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 <= 105
1 <= piles[i] <= 104
1 <= k <= 105
Solution: Greedy / Heap
Always choose the largest pile to remove.
Time complexity: O(n + klogn)
Space complexity: O(n)
C++
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// Author: Huahua class Solution { public: int minStoneSum(vector<int>& piles, int k) { int total = accumulate(begin(piles), end(piles), 0); priority_queue<int> q(begin(piles), end(piles)); while (k--) { int curr = q.top(); q.pop(); int remove = curr / 2; total -= remove; q.push(curr - remove); } return total; } }; |
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