1518 Water Bottles

Intuition

We are required to compute the number of bottles that is drinkable. A twist in the problem is that after every session of drinking all the bottles, we can exhange the empty bottles for more bottles (depending on the numExchange).

Code

Python3

def numWaterBottles(self, numBottles: int, numExchange: int) -> int:
    ret = 0
 
    while numBottles >= numExchange:
        numBottles, carry = divmod(numBottles, numWater)
        ret += numBottles
        numBottles += carry
 
    return ret

Big O Analysis

• Runtime

  The runtime complexity here is $O(N)$ where N = numBottles since we would be iterating the entire sample set of number of bottles.

• Memory

  The memory usage is $O(1)$ since we are not using any extra data structure.

— A

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