1046. Last Stone Weight

Problem

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

Solution (Java)

class Solution {
    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> heap = new PriorityQueue<>((a, b) -> b - a);
        for (int stone : stones) {
            heap.offer(stone);
        }
        while (!heap.isEmpty()) {
            if (heap.size() >= 2) {
                int one = heap.poll();
                int two = heap.poll();
                int diff = one - two;
                heap.offer(diff);
            } else {
                return heap.poll();
            }
        }
        return -1;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).