Last Stone Weight

Problem

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.

Examples

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

Solution

Method 1 - Using Max Heap

public int lastStoneWeight(int[] stones) {
	PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Comparator.reverseOrder());
	for (int st: stones) {
		maxHeap.offer(st);
	}
	while (maxHeap.size() > 1) {
		int diff = maxHeap.poll() - maxHeap.poll();
		if (diff > 0) {
			maxHeap.offer(diff);
		}

	}
	return maxHeap.isEmpty() ? 0 : maxHeap.peek();
}

Complexity

• ⏰ Time complexity: O(n log n)
• 🧺 Space complexity: O(n)