LeetCode Challenge #1046. Last Stone Weight

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 weight of the last remaining stone. If there are no stones left, return `0`.

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
```

Constraints:

• `1 <= stones.length <= 30`
• `1 <= stones[i] <= 1000`
Video Solution
Java Solution
```				```
class Solution {
public int lastStoneWeight(int[] stones) {
//max priority queue
PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());

for(int ele : stones){ // adding the elements of array to pq
pq.add(ele);
}

while(pq.size()>1){
int max = pq.remove();
int smax = pq.remove();

int nstone = max - smax ;

if(nstone!=0){
pq.add(nstone);
}
}

if(pq.size()==0){
return 0 ;
}else{
return pq.remove();
}
}
}
```
```