main.md

Question

Approach

I drink all the initial bottles first, keeping track of empties. Then, as long as I have at least numExchange empties, I perform a single exchange: I trade the empties for one full bottle, drink it, update the empties, and increase numExchange by one since future exchanges are harder. I repeat this process until I can no longer exchange.

Implementation

class Solution:
    def maxBottlesDrunk(self, numBottles: int, numExchange: int) -> int:
        total = numBottles
        empty = numBottles
        
        while empty >= numExchange:
            empty -= numExchange
            numBottles = 1
            total += 1
            empty += numBottles
            numExchange += 1
        
        return total

Complexities

• Time: O(numBottles + numExchange) in the worst case (each loop step increases numExchange by 1)
• Space: O(1)