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# Remove Stones to minimize the Total(code and Explanation)

PROBLEM STATEMENT
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.

EXPLANATION
Everytime you want the maximum value from the pile to be divided.So we will use a maxheap to store the elements in
greatest to least order and after dividing we push the resultant value again inside the heap
CODE

``````from heapq import *
class Solution:
def minStoneSum(self, piles: List[int], k: int) -> int:
maxHeap = []
for i in range(len(piles)):
heappush(maxHeap, -piles[i])

i = k
while i > 0:
temp = heappop(maxHeap)
x = floor(temp/2)
heappush(maxHeap,x)
i -= 1
return -sum(maxHeap)
``````

Time Complexity
O(KlogN)
Space Complexity
O(N) for storing N elements in heap

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