Solution Explanation

The problem involves calculating the maximum number of water bottles you can drink given an initial number of bottles and the exchange rate for empty bottles. The solution utilizes a simple iterative approach.

Approach:

1. Initialization: The total number of bottles drunk (ans) is initialized to the initial number of full bottles (numBottles).

2. Iteration: The loop continues as long as the number of empty bottles is greater than or equal to the exchange rate (numExchange). In each iteration:

   • We calculate the number of additional bottles obtained from exchanging empty bottles: numBottles / numExchange.
   • The total number of drunk bottles (ans) is increased by this number.
   • The number of full bottles is updated by subtracting the exchanged bottles and adding the newly obtained bottles.

3. Return Value: The function returns ans, representing the total number of water bottles drunk.

Code Explanation (Python Example):

class Solution:
    def numWaterBottles(self, numBottles: int, numExchange: int) -> int:
        ans = numBottles
        while numBottles >= numExchange:
            empty_bottles = numBottles // numExchange  # Integer division to find the number of new bottles
            ans += empty_bottles # Add the number of new bottles to the total
            numBottles = numBottles % numExchange + empty_bottles # Update the number of bottles.
        return ans
 

Time and Space Complexity Analysis:

• Time Complexity: O(log n), where n is numBottles. The loop iterates until numBottles becomes less than numExchange. In each iteration, numBottles is reduced significantly, resulting in a logarithmic time complexity. In the worst case, it will be O(n) if numExchange is 2.

• Space Complexity: O(1). The algorithm uses only a constant amount of extra space to store variables, regardless of the input size.

Example Walkthrough (Python):

Let's say numBottles = 9 and numExchange = 3.

1. ans is initialized to 9.
2. The loop starts:
   • Iteration 1: numBottles (9) >= numExchange (3). empty_bottles = 9 // 3 = 3. ans becomes 9 + 3 = 12. numBottles becomes 9 % 3 + 3 = 3.
   • Iteration 2: numBottles (3) >= numExchange (3). empty_bottles = 3 // 3 = 1. ans becomes 12 + 1 = 13. numBottles becomes 3 % 3 + 1 = 1.
   • The loop terminates because numBottles (1) < numExchange (3).
3. The function returns ans, which is 13.

The provided code in other languages follows the same logic and complexity. The only differences are syntactic variations specific to each programming language.