LeetCode Challenge #1046. Last Stone Weight

Day 42 of 100 Days LeetCode Challenge

LeetCode Challenge #1046. Last Stone Weight

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

Video Solution

Java Solution

				
					class Solution {
    public int lastStoneWeight(int[] stones) {
        //max priority queue 
        PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());

        for(int ele : stones){ // adding the elements of array to pq 
            pq.add(ele);
        }

        while(pq.size()>1){
            int max = pq.remove();
            int smax = pq.remove();

            int nstone = max - smax ;

            if(nstone!=0){
                pq.add(nstone);
            }
        }

        if(pq.size()==0){
            return 0 ;
        }else{
            return pq.remove();
        }
    }
}

Day 42 of 100 Days LeetCode Challenge