problem

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 largest 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

submission

use std::collections::BinaryHeap;

// just use binary heap
impl Solution {
    pub fn min_stone_sum(piles: Vec<i32>, k: i32) -> i32 {
        let mut heap = BinaryHeap::from(piles);
        for _ in 0..k {
            let Some(pile) = heap.pop() else {
                continue;
            };
            heap.push(pile - pile / 2);
        }
        heap.into_iter().sum()
    }
}