Remove Stones to Minimize the Total

Description

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

Code

Time Complexity: $O(n\log n)$, Space Complexity: $O(n)$

class Solution {
public:
    int minStoneSum(vector<int>& piles, int k) {
        priority_queue<int> max_heap;
        int total = 0;
        for(auto& p: piles) {
            max_heap.push(p);
            total += p;
        }
 
        for(int i = 0; i < k; i++) {
            auto n = max_heap.top();
            max_heap.pop();
            total -= n / 2;
            max_heap.push(n - n / 2);
        }
 
        return total;
    }
};

Source

• Remove Stones to Minimize the Total - LeetCode