README_EN.md

1962. Remove Stones to Minimize the Total

中文文档

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

Solutions

Python3

class Solution:
    def minStoneSum(self, piles: List[int], k: int) -> int:
        h = []
        for p in piles:
            heappush(h, -p)
        for _ in range(k):
            p = -heappop(h)
            heappush(h, -((p + 1) >> 1))
        return -sum(h)

Java

class Solution {
    public int minStoneSum(int[] piles, int k) {
        PriorityQueue<Integer> q = new PriorityQueue<>((a, b) -> (b - a));
        for (int p : piles) {
            q.offer(p);
        }
        while (k-- > 0) {
            int p = q.poll();
            q.offer((p + 1) >> 1);
        }
        int ans = 0;
        while (!q.isEmpty()) {
            ans += q.poll();
        }
        return ans;
    }
}

C++

class Solution {
public:
    int minStoneSum(vector<int>& piles, int k) {
        priority_queue<int> q;
        for (int& p : piles) q.push(p);
        while (k--) {
            int p = q.top();
            q.pop();
            q.push((p + 1) >> 1);
        }
        int ans = 0;
        while (!q.empty()) {
            ans += q.top();
            q.pop();
        }
        return ans;
    }
};

Go

func minStoneSum(piles []int, k int) int {
    q := &hp{piles}
    heap.Init(q)
    for k > 0 {
        p := q.pop()
        q.push((p + 1) >> 1)
        k--
    }
    ans := 0
    for q.Len() > 0 {
        ans += q.pop()
    }
    return ans
}

type hp struct{ sort.IntSlice }

func (h hp) Less(i, j int) bool  { return h.IntSlice[i] > h.IntSlice[j] }
func (h *hp) Push(v interface{}) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() interface{} {
	a := h.IntSlice
	v := a[len(a)-1]
	h.IntSlice = a[:len(a)-1]
	return v
}
func (h *hp) push(v int) { heap.Push(h, v) }
func (h *hp) pop() int   { return heap.Pop(h).(int) }

...