Maximum litres of water that can be bought with N Rupees
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: 2
One glass bottle can be bought and then can be returned to buy one more glass bottleInput: 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<bits/stdc++.h>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<<ans<<endl; } // if only plastic bottles need to be bought else cout<<(budget / plastic)<<endl;} // Driver Codeint main(){ int budget = 10, plastic=11, glass=9, refund = 8; maxLitres(budget, plastic, glass, refund);}// This code is contributed by// Surendra_Gangwar |
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 approachdef 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 Codebudget, plastic, glass, refund = 10, 11, 9, 8maxLitres(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
<?php// PHP implementation of the above approachfunction maxLitres($budget, $plastic, $glass, $refund){ // if buying glass bottles is profitable if ($glass - $refund < $plastic) { // Glass bottles that can be bought $ans = max((int)($budget - $refund) / ($glass - $refund), 0); // Change budget according the bought bottles $budget -= $ans * ($glass - $refund); // Plastic bottles that can be bought $ans += (int)($budget / $plastic); echo $ans . "\n"; } // if only plastic bottles need to be bought else echo (int)($budget / $plastic) . "\n";}// Driver Code$budget = 10;$plastic = 11;$glass = 9;$refund = 8;maxLitres($budget, $plastic, $glass, $refund);// This code is contributed by mits?> |
Javascript
<script>// Javascript implementation of the above approach function maxLitres(budget, plastic, glass, refund) { // if buying glass bottles is profitable if (glass - refund < plastic) { // Glass bottles that can be bought let 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 += Math.floor(budget / plastic); document.write(ans); } // if only plastic bottles need to be bought else { document.write(Math.floor(budget / plastic)); } } // Driver code let budget = 10, plastic = 11, glass = 9, refund = 8; maxLitres(budget, plastic, glass, refund); </script> |
Output
2
Time complexity: O(1)
Auxiliary space: O(1)