# LeetCode 1046. Last Stone Weight

## Description

https://leetcode.com/problems/last-stone-weight/

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` and `y` with `x <= y`.  The result of this smash is:

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

At the end, there is at most 1 stone left.  Return the weight of this stone (or 0 if there are no stones left.)

Example 1:

```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.```

Note:

1. `1 <= stones.length <= 30`
2. `1 <= stones[i] <= 1000`

## Explanation

just simulate the process to find the last stone weight

## Python Solution

``````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).
• Space complexity: O(1).