## Remove Stones to Minimize the Total solution leetcode

You are given a **0-indexed** integer array `piles`

, where `piles[i]`

represents the number of stones in the `i`

pile, and an integer ^{th}`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: Remove Stones to Minimize the Total solution leetcode**

Input:piles = [5,4,9], k = 2Output:12Explanation: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: Remove Stones to Minimize the Total solution leetcode**

Input:piles = [4,3,6,7], k = 3Output:12Explanation: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: Remove Stones to Minimize the Total solution leetcode**

`1 <= piles.length <= 10`

^{5}`1 <= piles[i] <= 10`

^{4}`1 <= k <= 10`

^{5}

# Solution: Click here

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