Python3   ||    one line w/explanation   ||  T/S: 90% / 82%
Here\'s the intuition:\n\nBecause only one of any type can be filled per second, the minimum cannot be less than the max(amount)-- see Ex 1.\n\n The minimum als
Here\'s the intuition:\n\nBecause only one of any type can be filled per second, the minimum cannot be less than the max(amount)-- see Ex 1.\n\n The minimum also cannot be less than `ceil(sum(amount)/2)`--see Ex 2.\n                                    \n The mminimumin cannot be greater than both of these two\n quantities, so... tada\n```\nclass Solution:\n    def fillCups(self, amount: List[int]) -> int:\n        return max(max(amount), ceil(sum(amount)/2))\n```\n[https://leetcode.com/problems/minimum-amount-of-time-to-fill-cups/submissions/1286771753/](https://leetcode.com/problems/minimum-amount-of-time-to-fill-cups/submissions/1286771753/)\n\nI could be wrong, but I think that time complexity is *O*(*N*) and space complexity is *O*(1), in which *N* ~ `len(amount)`.
['Python3']