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tags: data_structure_python
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# Last Stone Weight <img src="https://img.shields.io/badge/-easy-brightgreen">

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.)

<ins>**Example 1:**</ins>

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 [1] then that's the value of last stone.

**Note:**

- 1 <= stones.length <= 30
- 1 <= stones[i] <= 1000

# Solution

class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        
        while len(stones) > 1:
            stones = sorted(stones)
            x = stones.pop()
            y = stones.pop()
            
            if x != y:
                stones.append(abs(y-x))
            
        return stones[0] if len(stones) == 1 else 0