main.cpp

// Source: https://leetcode.com/problems/last-stone-weight
// Title: Last Stone Weight
// Difficulty: Easy
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// 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 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`
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

#include <functional>
#include <queue>
#include <vector>

using namespace std;

// Heap
class Solution {
 public:
  int lastStoneWeight(vector<int>& stones) {
    auto heap = priority_queue(less<int>(), std::move(stones));  // max-heap

    while (heap.size() > 1) {
      auto y = heap.top();
      heap.pop();
      auto x = heap.top();
      heap.pop();
      if (x != y) heap.push(y - x);
    }

    return heap.empty() ? 0 : heap.top();
  }
};