Problem Description

You are given two integers numBottles and numExchange.

• numBottles is the initial number of full water bottles you have.
• You can perform the following actions:
  • Drink: Drink any number of full water bottles, turning them into empty bottles.
  • Exchange: Exchange numExchange empty bottles for one full water bottle. Each successful exchange increases numExchange by one.

Note: You can't perform multiple exchanges with the same numExchange value in one operation sequence, i.e., numExchange must increment after each exchange.

Your task is to determine the maximum number of water bottles you can drink.

Intuition

The core idea of solving this problem involves simulating the process of drinking and exchanging bottles efficiently. Here's how we think about the solution:

1. Initial Drinking: Start by drinking all the full water bottles you have initially (numBottles). This gives your initial count of bottles drunk.

2. Exchange and Drink:

   • As long as the number of empty bottles is numExchange or more, you can exchange them for a new full bottle.
   • Each exchange reduces the empty bottles by numExchange but then you drink the new full bottle, which returns one empty.

3. Increment and Continue: After each successful exchange and drinking:

   • Increment numExchange by one, following the rule specified.

4. Check Condition: Continue this cycle until you can no longer exchange the required number of empty bottles for a full one.

Intuitively, the problem is about maximizing drinkable bottles through strategic exchanges until the conditions no longer permit further actions. This simulation ensures that each step adheres to the rules given, particularly the increment of numExchange after each valid exchange.

Learn more about Math patterns.

Solution Approach

The solution utilizes a simulation approach to maximize the number of water bottles drunk by leveraging the exchange operation optimally.

1. Initialize:

   • Start by setting ans to numBottles, as you can initially drink all the bottles you have.

2. Loop Through Exchanges:

   • Enter a loop that continues while the number of empty bottles is greater than or equal to numExchange.
   • In each iteration:
     • Exchange Operation: Reduce the number of empty bottles by numExchange to simulate trading them in for one full bottle.
     • Increment Exchange Requirement: Increase numExchange by 1 to simulate the rule where each successful exchange increases the exchange requirement.
     • Drink the New Full Bottle: Increase ans by 1 because you drink the newly acquired full bottle.
     • Update Empty Bottles: Add back one to the empty bottles count because drinking the exchanged bottle creates one additional empty bottle.

3. Return the Total:

   • Once the loop can no longer proceed (when empty bottles are fewer than numExchange), the total number of bottles drunk (ans) is returned as the result.

This approach ensures each step follows the problem constraints and maximizes the total possible number of bottles you can drink.

Here's the solution code implemented using this logic:

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

This code reflects the outlined algorithm, iterating over possible exchanges and drinking while adhering to the increasing numExchange constraint. The straightforward while-loop handles both the exchange and the tallying of the bottles drunk in a systematic manner.

Example Walkthrough

Let's explore the solution using a concrete example:

Assume numBottles = 9 and numExchange = 3.

Step 1: Initial Drinking

• Start by drinking all 9 full bottles you have initially.
• Total bottles drunk so far: 9.
• Empty bottles available after drinking: 9.

Step 2: Loop Through Exchanges

• While there are at least 3 empty bottles (because numExchange = 3), we can exchange them for a full bottle.

  • Iteration 1:

    • Current empty bottles: 9
    • Exchange 3 empty bottles for 1 full bottle.
    • Total bottles drunk: 9 + 1 = 10
    • Remaining empty bottles: 9 - 3 + 1 (from drinking the new full bottle) = 7
    • Increase numExchange by 1: Now numExchange = 4

  • Iteration 2:

    • Current empty bottles: 7
    • Exchange 4 empty bottles for 1 full bottle.
    • Total bottles drunk: 10 + 1 = 11
    • Remaining empty bottles: 7 - 4 + 1 = 4
    • Increase numExchange by 1: Now numExchange = 5

  • Iteration 3:

    • Current empty bottles: 4
    • Exchange 5 empty bottles for 1 full bottle is not possible as only 4 empty bottles are left.
    • Exit the loop.

Step 3: Return the Total

• Since there are no further exchanges possible, the maximum number of bottles you can drink is 11.

This walkthrough illustrates the iterative process of drinking and exchanging bottles, highlighting how the numExchange increment impacts each cycle. By leveraging all available exchanges while adhering to the rule of increasing exchange challenges, the solution maximizes the number of bottles drunk.

Solution Implementation

Time and Space Complexity

The time complexity of the code is actually O(numBottles) instead of O(\sqrt{numBottles}). This is because in the worst-case scenario, the loop will iterate as long as numBottles is greater than or equal to numExchange. In each iteration, one bottle is exchanged, and this process continues until there are insufficient bottles to exchange. Therefore, the loop depends linearly on the total number of initial bottles, numBottles.

The space complexity of the code is O(1). This is because the algorithm uses a constant amount of extra space regardless of the input size. All operations and variable updates happen in-place, without requiring additional data structures that scale with the input.

Learn more about how to find time and space complexity quickly.