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# 1046. Last Stone Weight

We have a collection of rocks, each rock has a positive integer weight.

Each turn, we choose the two **heaviest** rocks 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 [1] then that's the value of last stone.
```

**Note:**

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

```cpp
int lastStoneWeight(vector<int>& stones) { // time: O(nlogn); space: O(n)
    priority_queue<int> pq(stones.begin(), stones.end());
    int x, y;
    while (pq.size() > 1) {
        y = pq.top(); pq.pop();
        x = pq.top(); pq.pop();
        if (y > x) pq.push(y - x);
    }
    return pq.empty() ? 0 : pq.top();
}
```


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