#1049

Last Stone Weight II

specialist · 720 · lc medium +30 · verified · 59.3% accepted · 3,371 likes · top 57%

Description

You are given an array stones where stones[i] is the weight of the i^th stone. On each turn, choose any two stones (weights x <= y) and smash them: if equal both vanish; otherwise the stone of weight x is destroyed and y becomes y - x.

Return the smallest possible weight of the last stone. Return 0 if no stones remain.

Example 1:

Input: stones = [2,7,4,1,8,1]

Output: 1

Explanation:

We can combine 2 and 4 to get 2, so the array converts to [2,7,1,8,1] then,

we can combine 7 and 8 to get 1, so the array converts to [2,1,1,1] then,

we can combine 2 and 1 to get 1, so the array converts to [1,1,1] then,

we can combine 1 and 1 to get 0, so the array converts to [1], then that's the optimal value.

Example 2:

Input: stones = [31,26,33,21,40]

Output: 5

Code

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class Solution:
    def lastStoneWeightII(self, stones: List[int]) -> int:

class Solution:
    def lastStoneWeightII(self, stones: List[int]) -> int: