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 ktimes:
- 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 koperations.
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 3. The resulting piles are [4,3,3,7].
- Apply the operation on pile 4. 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