_1046.java

package com.fishercoder.solutions;

import java.util.PriorityQueue;

/**
 * 1046. Last Stone Weight
 *
 * We have a collection of rocks, each rock has a positive integer weight.
 *
 * Each turn, we choose the two heaviest rocks and smash them together.
 * Suppose the stones have weights x and y with x <= y.  The result of this smash is:
 *
 * If x == y, both stones are totally destroyed;
 * If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.
 * At the end, there is at most 1 stone left.  Return the weight of this stone (or 0 if there are no stones left.)
 *
 * Example 1:
 * Input: [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 last stone.
 *
 * Note:
 * 1 <= stones.length <= 30
 * 1 <= stones[i] <= 1000
 * */
public class _1046 {
    public static class Solution1 {
        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;
        }
    }
}