You are given an array of integers
stones[i] is the weight of the
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 <= y. The result of this smash is:
x == y, both stones are destroyed, and
x != y, the stone of weight
xis destroyed, and the stone of weight
yhas 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
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  then that's the value of the last stone.
Input: stones =  Output: 1
1 <= stones.length <= 30
1 <= stones[i] <= 1000
class Solution: def lastStoneWeight(self, stones: List[int]) -> int: stones.sort(reverse=True) while 1: if len(stones)<=1: break if stones>stones: stones.append(stones-stones) stones.remove(stones) stones.remove(stones) stones.sort(reverse=True) if len(stones)==1: return(stones) else: return(0)