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 smallest possible weight of the left 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
- code
class Solution:
def lastStoneWeight(self, stones: List[int]) -> int:
hp = [-s for s in stones] # max heap
heapq.heapify(hp)
while len(hp) > 1:
heavy = -heapq.heappop(hp)
light = -heapq.heappop(hp)
if heavy != light:
heapq.heappush(hp, - heavy + light)
if not hp: return 0
return -hp[0]