1046. Last Stone Weight.cpp

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 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
********************************************************************************










# Approach 1:

class Solution {
public:
    int lastStoneWeight(vector<int>& stones) {
        priority_queue <int> pq;

        for (auto &x: stones) pq.push(x);

        while (pq.size() > 1) {
            int y = pq.top();
            pq.pop();
            int x = pq.top();
            pq.pop();
            if (x != y) pq.push(y - x);
        }
        return pq.size() == 0 ? 0 : pq.top();
    }
};

TC -> O(nlogn), n is the size of the stones
SC -> O(n), n is the size of the stones