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1046. Last Stone Weight

We have a collection of stones, each stone has a positive integer weight.

Each turn, we choose the two heaviest stones and smash them together. Suppose the stones have weights x and y with x <= y. The result of this smash is:

If x == y, both stones are totally destroyed; If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x. At the end, there is at most 1 stone left. Return the weight of this stone (or 0 if there are no stones left.)

Example 1:

Input: [2,7,4,1,8,1]
Output: 1
Explanation: 
We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of last stone.

Note:

1 <= stones.length <= 30
1 <= stones[i] <= 1000

Solutions

from heapq import heapify, heappush, heappop
class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        if len(stones) == 0:
            return 0
        h = [-s for s in stones]
        heapify(h)
        while len(h) >= 2:
            x = -heappop(h)
            y = -heappop(h)
            remain = abs(x - y)
            heappush(h, -remain)
        return -h[0]

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