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# Leetcode Solutions: 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.

``````class Solution(object):
def lastStoneWeight(self, stones):
"""
:type stones: List[int]
:rtype: int
"""

stones = [-stone for stone in stones]

heapq.heapify(stones)

while len(stones) > 1:
first  = heapq.heappop(stones)
second = heapq.heappop(stones)

if second > first:
heapq.heappush(stones, first-second)

return -stones[0] if len(stones) else 0

``````