# Maximum litres of water that can be bought with N Rupees

• Difficulty Level : Hard
• Last Updated : 29 Sep, 2022

Given Rupees. A liter plastic bottle of water costs Rupees and a litre of glass bottle of water costs Rupees. But the empty glass bottle after buying can be exchanged for Rupees. Find the maximum liters of water which can be bought with Rupees.

Examples:

Input: N = 10 , A = 11 , B = 9 , C = 8
Output:
One glass bottle can be bought and then can be returned to buy one more glass bottle

Input: N = 15 , A = 6 , B = 4 , C = 3
Output: 12

Approach: If we have at least money then cost of one glass bottle is b – c. This means that if a ≤ (b – c) then we don’t need to buy glass bottles, only plastic ones, and the answer will be floor(n / a). Otherwise we need to buy glass bottles while we can.

So, if we have at least money, then we will buy floor((n – c) / (b – c)) glass bottles and then spend rest of the money on plastic ones.

Below is the implementation of the above approach:

## C++

 `// CPP implementation of the above approach``#include``using` `namespace` `std;`  `void` `maxLitres(``int` `budget,``int` `plastic,``int` `glass,``int` `refund)``{` `    ``// if buying glass bottles is profitable``    ``if` `(glass - refund < plastic)``       ``{``           ``// Glass bottles that can be bought``        ``int` `ans = max((budget - refund) / (glass - refund), 0);` `        ``// Change budget according the bought bottles``        ``budget -= ans * (glass - refund);` `        ``// Plastic bottles that can be bought``        ``ans += budget / plastic;``        ``cout<

## Java

 `// Java implementation of the above approach``class` `GFG``{` `    ``static` `void` `maxLitres(``int` `budget, ``int` `plastic,``                            ``int` `glass, ``int` `refund)``    ``{` `        ``// if buying glass bottles is profitable``        ``if` `(glass - refund < plastic)``        ``{``            ` `            ``// Glass bottles that can be bought``            ``int` `ans = Math.max((budget - refund) / (glass - refund), ``0``);` `            ``// Change budget according the bought bottles``            ``budget -= ans * (glass - refund);` `            ``// Plastic bottles that can be bought``            ``ans += budget / plastic;``            ``System.out.println(ans);``        ``}``        ` `        ``// if only plastic bottles need to be bought``        ``else``        ``{``            ``System.out.println((budget / plastic));``        ``}` `    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `budget = ``10``, plastic = ``11``, glass = ``9``, refund = ``8``;``        ``maxLitres(budget, plastic, glass, refund);``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python3 implementation of the above approach` `def` `maxLitres(budget, plastic, glass, refund):` `    ``# if buying glass bottles is profitable``    ``if` `glass ``-` `refund < plastic:` `        ``# Glass bottles that can be bought``        ``ans ``=` `max``((budget ``-` `refund) ``/``/` `(glass ``-` `refund), ``0``)` `        ``# Change budget according the bought bottles``        ``budget ``-``=` `ans ``*` `(glass ``-` `refund)` `        ``# Plastic bottles that can be bought``        ``ans ``+``=` `budget ``/``/` `plastic``        ``print``(ans)` `    ``# if only plastic bottles need to be bought``    ``else``:``        ``print``(budget ``/``/` `plastic)` `# Driver Code``budget, plastic, glass, refund ``=` `10``, ``11``, ``9``, ``8``maxLitres(budget, plastic, glass, refund)`

## C#

 `// C# implementation of the above approach``using` `System;   ` `class` `GFG``{` `    ``static` `void` `maxLitres(``int` `budget, ``int` `plastic,``                            ``int` `glass, ``int` `refund)``    ``{` `        ``// if buying glass bottles is profitable``        ``if` `(glass - refund < plastic)``        ``{``            ` `            ``// Glass bottles that can be bought``            ``int` `ans = Math.Max((budget - refund) / (glass - refund), 0);` `            ``// Change budget according the bought bottles``            ``budget -= ans * (glass - refund);` `            ``// Plastic bottles that can be bought``            ``ans += budget / plastic;``            ``Console.WriteLine(ans);``        ``}``        ` `        ``// if only plastic bottles need to be bought``        ``else``        ``{``            ``Console.WriteLine((budget / plastic));``        ``}` `    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int` `budget = 10, plastic = 11, glass = 9, refund = 8;``        ``maxLitres(budget, plastic, glass, refund);``    ``}``}` `// This code contributed by Rajput-Ji`

## PHP

 ``

## Javascript

 ``

Output

`2`

Time complexity: O(1)
Auxiliary space: O(1)

My Personal Notes arrow_drop_up