1962. Remove Stones to Minimize the Total

Problem Link

Description

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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

Solution

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Python3

class Solution:
    def minStoneSum(self, piles: List[int], k: int) -> int:
        heap = [-x for x in piles]
        heapify(heap)
 
        for _ in range(k):
            x = -heappop(heap)
            x = ceil(x / 2)
 
            heappush(heap, -x)
            
        return sum(-x for x in heap)