# 1046. Last Stone Weight

#### Easy

***

You are given an array of integers `stones` where `stones[i]` is the weight of the `ith` stone.

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 smallest possible weight of the left stone*. If there are no stones left, return `0`.

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**Example 1:**

```
Input: stones = [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 the last stone.
```

**Example 2:**

```
Input: stones = [1]
Output: 1
```

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**Constraints:**

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

```python
class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        arr = []
        for stone in stones:
            heapq.heappush(arr, (-stone, stone))
        while arr:
            if len(arr) == 1:
                _, stone = heapq.heappop(arr)
                return stone
            _, stone1 = heapq.heappop(arr)
            _, stone2 = heapq.heappop(arr)
            if stone1 == stone2:
                pass
            elif stone1 != stone2:
                left = abs(stone1-stone2)
                heapq.heappush(arr, (-left, left))
        return 0
```