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Last Stone Weight

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`. If `x == y`, both stones are destroyed. 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`.

Examples

Input:stones = [2,7,4,1,8,1]
Output:1
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.
Input:stones = [1]
Output:1

Constraints

  • 1 <= stones.length <= 30
  • 1 <= stones[i] <= 1000

Max Heap

Approach

Since we always need to find and remove the two largest elements, a Max Heap is the perfect data structure. Add all stone weights to a Max Heap. While the heap has more than one stone, pop the top two stones. If they are not equal, push their difference (`y - x`) back into the heap. Finally, return the remaining stone's weight if the heap is not empty, otherwise return 0. In Python, use a Min Heap with negative values to simulate a Max Heap.

Complexity Analysis

Time Complexity
O(N log N)
Space Complexity
O(N)

N is the number of stones. Building the heap takes O(N) or O(N log N) depending on the implementation. Each of the N iterations does two pops and one push, taking O(log N).

Solution.java
class Solution {    public int lastStoneWeight(int[] stones) {        // Max Heap (reverse order)        PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());        for (int stone : stones) {            maxHeap.offer(stone);        }                while (maxHeap.size() > 1) {            int y = maxHeap.poll(); // Heaviest            int x = maxHeap.poll(); // Second heaviest                        if (y > x) {                maxHeap.offer(y - x);            }        }                return maxHeap.isEmpty() ? 0 : maxHeap.poll();    }}