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 <= y. The result of this smash is:
x == y, both stones are totally destroyed;
x != y, the stone of weight
xis totally destroyed, and the stone of weight
yhas new weight
At the end, there is at most 1 stone left. Return the weight of this stone (or 0 if there are no stones left.)
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  then that's the value of last stone.
1 <= stones.length <= 30
1 <= stones[i] <= 1000
just simulate the process to find the last stone weight
class Solution: def lastStoneWeight(self, stones: List[int]) -> int: while (len(stones) > 1): max_stone = max(stones) stones.remove(max_stone) second_max_stone = max(stones) stones.remove(second_max_stone) new_stone = max_stone - second_max_stone if new_stone != 0: stones.append(new_stone) if (len(stones) == 1): return stones return 0
- Time complexity: O(N^2).
- Space complexity: O(1).