DAY 1

# LeetCode 3000 – Water Bottles II

## Problem
You are given two integers:  
- **numBottles** → initial number of full water bottles  
- **numExchange** → number of empty bottles required to exchange for one full bottle  

### Rules
- You can drink any number of full bottles, turning them into empty ones.  
- You can exchange **numExchange** empty bottles for one full bottle. After each exchange, `numExchange` increases by 1.  
- You cannot exchange multiple batches at the same exchange value.  

**Goal:** Find the maximum number of water bottles you can drink.  

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## Examples
**Example 1**  
Input: `numBottles = 13, numExchange = 6`  
Output: `15`  

**Example 2**  
Input: `numBottles = 10, numExchange = 3`  
Output: `13`  

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## Key Insights
- Every bottle drunk increases the number of empty bottles.  
- After each exchange, the requirement (`numExchange`) becomes harder by +1.  
- Greedy strategy: always exchange whenever possible.  
- Stop when there are not enough empty bottles to perform another exchange.  

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## Approach
1. Start with `drunk = numBottles` (drink all initial bottles).  
2. Convert them to `empty = numBottles`.  
3. While `empty >= numExchange`:  
   - Use `numExchange` empty bottles → gain 1 full bottle.  
   - Increase `numExchange` by 1.  
   - Drink the new bottle → increment `drunk`.  
   - Add the new empty bottle back to `empty`.  
4. Return `drunk`.  

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## Complexity
- **Time Complexity:** `O(numBottles + numExchange)` → at most 200 steps since both ≤ 100.  
- **Space Complexity:** `O(1)` → only variables used.  

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## Python Solution
```python
class Solution:
    def maxBottlesDrunk(self, numBottles: int, numExchange: int) -> int:
        # Drink all initial bottles
        drunk = numBottles
        # They all become empty
        empty = numBottles

        # Continue while we can exchange
        while empty >= numExchange:
            # Exchange empty bottles for one full bottle
            empty -= numExchange
            # Increase exchange requirement
            numExchange += 1
            # Drink the new bottle
            drunk += 1
            # That bottle becomes empty
            empty += 1

        return drunk