Tags: "leetcode", access_time 2-min read

Last Stone Weight

Created: April 12, 2020 by [lek-tin]

Last updated: April 12, 2020

We have a collection of stones, each stone has a positive integer weight.

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

1. If x == y, both stones are totally destroyed;
2. If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.
3. 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

Solution

import heapq

class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        max_heap = [ -w for w in stones]
        heapq.heapify(max_heap)

        for i in range(len(stones)-1):
            remained = -heapq.heappop(max_heap) + heapq.heappop(max_heap)
            heapq.heappush(max_heap, -remained)

        return -max_heap[0]