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Minimum Amount of Time to Fill Cups

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Statistics

Time Spent Coding
15 minutes

Time Complexity
O(1) - Since the amount list is always 3 elements the growth rate is constant, resulting in the O(1) time complexity.

Space Complexity
O(1) - You must always store 3 elements, which is a constant, resulting in the O(1) space complexity.

Runtime Beats
75.60% of other submissions

Memory Beats
71.50% of other sumbissions

Solution

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class Solution:
    def fillCups(self, amount: List[int]) -> int:
        # O(3 log 3) == O(1), since there are never more than 3 elements
        a, b, c = sorted(amount)

        # If the two smallest sum to less than c
        # Then they would either cancel out (a == b) 
        # or 
        # the remaining of b would be able to be dispensed in parallel with c
        if a + b <= c: return c
        

        # Otherwise the time is less than c, becuase you can do more than 
        # c in parallel.
        return ceil((a+b+c)/2)
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