source: https://leetcode.com/problems/last-stone-weight/description/
Last Stone Weight
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 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
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class Solution { public int lastStoneWeight(int[] stones) { if(stones == null || stones.length == 0) return 0; if(stones.length == 1) return stones[0]; if(stones.length == 2) return stones[0] == stones[1]?0:(int)Math.abs(stones[0]-stones[1]); //max heap PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder()); for(int stone : stones){ pq.offer(stone); } while(pq.size() > 1){ //get two heaviest stone and smash them int x = pq.poll(); int y = pq.poll(); int z = x-y; if(z > 0) pq.offer(z); } return pq.isEmpty()?0:pq.poll(); } } |
Let’s say n is the number of stones in an array
Time Complexity
O(nlogn)
Space Complexity
O(n)