Introduction

In this blog post, we will explore how to solve the LeetCode problem "1962" 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 a 0-indexed integer array piles, where piles[i] represents the number of stones in the ith pile, and an integer k. You should apply the following operation exactly k times: Choose any piles[i] and remove floor(piles[i] / 2) stones from it. Notice that you can apply the operation on the same pile more than once. Return the minimum possible total number of stones remaining after applying the k operations. floor(x) is the greatest integer that is smaller than or equal to x (i.e., rounds x down). Example 1: Input: piles = [5,4,9], k = 2 Output: 12 Explanation: Steps of a possible scenario are: - Apply the operation on pile 2. The resulting piles are [5,4,5]. - Apply the operation on pile 0. The resulting piles are [3,4,5]. The total number of stones in [3,4,5] is 12. Example 2: Input: piles = [4,3,6,7], k = 3 Output: 12 Explanation: Steps of a possible scenario are: - Apply the operation on pile 2. The resulting piles are [4,3,3,7]. - Apply the operation on pile 3. The resulting piles are [4,3,3,4]. - Apply the operation on pile 0. The resulting piles are [2,3,3,4]. The total number of stones in [2,3,3,4] is 12. Constraints: 1 <= piles.length <= 105 1 <= piles[i] <= 104 1 <= k <= 105

Explanation

Here's the breakdown of the solution:

• Greedy Approach with Priority Queue: The core idea is to use a greedy approach. At each step, we want to remove the maximum number of stones possible. A priority queue (max heap) helps us efficiently find the pile with the most stones.
• Iterative Reduction: We iterate k times. In each iteration, we extract the largest pile, remove floor(pile / 2) stones, and re-insert the reduced pile back into the priority queue.

• Summation: Finally, we sum the remaining stones in all piles to get the minimum possible total.

• Runtime Complexity: O(n + k * log n), where n is the number of piles. Building the heap takes O(n) time, and performing k heap operations (pop and push) takes O(k * log n) time. Storage Complexity: O(n) for the heap.

Code

    import heapq
import math

def minStoneSum(piles, k):
    """
    Calculates the minimum possible total number of stones remaining after applying k operations.

    Args:
        piles: A list of integers representing the number of stones in each pile.
        k: The number of operations to perform.

    Returns:
        The minimum possible total number of stones remaining.
    """

    # Convert piles to a max heap by negating the values
    heap = [-pile for pile in piles]
    heapq.heapify(heap)

    for _ in range(k):
        largest_pile = -heapq.heappop(heap)
        reduction = math.floor(largest_pile / 2)
        heapq.heappush(heap, -(largest_pile - reduction))

    return -sum(heap)