1962. Remove Stones to Minimize the Total

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 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

Solution (Java)

class Solution {
    public int minStoneSum(int[] piles, int k) {
        PriorityQueue<Integer> descendingQueue = new PriorityQueue<>(Collections.reverseOrder());
        int sum = 0;
        int newValue;
        int currentValue;
        int half;
        for (int stones : piles) {
            sum += stones;
            descendingQueue.offer(stones);
        }
        while (k > 0) {
            currentValue = descendingQueue.poll();
            half = currentValue / 2;
            newValue = currentValue - half;
            descendingQueue.offer(newValue);
            sum -= half;
            k--;
        }
        return sum;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).