Skip to content

✅ 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

Solution:

class Solution {
    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> maxHeap = new PriorityQueue<Integer>(Collections.reverseOrder());
        for (int i = 0; i < stones.length; i++){
            maxHeap.offer(stones[i]);
        }

        while(maxHeap.size() >= 2){
            int y = maxHeap.poll();
            int x = maxHeap.poll();

            if (x != y){
                int newWeight = y - x;
                maxHeap.offer(newWeight);
            }
        }

        if (maxHeap.size() == 0){
            return 0;
        }else{
            return maxHeap.poll();
        }

    }
}

// TC: O(nlogn)
// SC: O(n)