Last Stone Weight

Problem Statement - link #

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.

Examples #

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

Solutions #

class Solution {
public:
    int lastStoneWeight(vector<int>& stones) {
        priority_queue<int> pq;
        for(auto stone: stones) pq.push(stone);
        int lastWeight=0;
        while(!pq.empty()){
            if(pq.size()==1) {
                lastWeight = pq.top();
                break;
            }
            int x = pq.top(); pq.pop();
            int y = pq.top(); pq.pop();
            if(x!=y) pq.push(abs(x-y));
        }
        return lastWeight;
    }
};