Introduction

In this blog post, we will explore how to solve the LeetCode problem "1103" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

We distribute some number of candies, to a row of n = num_people people in the following way: We then give 1 candy to the first person, 2 candies to the second person, and so on until we give n candies to the last person. Then, we go back to the start of the row, giving n + 1 candies to the first person, n + 2 candies to the second person, and so on until we give 2 * n candies to the last person. This process repeats (with us giving one more candy each time, and moving to the start of the row after we reach the end) until we run out of candies. The last person will receive all of our remaining candies (not necessarily one more than the previous gift). Return an array (of length num_people and sum candies) that represents the final distribution of candies. Example 1: Input: candies = 7, num_people = 4 Output: [1,2,3,1] Explanation: On the first turn, ans[0] += 1, and the array is [1,0,0,0]. On the second turn, ans[1] += 2, and the array is [1,2,0,0]. On the third turn, ans[2] += 3, and the array is [1,2,3,0]. On the fourth turn, ans[3] += 1 (because there is only one candy left), and the final array is [1,2,3,1]. Example 2: Input: candies = 10, num_people = 3 Output: [5,2,3] Explanation: On the first turn, ans[0] += 1, and the array is [1,0,0]. On the second turn, ans[1] += 2, and the array is [1,2,0]. On the third turn, ans[2] += 3, and the array is [1,2,3]. On the fourth turn, ans[0] += 4, and the final array is [5,2,3]. Constraints: 1 <= candies <= 10^9 1 <= num_people <= 1000

Explanation

Here's the breakdown:

• Calculate complete rounds: Determine how many full cycles of distributing candies to all people can be completed. This involves solving a quadratic equation.
• Distribute candies in complete rounds: Calculate the candies distributed in these full cycles and update the result array accordingly.

• Distribute remaining candies: Handle the candies left after the complete rounds by iterating through the people and distributing the appropriate number of candies until none remain.

• Runtime Complexity: O(n), where n is num_people. Storage Complexity: O(n)

Code

    def distributeCandies(candies: int, num_people: int) -> list[int]:
    """
    Distributes candies to people in a row until no candies remain.

    Args:
        candies: The total number of candies to distribute.
        num_people: The number of people in the row.

    Returns:
        A list representing the final distribution of candies to each person.
    """

    result = [0] * num_people
    n = num_people

    # Calculate the number of complete rounds
    rounds = 0
    total_candies_in_rounds = 0

    k = 0
    while True:
      total_candies_in_rounds = (n * (2 * (rounds * n + 1) + (n - 1))) // 2

      if total_candies_in_rounds <= candies:
        rounds += 1
      else:
        rounds -=1
        break

    if rounds > 0:
        total_candies_in_rounds = (n * (2 * ((rounds-1) * n + 1) + (n - 1))) // 2
    else:
        total_candies_in_rounds = 0


    # Distribute candies in complete rounds
    for i in range(n):
        result[i] += rounds * (i + 1) + (n * (rounds * (rounds - 1) // 2))

    remaining_candies = candies - total_candies_in_rounds

    # Distribute remaining candies
    person_index = 0
    candy_to_give = rounds * n + 1
    while remaining_candies > 0:
        if remaining_candies >= candy_to_give:
            result[person_index] += candy_to_give
            remaining_candies -= candy_to_give
        else:
            result[person_index] += remaining_candies
            remaining_candies = 0

        candy_to_give += 1
        person_index = (person_index + 1) % n

    return result