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

Note:

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

Solution:

```class Solution {
public int lastStoneWeight(int[] stones) {
PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
for (int a : stones)
pq.offer(a);
while (pq.size() > 1)
pq.offer(pq.poll() - pq.poll());
return pq.poll();
}
}```