LastStoneWeight.java

/*
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.
*/ 

class Solution {
    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> maxPQ = new PriorityQueue<>(Collections.reverseOrder());

        for(int stone : stones)
            maxPQ.offer(stone);
        
        int size = maxPQ.size();

        while(size != 0 && size != 1){
            int x = maxPQ.poll();
            int y = maxPQ.poll();
            if(x < y){
                maxPQ.offer(y-x);
            }
            else if(x > y){
                maxPQ.offer(x-y);
            }
            size = maxPQ.size();
        }

        return size == 0 ? 0 : maxPQ.peek();
    }
}