algoadvance

1. Water Bottles

Given numBottles full water bottles, you can exchange numExchange empty water bottles for one full water bottle.

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

Clarifying Questions

1. Input Constraints: Are there any constraints on the values of numBottles and numExchange (e.g., maximum values)?
2. Outputs: Should the output be just the number of water bottles that can be drank?
3. Exchanges: Is it safe to assume that the exchange process can continue until no more full bottles can be obtained?

Assuming no additional constraints and the typical input-output format for LeetCode, I will proceed to solve the problem.

Strategy

1. Drink and Exchange: Start by drinking all the initial full water bottles. This will give you the same number of empty bottles.
2. Iterate Until Exhausted: Repeatedly check if the number of empty bottles is enough to exchange for new full bottles. Perform the exchange and update the count of full and empty bottles accordingly.
3. Accumulate Total: Keep a running total of the number of water bottles consumed.

Code

def numWaterBottles(numBottles: int, numExchange: int) -> int:
    total_bottles_drunk = numBottles
    empty_bottles = numBottles
    
    while empty_bottles >= numExchange:
        new_bottles = empty_bottles // numExchange
        empty_bottles = empty_bottles % numExchange + new_bottles
        total_bottles_drunk += new_bottles
        
    return total_bottles_drunk

# Example Usage
# numBottles = 9, numExchange = 3 => Output should be 13
print(numWaterBottles(9, 3))  # Output: 13

Time Complexity

• Time Complexity: O(numBottles). In the worst-case scenario, we may need to iterate approximately numBottles times to keep exchanging until the number of bottles goes below the exchange threshold.
• Space Complexity: O(1). The solution uses a constant amount of extra space regardless of the input size.

Explanation

1. Start by initializing total_bottles_drunk with the initial number of full water bottles.
2. Use a while loop to repeatedly exchange empty bottles for full ones as long as the number of empty bottles is sufficient (>= numExchange).
3. Calculate the number of new full bottles you can get and update the count of empty bottles accordingly after each exchange.
4. Accumulate the total count of full bottles drunk during each cycle of exchange.
5. Return the total count as the result.

This approach ensures that you maximize the total number of water bottles you can drink based on the given exchange rate.

Try our interview co-pilot at AlgoAdvance.com

• Got blindsided by a question you didn’t expect?

• Spend too much time studying?

• Or simply don’t have the time to go over all 3000 questions?