## Question

You are given an array of integers `stones`

where `stones[i]`

is the weight of the `i`

stone.^{th}

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:1Explanation: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`

## Python Solution

class Solution: def lastStoneWeight(self, stones: List[int]) -> int: stones.sort(reverse=True) while 1: if len(stones)<=1: break if stones[0]>stones[1]: stones.append(stones[0]-stones[1]) stones.remove(stones[0]) stones.remove(stones[0]) stones.sort(reverse=True) if len(stones)==1: return(stones[0]) else: return(0)