Introduction

In this blog post, we will explore how to solve the LeetCode problem "2558" 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

You are given an integer array gifts denoting the number of gifts in various piles. Every second, you do the following: Choose the pile with the maximum number of gifts. If there is more than one pile with the maximum number of gifts, choose any. Reduce the number of gifts in the pile to the floor of the square root of the original number of gifts in the pile. Return the number of gifts remaining after k seconds. Example 1: Input: gifts = [25,64,9,4,100], k = 4 Output: 29 Explanation: The gifts are taken in the following way: - In the first second, the last pile is chosen and 10 gifts are left behind. - Then the second pile is chosen and 8 gifts are left behind. - After that the first pile is chosen and 5 gifts are left behind. - Finally, the last pile is chosen again and 3 gifts are left behind. The final remaining gifts are [5,8,9,4,3], so the total number of gifts remaining is 29. Example 2: Input: gifts = [1,1,1,1], k = 4 Output: 4 Explanation: In this case, regardless which pile you choose, you have to leave behind 1 gift in each pile. That is, you can't take any pile with you. So, the total gifts remaining are 4. Constraints: 1 <= gifts.length <= 103 1 <= gifts[i] <= 109 1 <= k <= 103

Explanation

Here's the solution to the problem:

• Key Idea: Use a max heap (priority queue) to efficiently track the pile with the maximum number of gifts. Each second, extract the maximum, update it, and re-insert it.
• Optimization: Avoid unnecessary heap operations by only updating and re-inserting when the square root is different from the original value.

• Final Sum: After k iterations, sum the remaining gifts in the heap to get the result.

• Complexity:

  • Runtime: O(k log n), where n is the number of piles (gifts.length) and k is the number of seconds.
  • Storage: O(n) to store the heap.

Code

    import heapq
import math

def pickGifts(gifts, k):
    """
    Calculates the number of gifts remaining after k seconds.

    Args:
        gifts: A list of integers representing the number of gifts in each pile.
        k: The number of seconds.

    Returns:
        The total number of gifts remaining.
    """

    # Convert the gifts array into a max heap by negating the values.
    # This is because Python's heapq is a min heap.
    heap = [-gift for gift in gifts]
    heapq.heapify(heap)

    for _ in range(k):
        # Get the pile with the most gifts.
        max_gifts = -heapq.heappop(heap)

        # Calculate the new number of gifts after taking sqrt(max_gifts).
        new_gifts = math.floor(math.sqrt(max_gifts))

        # Only push back if the number of gifts changed
        if new_gifts != max_gifts:
            heapq.heappush(heap, -new_gifts)
        else:
            heapq.heappush(heap, -new_gifts)

    # Calculate the sum of the remaining gifts.
    remaining_gifts = sum([-gift for gift in heap])

    return remaining_gifts