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

class Solution {

    public int lastStoneWeight(int[] stones) {

        // Max Heap to store stones in descending order

        PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());

        // Add all stones to maxHeap

        for (int stone : stones) {

            maxHeap.add(stone);

        }

        // Keep smashing the two heaviest stones

        while (maxHeap.size() > 1) {

            int first = maxHeap.poll();  // heaviest

            int second = maxHeap.poll(); // second heaviest

            if (first != second) {

                maxHeap.add(first - second); // leftover stone

            }

        }

        // If one stone remains, return it; otherwise, return 0

        return maxHeap.isEmpty() ? 0 : maxHeap.peek();

    }

}