Water Bottles

The task is to find out the maximum number of water bottles we can drink given the starting number of full water bottles and the number of empty water bottles required for an exchange.

We start with the initial number of full water bottles, numBottles, which we can drink. Once we drink a bottle, it becomes empty and can potentially be exchanged for another full bottle.

This gives us a simple strategy to tackle the problem:

1. We initialize total to numBottles, which are the bottles we can drink initially.
2. While numBottles is greater than or equal to numExchange, we can exchange our empty bottles for new full bottles.
   • To calculate the number of new bottles we get, we use integer division numBottles // numExchange. We add these new bottles to our total.
   • We then update numBottles to the number of new bottles plus the remaining bottles (numBottles % numExchange) that we couldn’t exchange.

Python implementation:

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class Solution:
    def numWaterBottles(self, numBottles: int, numExchange: int) -> int:
        total = numBottles
        while numBottles >= numExchange:
            new_bottles = numBottles // numExchange
            total += new_bottles
            numBottles = new_bottles + numBottles % numExchange
        return total

The time complexity of this solution is O(numBottles), and the space complexity is O(1).

Using an Equation

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class Solution:
    def numWaterBottles(self, numBottles: int, numExchange: int) -> int:
        return (numBottles*numExchange-1)//(numExchange-1)

The Python function numWaterBottles is designed to calculate the maximum number of water bottles you can drink given numBottles initial full water bottles and a rule that numExchange empty water bottles can be exchanged for one full water bottle.

Here’s how the function works:

1. The formula (numBottles * numExchange - 1) // (numExchange - 1) calculates the total number of bottles you can drink. Let’s break down this formula:

   • numBottles * numExchange: This is the maximum number of bottles you could have if you were able to exchange every bottle without any leftover. However, this isn’t possible because at the end you won’t have enough empty bottles to make another exchange.

   • - 1: We subtract one because we can’t make the last exchange (we will have less than numExchange bottles at the end).

   • // (numExchange - 1): This calculates how many full cycles of drinking and exchanging we can make. In each cycle, you drink numExchange bottles and get 1 bottle in return from the exchange, effectively making it as if you drank numExchange - 1 bottles. So, dividing by numExchange - 1 gives us the total number of bottles we can drink.

2. Finally, it returns the result of the calculation, which is the maximum number of water bottles you can drink.

This function uses a mathematical formula to solve the problem in constant time, resulting in a time complexity of O(1) and a space complexity of O(1).

Abstract Representation of the Problem

This problem can be considered as an instance of resource management and optimization problem, where the resource in question is water bottles.

The problem can be abstractly represented as follows:

Given an initial resource ‘R’ and a conversion factor ‘C’ which states that C units of consumed resources can be recycled to get one new unit of the resource, we are required to find out the maximum utilization of the initial resource.

In each iteration, we consume the resource and then recycle a part of the consumed resource into new resource units using the conversion factor. The process continues until no further recycling can be done.

The goal is to maximize the total consumption of the resource. This includes the initial resource units and any additional units gained through the recycling process.