1046. Last Stone Weight

Easy

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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 smallest possible weight of the left 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

class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        arr = []
        for stone in stones:
            heapq.heappush(arr, (-stone, stone))
        while arr:
            if len(arr) == 1:
                _, stone = heapq.heappop(arr)
                return stone
            _, stone1 = heapq.heappop(arr)
            _, stone2 = heapq.heappop(arr)
            if stone1 == stone2:
                pass
            elif stone1 != stone2:
                left = abs(stone1-stone2)
                heapq.heappush(arr, (-left, left))
        return 0