README.md

1962. Remove Stones to Minimize the Total

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 (Rust)

1. Solution

use std::collections::BinaryHeap;

impl Solution {
    pub fn min_stone_sum(piles: Vec<i32>, k: i32) -> i32 {
        let mut piles = BinaryHeap::from(piles);

        for _ in 0..k {
            let x = piles.pop().unwrap();

            piles.push((x + 1) / 2);
        }

        piles.into_iter().sum()
    }
}