1046

1046. Last Stone Weight (Easy)

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

Companies:
Visa, Nvidia, Amazon

Related Topics:
Array, Heap (Priority Queue)

Solution 1. Max-root Heap

// OJ: https://leetcode.com/problems/last-stone-weight/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    int lastStoneWeight(vector<int>& A) {
        priority_queue<int> pq(begin(A), end(A));
        while (pq.size() > 1) {
            int a = pq.top(); pq.pop();
            int b = pq.top(); pq.pop();
            int r = abs(a - b);
            if (r) pq.push(r);
        }
        return pq.size() ? pq.top() : 0;
    }
};