Water Bottles - Leetcode Solution

The Water Bottles problem on LeetCode involves calculating the maximum number of water bottles that can be drank given the number of empty bottles and the number of bottle caps that can be exchanged for a new bottle of water.

Based on the problem statement, we can assume that each empty bottle can be exchanged for one new water bottle, and each bottle cap can also be exchanged for one new water bottle. Additionally, the problem specifies that it is possible to drink all the available bottles, so there is no limit to the number of exchanges that can be made.

To solve this problem, we can start by initializing a variable to track the total number of bottles drank. We can then use a loop to repeatedly exchange the empty bottles and bottle caps for new water bottles until it is no longer possible to get any more. Within the loop, we can use two additional variables to keep track of the number of empty bottles and bottle caps that are currently available.

Here is the algorithm for solving the Water Bottles problem:

1. Initialize a variable "totalDrank" to 0.
2. Use a loop to repeatedly exchange empty bottles and bottle caps for new water bottles until it is no longer possible to get any more.
3. Within the loop, do the following: a. Calculate the number of new water bottles that can be obtained based on the current number of empty bottles and bottle caps (i.e., the minimum of the two values). b. Subtract the number of empty bottles and bottle caps used to obtain the new water bottles from the current number of empty bottles and bottle caps. c. Add the number of new water bottles obtained to the "totalDrank" variable. d. Update the number of empty bottles and bottle caps to reflect the new bottles obtained.
4. After the loop terminates, return the value of "totalDrank".

Here is the Python code for implementing this algorithm:

def numWaterBottles(numBottles: int, numExchange: int) -> int: # Initialize variables totalDrank = numBottles emptyBottles = numBottles bottleCaps = numBottles

# Exchange empty bottles and bottle caps for new water bottles
while emptyBottles >= numExchange or bottleCaps >= numExchange:
    # Calculate the number of new water bottles that can be obtained
    newBottles = min(emptyBottles // numExchange, bottleCaps // numExchange)
    
    # Subtract the number of empty bottles and bottle caps used to obtain the new water bottles
    emptyBottles -= newBottles * numExchange
    bottleCaps -= newBottles * numExchange
    
    # Add the number of new water bottles obtained to the total
    totalDrank += newBottles
    
    # Update the number of empty bottles and bottle caps
    emptyBottles += newBottles
    bottleCaps += newBottles

return totalDrank

In this code, the input "numBottles" represents the number of water bottles initially available, and "numExchange" represents the number of empty bottles or bottle caps that can be exchanged for a new water bottle. The function returns the maximum number of water bottles that can be drank.

This implementation should run in O(logN) time complexity because we reduce the number of empty bottles and bottle caps by the division of numExchange.

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