## 1046. 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`

## Rust Solution

``````struct Solution;

use std::collections::BinaryHeap;

impl Solution {
fn last_stone_weight(stones: Vec<i32>) -> i32 {
let mut pq: BinaryHeap<i32> = BinaryHeap::from(stones);
while let Some(a) = pq.pop() {
if let Some(b) = pq.pop() {
if a - b != 0 {
pq.push(a - b);
}
} else {
return a;
}
}
0
}
}

#[test]
fn test() {
let stones = vec![2, 7, 4, 1, 8, 1];
assert_eq!(Solution::last_stone_weight(stones), 1);
}
``````

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