Last Stone Weight💧

Question

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

Java Solution

Two different solutions with the same time complexity but one solution is using the build-in library function and in the other, I have written my own implementation of that function.
Ps: PriorityQueue is the function we are taking here.

Time Complexity
The time complexity of the above solutions is O(n log n), Once we remove a top element from the list we adjust it by applying max heap operation which takes Log n time. So for each n we are doing Log n operation i.e. O(n log n).

solution 1
Code

Solution 2
Code