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 <= 1051 <= piles[i] <= 1041 <= k <= 105
[Array] [Heap (Priority Queue)]