One of the strengths of programming is its ability to model and solve real-world problems, and the “Water Bottles” problem on Leetcode is no exception. This problem is deceptively simple but can be approached in a variety of ways, each with its own set of trade-offs. In this in-depth article, we’ll explore the problem statement, various methods to solve it, analyze the time and space complexities.

## Problem Statement

Given `numBottles`

full water bottles, you can exchange `numExchange`

empty water bottles for one full water bottle. The task is to find the maximum number of water bottles you can drink.

#### Example

```
Input: numBottles = 9, numExchange = 3
Output: 13
Explanation: You can exchange 3 empty bottles to get 1 full water bottle.
Number of water bottles you can drink: 9 + 3 + 1 = 13.
```

## The Iterative Approach

A simple iterative solution is the most intuitive way to approach this problem. You can keep drinking the water bottles and exchanging them until you can’t exchange any more.

Python Code:

```
def numWaterBottles(numBottles, numExchange):
total = numBottles
empty = numBottles
while empty >= numExchange:
new_bottles = empty // numExchange
total += new_bottles
empty = new_bottles + empty % numExchange
return total
# Test the function
print(numWaterBottles(9, 3)) # Output should be 13
```

### Time Complexity

The time complexity for this approach is O(numBottles).

### Space Complexity

The space complexity is O(1) as we’re using a constant amount of extra space.

## The Recursive Approach

The problem can also be solved recursively, which might make it easier to understand for those who are more comfortable with recursion.

Python Code:

```
def numWaterBottles(numBottles, numExchange):
if numBottles < numExchange:
return numBottles
new_bottles = numBottles // numExchange
remaining_bottles = numBottles % numExchange
return numBottles + numWaterBottles(new_bottles + remaining_bottles, numExchange) - remaining_bottles
# Test the function
print(numWaterBottles(9, 3)) # Output should be 13
```

### Time Complexity

The time complexity is the same as the iterative approach, O(numBottles).

### Space Complexity

The space complexity here would be O(numBottles) due to the recursive stack.

## The Mathematical Approach

If you analyze the problem carefully, you can deduce a mathematical formula to solve it. However, finding that formula is non-trivial and may not be the best approach for a timed coding interview.

### Time and Space Complexity

The mathematical approach, if formulated, would have a time and space complexity of O(1).

## Edge Cases and Further Considerations

**Zero Bottles**: What if`numBottles`

is zero? In this case, the answer should be zero as well.**One Bottle**: What if`numBottles`

is one? In this case, you cannot make any exchanges, and you can drink only that one bottle.**Exchange Requirement**: What if`numExchange`

is greater than`numBottles`

? You won’t be able to make an initial exchange but can still drink`numBottles`

.

## Conclusion

The “Water Bottles” problem is a great exercise for thinking about problem-solving from multiple perspectives. Whether it’s employing straightforward iterative methods, using recursion for more elegant code, or delving into a potential mathematical solution, each approach has its pros and cons. This problem not only allows you to practice coding but also prompts you to think deeply about optimization, real-world applications, and edge cases.