Last Stone Weight

You are given an array of integers stones where stones[i] is the weight of the ith stone.

We are playing a game with the stones. On each turn, we choose the heaviest two stones and smash them together. Suppose the heaviest two stones have weights x and y with x <= y. The result of this smash is:

• If x == y, both stones are destroyed, and
• If x != y, the stone of weight x is destroyed, and the stone of weight y has new weight y - x.

At the end of the game, there is at most one stone left.

Return the weight of the last remaining stone. If there are no stones left, return 0.

Example 1:

Input: stones = [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 the last stone.

Example 2:

Input: stones = [1]
Output: 1

Constraints:

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

SOLUTION:

import heapq

class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        heap = []
        for s in stones:
            heapq.heappush(heap, -s)
        while len(heap) > 1:
            y = -heapq.heappop(heap)
            x = -heapq.heappop(heap)
            if x != y:
                heapq.heappush(heap, x - y)
        if len(heap) == 1:
            return -heap[0]
        return 0