## 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: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 last stone.

**Note:**

`1 <= stones.length <= 30`

`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[0]
return 0
```

- Time complexity: O(N^2).
- Space complexity: O(1).

## One Thought to “LeetCode 1046. Last Stone Weight”