Last Stone Weight

LeetCode Q 1046 - 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 y with 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 <= stones.length <= 30
• 1 <= stones[i] <= 1000

Solution

Code:

public int lastStoneWeight(int[] stones) {

	if (stones.length == 1) return stones[0];

	PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> (b - a));

	for (int stone: stones) pq.offer(stone);

	while (pq.size() >= 2) {
		int first = pq.poll(), second = pq.poll();
		if (first != second)
			pq.offer(first - second);
	}

	return pq.size() == 0 ? 0 : pq.peek();
}