Today I have done leetcode's April LeetCoding Challenge with cpp.

April LeetCoding Challenge 7

Description

Last Stone Weight

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 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

Solution

class Solution {
public:
  int lastStoneWeight(vector<int>& stones) {
    priority_queue<int> pq;
    for(auto stone : stones) {
      pq.push(stone);
    }
    while(pq.size() > 1) {
      auto t = pq.top();
      pq.pop();
      auto s = pq.top();
      pq.pop();
      if(t > s) {
        pq.push(t - s);
      }
    }
    return pq.size() ? pq.top() : 0;
  }
};

// Accepted
// 70/70 cases passed (4 ms)
// Your runtime beats 37.54 % of cpp submissions
// Your memory usage beats 78.96 % of cpp submissions (7.6 MB)