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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1001_1100.s1046_last_stone_weight;

// #Easy #Array #Heap_Priority_Queue #Level_1_Day_15_Heap
// #2022_02_27_Time_2_ms_(73.81%)_Space_42.3_MB_(5.13%)

import java.util.PriorityQueue;

/**
 * 1046 - Last Stone Weight\.
 *
 * Easy
 *
 * You are given an array of integers `stones` where `stones[i]` is the weight of the i^th 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 smallest possible weight of the left 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`
**/
public class Solution {
    public int lastStoneWeight(int[] stones) {
        PriorityQueue 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;
    }
}