LeetCode 1046 - Last Stone Weight

https://leetcode.com/problems/last-stone-weight/

We have a collection of rocks, each rock has a positive integer weight.

Each turn, we choose the two heaviest rocks and smash them together.  Suppose the stones have weights x and ywith x <= y.  The result of this smash is:

• If x == y, both stones are totally destroyed;
• If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.

At the end, there is at most 1 stone left.  Return the weight of this stone (or 0 if there are no stones left.)

Example 1:

Input: [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 last stone.

Note:

1. 1 <= stones.length <= 30
2. 1 <= stones[i] <= 1000

X. PriorityQueue
https://leetcode.com/problems/last-stone-weight/discuss/294956/JavaPython-O(nlogn)

    public int lastStoneWeight(int[] A) {
        PriorityQueue<Integer> pq = new PriorityQueue<>((a, b)-> b - a);
        for (int a : A)
            pq.offer(a);
        for (int i = 0; i < A.length - 1; ++i)
            pq.offer(pq.poll() - pq.poll());
        return pq.poll();
    }

https://leetcode.com/problems/last-stone-weight/discuss/294993/Java-easy-code-using-PriorityQueue-w-brief-explanation-and-analysis.

Sort stones descendingly in PriorityQueue, then pop out pair by pair, compute the difference between them and add back to PriorityQueue.

Note: since we already know the first poped out is not smaller, it is not necessary to use Math.abs().

    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> q = new PriorityQueue<>(Comparator.reverseOrder());
        for (int st : stones) { q.offer(st); }
        while (q.size() > 1) {
            q.offer(q.poll() - q.poll());
        }
        return q.peek();
    }

Analysis:

Time: O(nlogn), space: O(n), where n = stones.length.

Q & A:

Q: If not adding zeroes in the queue when polling out two elements are equal, is the result same as the above code?

A: Yes. 0s are always at the end of the PriorityQueue. No matter a positive deduct 0 or 0 deduct 0, the result is same as NOT adding 0s into the PriorityQueue.