Last Stone Weight
We have a collection of rocks, each rock has a positive integer weight. Each turn, we choose the two heaviest rocks and smash them together. Suppose the stones have weights x and y with x <= y. The result of this smash is: If x == y, both stones are totally destroyed; If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x. 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
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.
Solution
class Solution: def lastStoneWeight(self, stones: List[int]) -> int: A = [-x for x in stones] heapq.heapify(A) while len(A)>1: x, y = -heapq.heappop(A), -heapq.heappop(A) heapq.heappush(A, -abs(x-y)) return -A[0]