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 removefloor(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 <= 10^51 <= piles[i] <= 10^41 <= k <= 10^5
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;
}
};