1962. Remove Stones to Minimize the Total

Medium

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 <= 10^5

• 1 <= piles[i] <= 10^4

• 1 <= k <= 10^5

解題

這題要用 Max Heap 或 Priority Queue 做，因為 Go 太麻煩所以寫 C++

class Solution {
public:
    int minStoneSum(vector<int>& piles, int k) {
        priority_queue <int> pq(piles.begin(), piles.end());
        int ans = accumulate(piles.begin(), piles.end(), 0);

        while (k > 0) {
            int t = pq.top();
            pq.pop();
            pq.push(t - t / 2);
            ans -= t / 2;
            k--;
        }

        return ans;
    }
};