minHeap: find last stone weight


Tue Aug 16 2022 06:14:12 GMT+0000 (UTC)

Saved by @bryantirawan #python #neetcode #minheap #maxheap #pop #push #heapify

class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        stones = [-s for s in stones] 
        while len(stones) > 1: 
            first = abs(heapq.heappop(stones)) #linear time for python 
            second = abs(heapq.heappop(stones))
            if first > second: 
                heapq.heappush(stones, second - first) #remember that since it's a minHeap we need negative numbers so it's second - first 
        #edge case to account if stones is empty whether originally or after smashing 
        return abs(stones[0])

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.