1046. Last Stone Weight
Easy
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 weightx
is destroyed, and the stone of weighty
has new weighty - 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
.
Solution
import heapq
class Solution:
def lastStoneWeight(self, stones: List[int]) -> int:
stones = [-s for s in stones]
heapq.heapify(stones)
while len(stones) > 1:
y = heapq.heappop(stones)
x = heapq.heappop(stones)
if x != y:
heapq.heappush(stones, y - x)
if len(stones) == 0:
return 0
return -heapq.heappop(stones)