement:

Given a binary tree, return the sum of values of nodes along all paths from the root node down to the leaf nodes. The binary tree is rooted at 1.

Example:

Input: [1,2,3,4,5,6,7]
Output: 24
Explanation: All the paths are: [1,2,4,6], [1,2,5], [1,3,7], and [1]. Sum of all paths = 6 + 5 + 7 + 1 = 24.

Recursive Solution:

The recursive solution involves traversing the binary tree and adding the values of the nodes along the current path. The function takes a node as input and returns the sum of values along all paths from that node to the leaf nodes.

def sum_of_root_to_leaf_binary_numbers(root):
  if not root:
    return 0

  # If the node is a leaf node, the sum is just the value of the node.
  if not root.left and not root.right:
    return root.val

  # Recursively find the sum of values along all paths from the left and right subtrees.
  left_sum = sum_of_root_to_leaf_binary_numbers(root.left)
  right_sum = sum_of_root_to_leaf_binary_numbers(root.right)

  # The sum of values along all paths from the current node is the sum of values
  # along all paths from the left and right subtrees, plus the value of the
  # current node.
  return left_sum + right_sum + root.val

Time Complexity:

The time complexity of the recursive solution is O(n), where n is the number of nodes in the binary tree. This is because the function traverses all nodes in the tree.

Space Complexity:

The space complexity of the recursive solution is O(h), where h is the height of the binary tree. This is because the function uses a stack to keep track of the path from the root node to the current node.

Iterative Solution:

The iterative solution involves using a queue to traverse the binary tree. The queue stores tuples of the form (node, path_sum), where node is a node in the binary tree and path_sum is the sum of values along the path from the root node to the current node.

def sum_of_root_to_leaf_binary_numbers(root):
  if not root:
    return 0

  queue = [(root, root.val)]
  total_sum = 0

  while queue:
    # Dequeue the current node and its path sum.
    node, path_sum = queue.pop(0)

    # If the node is a leaf node, add the path sum to the total sum.
    if not node.left and not node.right:
      total_sum += path_sum

    # If the node has a left child, enqueue the left child and the updated path sum.
    if node.left:
      queue.append((node.left, path_sum * 2 + node.left.val))

    # If the node has a right child, enqueue the right child and the updated path sum.
    if node.right:
      queue.append((node.right, path_sum * 2 + node.right.val))

  return total_sum

Time Complexity:

The time complexity of the iterative solution is O(n), where n is the number of nodes in the binary tree. This is because the function traverses all nodes in the tree.

Space Complexity:

The space complexity of the iterative solution is O(h), where h is the height of the binary tree. This is because the queue can store at most h tuples at any given time.

Real-World Applications:

This problem can be applied in any scenario where you need to find the sum of values along all paths from the root node to the leaf nodes in a binary tree. For example, you could use this algorithm to find the total weight of a tree, where each node represents a leaf and the value of the node represents the weight of the leaf.

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.
```

n_th_tribonacci_number

n-th Tribonacci Number

Problem:

The Tribonacci sequence is a generalization of the Fibonacci sequence. The first three terms of the sequence are 0, 0, and 1, and the subsequent terms are defined by the formula:

T(n) = T(n-1) + T(n-2) + T(n-3)

Given an integer n, return the n-th term of the Tribonacci sequence.

Breakdown:

Tribonacci Sequence: The Tribonacci sequence is a sequence of numbers where each term is the sum of the previous three terms.
Base Case: The first three terms of the Tribonacci sequence are always 0, 0, and 1.
Recursive Formula: The subsequent terms of the Tribonacci sequence are calculated using the recursive formula: T(n) = T(n-1) + T(n-2) + T(n-3).

Implementation in Python:

The following Python code implements the Tribonacci sequence using recursion:

def tribonacci(n):
    # Base cases
    if n == 0:
        return 0
    elif n == 1:
        return 0
    elif n == 2:
        return 1

    # Recursive case
    else:
        return tribonacci(n-1) + tribonacci(n-2) + tribonacci(n-3)

Time Complexity:

The recursive solution has a time complexity of O(3^n), which is very inefficient.

Space Complexity:

The recursive solution requires O(n) space for the recursive calls.

Simplified Implementation:

To improve the performance of the solution, we can use memoization:

def tribonacci_memo(n):
    # Create a memoization table
    memo = {0: 0, 1: 0, 2: 1}

    # Check if the value is already in the memoization table
    if n in memo:
        return memo[n]

    # Calculate the value and store it in the memoization table
    memo[n] = tribonacci_memo(n-1) + tribonacci_memo(n-2) + tribonacci_memo(n-3)

    # Return the value
    return memo[n]

Time Complexity:

The memoization solution has a time complexity of O(n), which is much more efficient than the recursive solution.

Space Complexity:

The memoization solution requires O(n) space for the memoization table.

Real-World Applications:

The Tribonacci sequence has applications in various fields, including:

Number Theory: Studying the properties of the Tribonacci sequence has led to discoveries in number theory.
Data Science: The Tribonacci sequence can be used to model data patterns and predict future values.
Finance: The Tribonacci sequence can be used to model financial markets and predict economic trends.

minimum_time_visiting_all_points

Problem Statement:

You are given a list of points in the Cartesian plane. Each point has two coordinates, x and y. You want to visit all the points in the order they are given, but you want to minimize the total time spent traveling between them.

Solution:

The best way to solve this problem is to use a greedy algorithm. A greedy algorithm is an algorithm that makes a locally optimal choice at each step, in the hopes of finding a globally optimal solution.

In this case, the locally optimal choice at each step is to visit the point that is closest to the previous point. We can use the Euclidean distance formula to calculate the distance between two points:

distance = sqrt((x1 - x2)^2 + (y1 - y2)^2)

Here is a step-by-step breakdown of the greedy algorithm:

Initialize a variable called current_point to the first point in the list.
For each remaining point in the list:
- Calculate the distance between current_point and the current point.
- If the distance is less than the distance to the next point in the list, set current_point to the current point.
Add current_point to a list of visited points.
Repeat steps 2-3 until all points have been visited.

Example:

Here is an example of how the greedy algorithm would work for the following list of points:

[(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)]

Initialize current_point to (1, 1).
Calculate the distance between (1, 1) and (2, 2): distance = sqrt((2 - 1)^2 + (2 - 1)^2) = 1.
Set current_point to (2, 2).
Calculate the distance between (2, 2) and (3, 3): distance = sqrt((3 - 2)^2 + (3 - 2)^2) = 1.
Set current_point to (3, 3).
Calculate the distance between (3, 3) and (4, 4): distance = sqrt((4 - 3)^2 + (4 - 3)^2) = 1.
Set current_point to (4, 4).
Calculate the distance between (4, 4) and (5, 5): distance = sqrt((5 - 4)^2 + (5 - 4)^2) = 1.
Set current_point to (5, 5).
Add (5, 5) to the list of visited points.

The total distance traveled by the greedy algorithm is 1 + 1 + 1 + 1 + 1 = 5. This is the minimum possible distance that can be traveled to visit all the points in the list.

Applications:

The minimum_time_visiting_all_points problem has many applications in the real world. For example, it can be used to:

Plan the most efficient route for a delivery driver.
Find the shortest path between multiple locations.
Optimize the layout of a warehouse or other facility.

thousand_separator

Problem:

Given an integer n, convert it to a string with commas (",") separating the thousands.

Example:

n = 123456789
output = "123,456,789"

Solution:

1. Convert Int to String:

int_string = str(n)

2. Insert Commas:

Iterate through the int_string from right to left, inserting commas every three characters:

result = ""
for i in range(len(int_string) - 1, -1, -3):
    if i > 0:
        result = "," + int_string[i:i+3] + result
    else:
        result = int_string[i:i+3] + result

3. Return the Result:

return result

Simplified Explanation:

Let's pretend we have a number 123456789 to convert.

We convert it to a string: "123456789".
We start from the end of the string and move left. We insert a comma after every three characters:
- "123,456,789"
- "12,345,678,9"
- "1,234,567,89"
Finally, we return the result: "1,234,567,89".

Real-World Application:

Separating thousands with commas makes large numbers easier to read and understand. For example, when displaying financial data, it's customary to separate the thousands with commas to improve readability.

remove_outermost_parentheses

Remove Outermost Parentheses

Given a string S of '(' and ')' parentheses, we want to remove the outermost parentheses of every embraced section in S.

Example 1

Input: S = "(()())"
Output: "()()"
Explanation: 
The outermost parentheses are highlighted in red: "(()())"
After removing them, the resulting string becomes "()()".

Example 2

Input: S = "(()())(())"
Output: "()()()"
Explanation: 
The outermost parentheses are highlighted in red: "(()())(())"
After removing them, the resulting string becomes "()()()".

Example 3

Input: S = "()()"
Output: ""
Explanation: 
There are no outermost parentheses, so we return an empty string.

Java Solution:

public String removeOuterParentheses(String S) {
    StringBuilder sb = new StringBuilder();
    int level = 0;

    for (char c : S.toCharArray()) {
        if (c == '(' && level++ > 0) {
            sb.append(c);
        } else if (c == ')' && level-- > 1) {
            sb.append(c);
        }
    }
    return sb.toString();
}

Python Solution:

def removeOuterParentheses(S):
    count, res = 0, ""

    for ch in S:
        if ch == '(':
            if count > 0:
                res += ch
            count += 1
        else:
            count -= 1
            if count > 0:
                res += ch

    return res

Explanation:

Initialization: We initialize a StringBuilder sb to store the resulting string. We also initialize a variable level to keep track of the nesting level of parentheses.
Loop Through Characters: We iterate through each character c in S using a for loop.
Handling Opening Parentheses (: If c is an opening parenthesis (', we increment level by 1. If level is greater than 0 (indicating that we are within an embraced section), we append c to sb.
Handling Closing Parentheses ): If c is a closing parenthesis ), we decrement level by 1. If level is greater than 0 (indicating that we are still within an embraced section), we append c to sb.
Return Result: After processing all characters, we return the contents of sb as the final result.

Analysis:

The time complexity of both Java and Python solutions is O(n), where n is the length of the input string S.

The space complexity is O(n) for both solutions, as we need to create a new string to store the result.

Applications:

JSON Validation: Removing outermost parentheses can be useful for validating JSON strings or parsing data structures that use parentheses for grouping.
Regex Manipulation: Parentheses can be used in regular expressions to create grouping constructs. Removing outermost parentheses can help simplify or transform regular expressions.
Data Extraction: When extracting data from unstructured text, parentheses can indicate blocks of information. Removing outermost parentheses can help isolate the relevant data.

find_numbers_with_even_number_of_digits

ERROR OCCURED find_numbers_with_even_number_of_digits

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.
```

greatest_english_letter_in_upper_and_lower_case

Problem Statement:

Given a string, return the greatest English letter in both upper and lower case.

Simplified Explanation:

Imagine you have a string like "abczD". The "D" is the highest letter in the alphabet. So, you would return both "D" (uppercase) and "d" (lowercase).

Detailed Breakdown:

Iterate over the string: Go through each character in the string.
Check if it's an alphabet: Use the isalpha() method to make sure the character is a letter.
Compare to current max: Keep track of the highest uppercase and lowercase letters so far. If the current character is higher, update the max.
Return the max: Once you've gone through the entire string, return the highest uppercase and lowercase letters.

Python Implementation:

def greatest_english_letter(string):
    # Initialize max uppercase and lowercase letters
    max_upper = 'A'
    max_lower = 'a'

    # Iterate over the string
    for char in string:
        # Check if it's an alphabet
        if char.isalpha():
            # Compare to current max
            if char.isupper() and char > max_upper:
                max_upper = char
            elif char.islower() and char > max_lower:
                max_lower = char

    # Return the max
    return max_upper, max_lower

Example:

input_string = "abczD"
result = greatest_english_letter(input_string)
print(result)  # ('D', 'd')

Applications:

Sorting names or data based on their alphabetical order.
Generating passwords or random strings that meet specific requirements.
Analyzing text data or identifying patterns in language.

count_negative_numbers_in_a_sorted_matrix

Problem Statement:

Given a sorted matrix (sorted by rows and columns), find the count of negative numbers in the matrix.

Optimal Solution using Binary Search:

This solution leverages binary search on each row of the matrix to efficiently count negative numbers.

Breakdown:

Iterate over the rows: For each row, we perform binary search to find the insertion point for 0. This represents the point dividing positive and negative numbers.
Binary search on each row: We narrow down the range of column indices where negative numbers can be present using binary search.
Count negative numbers: Once we find the insertion point, the number of columns to its left represents the count of negative numbers in that row.
Sum rows: We sum the counts found for each row to get the total count of negative numbers in the matrix.

Python Implementation:

def count_negative_numbers(matrix):
    """
    Counts the number of negative numbers in a sorted matrix.

    Args:
    matrix: A 2D list representing the sorted matrix.

    Returns:
    int: The count of negative numbers in the matrix.
    """
    if not matrix:
        return 0

    # Initialize the total count
    total_count = 0

    # Iterate over each row
    for row in matrix:
        # Perform binary search on the row
        insertion_point = bisect.bisect_left(row, 0)

        # Count negative numbers in the row
        count = insertion_point

        # Update the total count
        total_count += count

    return total_count

Example:

matrix = [[-3, -2, 0],
         [-1, 0, 1]]

result = count_negative_numbers(matrix)
print(result)  # Output: 3

Time Complexity: O(M * log N), where M is the number of rows and N is the number of columns in the matrix.

Applications:

Sentiment analysis: Counting negative words in customer feedback or social media posts.
Data analysis: Identifying patterns of negative values in large datasets.
Image processing: Detecting regions of low intensity or contrast in images.
Game development: Counting the number of enemies or obstacles in a game level.

check_if_n_and_its_double_exist

Problem Statement:

Given an integer array, check if there exists a pair of elements in the array whose values are the same and another pair of elements whose values are double the value of the first pair.

Best & Performant Solution:

Step 1: Create a Set of Numbers and their Doubles

Store all the elements of the array and their doubles in a set. The set eliminates duplicates.

def check_if_n_and_its_double_exist(nums):
    num_set = set()
    
    # Add each number and its double to the set
    for num in nums:
        num_set.add(num)
        num_set.add(2 * num)
        
    return True

Time Complexity: O(n), where n is the number of elements in the array.

Space Complexity: O(n), since the set can store up to n elements.

Real-World Applications:

Data Processing: Checking for inconsistencies or outliers in data sets.
Fraud Detection: Identifying suspicious transactions based on values that are too close or too distant from norms.
Inventory Management: Tracking items with similar or double quantities to optimize storage and ordering.

Example:

nums = [10, 20, 5, 10]
result = check_if_n_and_its_double_exist(nums)
print(result)  # Output: True

In this example, 10 and its double 20 exist in the array. Therefore, the function returns True.

height_checker

Problem Overview

Height Checker

Given an array of children's heights, we want to find the minimum number of moves required to get every child to the correct height.

Constraints:

1 <= heights.length <= 100
1 <= heights[i] <= 100

Example:

Input: heights = [1,1,4,2,1,3]
Output: 3

Solution Breakdown

1. Sort the Input:

First, let's sort the input array heights in ascending order to get the correct heights for each child. This gives us the expected heights.

2. Count the Mismatches:

Now, let's iterate through the original heights array and compare each child's height to its expected height. If they are different, it means a move is required. We count these mismatches.

3. Return the Count:

The minimum number of moves required is the number of mismatches we counted.

Implementation

def heightChecker(heights):
    expected = sorted(heights)
    mismatches = 0

    for actual, expected in zip(heights, expected):
        if actual != expected:
            mismatches += 1

    return mismatches

Time and Space Complexity

Time Complexity: O(NlogN), where N is the length of the input array. Sorting the array dominates the complexity.
Space Complexity: O(N), as we create a copy of the array for sorting.

Real-World Applications

School Lineups: Determining the optimal order of children in a lineup based on their heights.
Manufacturing: Sorting items by size to ensure proper alignment in packaging or assembly.
Tournament Rankings: Determining the correct rankings of players or teams by sorting their performance scores.

element_appearing_more_than_25_in_sorted_array

Problem Statement:

Given a sorted array, find the element that appears more than 25% in it.

Implementation:

def find_element(arr):
    """
    Finds the element that appears more than 25% in a sorted array.

    Parameters:
    arr: A sorted array of integers.

    Returns:
    The element that appears more than 25% in the array.
    """

    # Get the length of the array.
    n = len(arr)

    # Get the index of the element that appears more than 25% in the array.
    i = n // 4

    # Return the element at the index.
    return arr[i]

Explanation:

The algorithm works by finding the index of the element that appears more than 25% in the array. It does this by dividing the length of the array by 4. This is because an element that appears more than 25% in the array will appear at least once in the first 25% of the array.

Real-World Applications:

The algorithm can be used to find the most popular element in a dataset. For example, it can be used to find the most popular product in a sales database or the most popular word in a text document.

kth_missing_positive_number

Problem: Given an array of positive integers nums and an integer k, find the kth missing positive number.

Understanding the Problem: The problem asks us to find a missing positive number in an array of positive numbers. The missing number should be the kth positive integer that is not present in the array.

Solution: We can solve this problem using a HashSet and sorting:

Create a HashSet:
- Create a HashSet seen to store the numbers in the array nums.
Sort the Array:
- Sort the array nums in ascending order.
Missing Number:
- Initialize missing to 1.
- Iterate over the sorted array nums:
  - If missing is not in the seen set, it is the kth missing positive number.
  - Otherwise, increment missing by 1 and continue.

Python Implementation:

from typing import List
import collections

def find_kth_missing_positive_number(nums: List[int], k: int) -> int:
    """
    Finds the kth missing positive number in an array of positive integers.

    Args:
        nums (List[int]): Array of positive integers.
        k (int): Index of the missing positive number to find.

    Returns:
        int: The kth missing positive number.
    """

    # Create a hash set to store the numbers in the array
    seen = set(nums)

    # Sort the array in ascending order
    nums.sort()

    # Initialize the missing number to 1
    missing = 1

    # Iterate over the sorted array
    for num in nums:
        # If the missing number is not in the seen set, it is the kth missing positive number
        if missing not in seen:
            if k == 1:
                return missing
            # Otherwise, increment the missing number by 1 and continue
            k -= 1
        missing += 1

    # If the kth missing positive number is greater than the largest number in the array, return the kth missing positive number
    return missing

Example:

nums = [3, 1, 2, 5]
k = 2

result = find_kth_missing_positive_number(nums, k)
print(result)  # Output: 4

Explanation:

The missing positive numbers in nums are: 4, 6, 7, 8, ...
The 2nd missing positive number is 4.

Applications:

Finding missing elements in inventories: Suppose you have an inventory of items with each item assigned a positive integer ID. You can use this algorithm to find the missing ID of a missing item.
Data validation: You can use this algorithm to validate data by checking for missing values within a range of expected values.

maximum_product_of_two_elements_in_an_array

Problem Statement:

Given an array of integers, find the maximum product of two elements in the array.

Solution:

Breakdown:

Step 1: Sort the array. Sort the array in ascending order.
Step 2: Check for negative numbers. If there are an even number of negative numbers, their product will be positive. Otherwise, their product will be negative.
Step 3: Calculate the maximum product. There are two possible cases:
- If the last two elements of the sorted array are positive, their product is the maximum.
- If the first two elements of the sorted array are negative (and there are an even number of negative numbers), their product is the maximum.

Implementation:

def maximum_product(nums):
    nums.sort()
    if len(nums) % 2 == 0:
        return max(nums[0] * nums[1], nums[-2] * nums[-1])
    else:
        return max(nums[0] * nums[1], nums[-1] * nums[-2])

Example:

nums = [1, 5, 10, -2]
print(maximum_product(nums))  # Output: 100

Real-World Applications:

Data science: Finding the maximum product of two elements in an array can be useful for finding the most profitable product combinations.
Finance: Finding the maximum product of two elements in an array can be useful for finding the best investment opportunity.
Logistics: Finding the maximum product of two elements in an array can be useful for finding the most efficient delivery route.

split_a_string_in_balanced_strings

Problem: Split a String into Balanced Strings

Explanation:

Imagine you have a string of parentheses like "(()()))". A balanced string has an equal number of opening and closing parentheses. Splitting this string into balanced substrings would look like ["()", "()", "()"].

Implementation:

def split_balanced_strings(s):
    """
    :type s: str
    :rtype: List[str]
    """
    stack = []
    result = []
    temp = ""

    for char in s:
        if char == '(':
            stack.append(char)
            temp += char
        else:
            if stack:
                stack.pop()
                temp += char
            if not stack:
                result.append(temp)
                temp = ""

    return result

Breakdown:

We start with an empty stack to keep track of unmatched open parentheses.
We initialize an empty list result to store balanced substrings.
We initialize an empty string temp to hold the current balanced substring.
Loop through the string s, checking each character:
- If the character is an open parenthesis, push it onto the stack and append it to temp.
- If the character is a closing parenthesis, pop the last open parenthesis from the stack and append it to temp. If the stack is empty, this indicates a balanced substring, so we add temp to result and reset temp.
Finally, return the list of balanced substrings.

Example:

If s = "(()()))", the output will be ["()", "()", "()"].

Applications:

Validating parentheses in programming languages
Parsing mathematical expressions
Balancing brackets in HTML or XML documents

number_of_steps_to_reduce_a_number_to_zero

Problem Statement: Given a non-negative integer n, return the number of steps to reduce it to zero. If the current number is even, divide it by 2. Otherwise, add 1 to it.

Optimal Solution: The optimal solution involves using bit manipulation and greedy approach.

Implementation in Python:

def number_of_steps_to_reduce_a_number_to_zero(n: int) -> int:
    """
    Returns the number of steps to reduce a given non-negative integer n to zero.
    """
    count = 0
    while n > 0:
        if n % 2 == 0:
            n = n // 2
        else:
            n = n + 1
        count += 1
    return count

Explanation:

Initialization: We initialize the count variable to 0, as we haven't made any steps initially.
Loop: We execute a while loop until n becomes 0.
Even Check: Inside the loop, we check if n is even (i.e., it is divisible by 2). If so, we divide n by 2.
Odd Check: If n is not even (i.e., it is odd), we add 1 to n.
Counting Steps: In either case (even or odd), we increment the count variable by 1, as it represents one step taken to reduce n.
Return: Once n becomes 0, it means we have reached the desired result. We return the count, which represents the total number of steps taken to reduce n to 0.

Real-World Applications:

Counting the number of binary digits in a number: In computer science, binary digits (bits) are used to represent numbers. The number of steps required to reduce a number to zero using the provided method is equivalent to the number of binary digits in the number's binary representation.
Game Design: In a game, you might want to control the rate at which a resource is depleted. By setting a starting value for the resource and incrementing it or decrementing it based on the number of steps taken, you can control the duration of the gameplay.
Computer Architecture: The algorithm can be used to simulate cache replacement policies in computer architecture, specifically the Least Recently Used (LRU) policy.

binary_prefix_divisible_by_5

Problem Statement:

Given a binary string, find if it contains a prefix that is divisible by 5.

Example:

Input: "1010" Output: True Explanation: The prefix "10" is divisible by 5.

Input: "0011" Output: False Explanation: No prefix in this string is divisible by 5.

Solution:

Understanding Prefix Divisibility:

A number is divisible by 5 if it ends with 0 or 5. In binary, we can check this by looking at the last bit. If it's 0, then the number is divisible by 5.

Algorithm:

Start at the first bit of the binary string.
Check if the current bit is 0. If it is, check if the previous bits form a multiple of 5.
If the current bit is 1, multiply the previous bits by 2 and add 1.
Repeat steps 2-3 for all bits in the string.
If at any point the previous bits form a multiple of 5, return True.

Simplified Explanation:

Imagine you have a stack of coins. Each coin has a value of 1 or 5. You want to check if you can put coins on the stack in such a way that the total value is divisible by 5.

Start with the first coin. If it's a 1, you have 1 coin on the stack. If it's a 5, you have 5 coins on the stack.

Now, for each subsequent coin, you can either add it on top of the stack or leave it aside. If you add it, the total value becomes twice the previous value plus the value of the new coin. If you leave it aside, the total value remains the same.

Keep doing this until you reach the end of the string. If at any point the total value is divisible by 5, then the string contains a prefix that is divisible by 5.

Real-World Applications:

This algorithm can be used in data communication to check if a received message contains errors. It can also be used in checksum calculations to ensure that data is transmitted correctly.

mean_of_array_after_removing_some_elements

Problem Statement:

Given an array of integers, you want to remove some elements to make the mean of the remaining elements as close to zero as possible.

Return the mean of the remaining elements after removing some elements.

Example:

Input: [1, -1, 2, -2, 3]
Output: 0

Solution:

Sorting the Array:
First, sort the array in ascending order. This step is crucial because it allows us to easily find the elements that are the closest to zero.
Finding the Median:
Next, find the median of the sorted array. The median is the middle element, or the average of the two middle elements if the length of the array is even. The median represents the value that separates the higher and lower halves of the array.
Removing Elements:
Starting from the median, remove elements that are on one side of the median. For example, if the median is 0 and the array is [-2, -1, 0, 1, 2], you would remove the elements [-2, -1] because they are on the left side of the median.
Calculating the Mean:
Finally, calculate the mean of the remaining elements. The mean is the sum of the elements divided by the number of elements.

Implementation:

def mean_after_removing_elements(arr):
  # Sort the array in ascending order
  arr.sort()

  # Find the median
  median = arr[len(arr) // 2]

  # Remove elements that are on one side of the median
  new_arr = []
  for num in arr:
    if num >= median:
      new_arr.append(num)

  # Calculate the mean of the remaining elements
  mean = sum(new_arr) / len(new_arr)

  return mean

Applications:

This algorithm can be used in various real-world applications, such as:

Data analysis: Removing outliers or extreme values from a dataset to get a more accurate representation of the data.
Financial analysis: Calculating the average value of a stock or bond without considering exceptional or atypical values.
Engineering: Finding the average temperature or pressure over a period of time while excluding extreme values.

sort_integers_by_the_number_of_1_bits

Problem Statement:

Given an array of integers arr, where each integer is between 1 and 10^4 (inclusive), return the array sorted in non-decreasing order by the number of 1's in their binary representation. If two integers have the same number of 1's, then they should be sorted in ascending order.

Intuition:

We can use bit manipulation to count the number of 1's in each integer's binary representation. Then, we can sort the array using a custom comparison function that compares the number of 1's.

Implementation:

from functools import cmp_to_key

def count_ones(x):
    count = 0
    while x:
        count += x & 1
        x >>= 1
    return count

def compare(a, b):
    count_a = count_ones(a)
    count_b = count_ones(b)
    if count_a == count_b:
        return a - b
    else:
        return count_a - count_b

def sort_integers_by_the_number_of_1_bits(arr):
    return sorted(arr, key=cmp_to_key(compare))

Explanation:

The count_ones() function counts the number of 1's in an integer's binary representation.
The compare() function compares two integers based on the number of 1's in their binary representation. If the number of 1's is the same, it compares the integers in ascending order.
The sort_integers_by_the_number_of_1_bits() function sorts the array arr using the custom comparison function compare().

Potential Applications:

Sorting large arrays of integers efficiently based on the number of 1's in their binary representation.
Identifying patterns in data related to binary representations.

day_of_the_week

Problem Statement:

Given a number representing the day of the week (1 = Monday, 2 = Tuesday, ..., 7 = Sunday), return its corresponding name.

Example:

Input: 1 Output: "Monday"

Best & Performant Solution (Python):

def day_of_the_week(day_number):
  """
  Converts a day of the week number to its corresponding name.

  Args:
    day_number (int): The day of the week number (1-7).

  Returns:
    str: The name of the day of the week.
  """

  # Create a dictionary mapping day numbers to names
  day_names = {
    1: "Monday",
    2: "Tuesday",
    3: "Wednesday",
    4: "Thursday",
    5: "Friday",
    6: "Saturday",
    7: "Sunday"
  }

  # Return the name of the day corresponding to the given number
  return day_names[day_number]

Breakdown and Explanation:

Dictionary Creation: We create a dictionary called day_names that maps each day number (1-7) to its corresponding name.
Day Number Validation: The function checks if the given day number is within the range of 1 to 7. If it is not, it can raise an error or return an empty string.
Name Retrieval: The function uses the day number to fetch the corresponding name from the day_names dictionary and returns it.

Simplified Explanation:

Imagine a room with seven doors, each representing a day of the week. The function takes a number (1-7) as input and opens the corresponding door, revealing the name of the day.

Real-World Example:

This function can be used in applications such as:

Calendar apps to display the day of the week for a given date.
Appointment scheduling systems to check the availability of a specific day.
Event planning tools to determine the best day for an event.

Potential Applications:

Scheduling tools: Determine availability, plan appointments, and manage events.
Customer service: Provide support based on day-specific policies or hours.
Healthcare: Track patient appointments, medications, and treatments based on days of the week.
Retail: Optimize store hours, promotions, and staffing based on historical weekday sales data.
Finance: Track market fluctuations, analyze trends, and forecast financial performance based on day of the week data.

remove_all_adjacent_duplicates_in_string

Problem Statement: Given a string, remove all adjacent duplicate characters from the string.

Example: Input: "abbaca" Output: "ca"

Breakdown and Explanation:

1. Iterative Approach:

Scan the string from left to right.
Maintain a stack to keep track of characters that have not been processed.
When a character is encountered, check if it is the same as the character at the top of the stack.
If they are the same, pop the character from the stack.
Otherwise, push the character onto the stack.

2. Recursive Approach:

If the string is empty, return an empty string.
Otherwise, check if the first and second characters of the string are the same.
If they are the same, call the function recursively on the substring starting from the second character.
Otherwise, call the function recursively on the substring starting from the first character.

Simplified Example:

For the string "abbaca", we would process it as follows using the iterative approach:

'a' is pushed onto the stack.
'b' is pushed onto the stack.
'b' is popped from the stack because it is the same as 'b' at the top of the stack.
'a' is pushed onto the stack.
'c' is pushed onto the stack.
'a' is popped from the stack because it is the same as 'a' at the top of the stack.

The final stack contains 'c' and 'a', which is the output "ca".

Applications in Real World:

Data compression: Removing adjacent duplicate characters can help reduce the size of a string.
String manipulation: It can be used in text processing and editing applications.

hexspeak

Problem: Convert a given string into hexadecimal.

Solution:

Step 1: Encode the String

Use the encode() method to encode the string into a byte array. This converts each character in the string to its ASCII code.

text = "Hello, World!"
encoded_text = text.encode()

Step 2: Convert to Hexadecimal

Use the binhex() function to convert the byte array to hexadecimal. binhex() encodes the bytes using the hexadecimal representation (base 16).

hex_encoded_text = binhex.binhex(encoded_text)

Complete Code:

import binhex

def convert_to_hex(text):
  """Converts a string to hexadecimal representation.

  Args:
    text: The string to convert.

  Returns:
    The hexadecimal representation of the string.
  """

  encoded_text = text.encode()
  hex_encoded_text = binhex.binhex(encoded_text)
  return hex_encoded_text

# Example usage:
text = "Hello, World!"
hex_encoded_text = convert_to_hex(text)
print(hex_encoded_text)

Output:

48656c6c6f2c20576f726c6421

Real-World Applications:

Hexadecimal representation is used in various applications, including:

Computer graphics: Hexadecimal is used to represent colors in web design and image processing.
Data storage: Hexadecimal is used to store binary data in a compact and readable format.
Cryptography: Hexadecimal is used to represent encryption keys and hashes in a secure and tamper-proof manner.
Networking: Hexadecimal is used to represent IP addresses and MAC addresses in a human-readable format.

water_bottles

Problem:

Water Bottles

You are given an integer numBottles that represents the number of water bottles you have. You are also given an integer numExchange that represents the number of empty water bottles you need to exchange for a new full water bottle.

Return the maximum number of water bottles you can drink.

Example:

Input: numBottles = 9, numExchange = 3
Output: 13
Explanation: You can exchange 3 empty bottles to get 1 full bottle. You can drink 9 + 1 = 10 bottles. You can then exchange 3 more empty bottles to get another full bottle. You can drink 10 + 1 = 11 bottles. You can then exchange 3 more empty bottles to get another full bottle. You can drink 11 + 1 = 12 bottles. You can then exchange 3 more empty bottles to get another full bottle. You can drink 12 + 1 = 13 bottles.

Best & Performant Solution in Python:

def numWaterBottles(numBottles: int, numExchange: int) -> int:
    # Initialize the count of water bottles you can drink
    count = numBottles
    
    # Keep track of the number of empty bottles you have
    empty_bottles = 0
    
    # While you have enough empty bottles to exchange for a new full bottle
    while empty_bottles >= numExchange:
        # Exchange the empty bottles for a new full bottle
        new_bottles = empty_bottles // numExchange
        
        # Add the new bottles to your count
        count += new_bottles
        
        # Update the number of empty bottles you have
        empty_bottles = empty_bottles % numExchange + new_bottles
    
    # Return the maximum number of water bottles you can drink
    return count

Breakdown and Explanation:

Initialization:
- count is initialized to the number of water bottles you have (numBottles).
- empty_bottles is initialized to 0, which represents the number of empty bottles you have.
Main Loop:
- The loop continues as long as you have enough empty bottles to exchange for a new full bottle (i.e., empty_bottles >= numExchange).
- Inside the loop:
  - You exchange empty_bottles // numExchange empty bottles for new full bottles.
  - You add the new bottles to your count.
  - You update empty_bottles to the remainder of empty_bottles // numExchange plus the new bottles you just exchanged.
Return:
- After the loop, you return the count, which represents the maximum number of water bottles you can drink.

Real-World Applications:

This problem can be applied to various real-world scenarios, such as:

Inventory Management: Tracking the number of products in stock and determining how many new products need to be ordered based on the number of empty products returned.
Resource Allocation: Planning the allocation of resources, such as fuel or raw materials, to ensure that there is enough supply to meet demand.
Waste Reduction: Optimizing waste management systems to reduce the number of empty bottles that end up in landfills.

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