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Find minimum number of currency notes and values that sum to given amount
• Difficulty Level : Basic
• Last Updated : 01 Aug, 2018

Given an amount, find the minimum number of notes of different denominations that sum upto the given amount. Starting from the highest denomination note, try to accommodate as many notes possible for given amount.

We may assume that we have infinite supply of notes of values {2000, 500, 200, 100, 50, 20, 10, 5, 1}

Examples:

```Input : 800
Output : Currency  Count
500 : 1
200 : 1
100 : 1

Input : 2456
Output : Currency  Count
2000 : 1
200 : 2
50 : 1
5 : 1
1 : 1
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

This problem is a simple variation of coin change problem. Here Greedy approach works as given system is canonical (Please refer this and this for details)

Below is the program implementation to find number of notes:

## C++

 `// C++ program to accept an amount``// and count number of notes``#include ``using` `namespace` `std;`` ` `// function to count and ``// print currency notes``void` `countCurrency(``int` `amount)``{``    ``int` `notes = { 2000, 500, 200, 100,``                     ``50, 20, 10, 5, 1 };``    ``int` `noteCounter = { 0 };``     ` `    ``// count notes using Greedy approach``    ``for` `(``int` `i = 0; i < 9; i++) {``        ``if` `(amount >= notes[i]) {``            ``noteCounter[i] = amount / notes[i];``            ``amount = amount - noteCounter[i] * notes[i];``        ``}``    ``}``     ` `    ``// Print notes``    ``cout << ``"Currency Count ->"` `<< endl;``    ``for` `(``int` `i = 0; i < 9; i++) {``        ``if` `(noteCounter[i] != 0) {``            ``cout << notes[i] << ``" : "` `                ``<< noteCounter[i] << endl;``        ``}``    ``}``}`` ` `// Driver function``int` `main()``{``    ``int` `amount = 868;``    ``countCurrency(amount);``    ``return` `0;``}`

## Python3

 `# Python3 program to accept an amount``# and count number of notes `` ` `# Function to count and print ``# currency notes``def` `countCurrency(amount):``     ` `    ``notes ``=` `[``2000``, ``500``, ``200``, ``100``,``               ``50``, ``20``, ``10``, ``5``, ``1``]``                ` `    ``noteCounter ``=` `[``0``, ``0``, ``0``, ``0``, ``0``,``                     ``0``, ``0``, ``0``, ``0``]``     ` `    ``print` `(``"Currency Count -> "``)``     ` `    ``for` `i, j ``in` `zip``(notes, noteCounter):``        ``if` `amount >``=` `i:``            ``j ``=` `amount ``/``/` `i``            ``amount ``=` `amount ``-` `j ``*` `i``            ``print` `(i ,``" : "``, j)`` ` `# Driver code``amount ``=` `868``countCurrency(amount)`

## Java

 `// Java program to accept an amount``// and count number of notes``import` `java.util.*;``import` `java.lang.*;`` ` `public` `class` `GfG{`` ` `    ``// function to count and ``    ``// print currency notes``    ``public` `static` `void` `countCurrency(``int` `amount)``    ``{``        ``int``[] notes = ``new` `int``[]{ ``2000``, ``500``, ``200``, ``100``, ``50``, ``20``, ``10``, ``5``, ``1` `};``        ``int``[] noteCounter = ``new` `int``[``9``];``      ` `        ``// count notes using Greedy approach``        ``for` `(``int` `i = ``0``; i < ``9``; i++) {``            ``if` `(amount >= notes[i]) {``                ``noteCounter[i] = amount / notes[i];``                ``amount = amount - noteCounter[i] * notes[i];``            ``}``        ``}``      ` `        ``// Print notes``        ``System.out.println(``"Currency Count ->"``);``        ``for` `(``int` `i = ``0``; i < ``9``; i++) {``            ``if` `(noteCounter[i] != ``0``) {``                ``System.out.println(notes[i] + ``" : "``                    ``+ noteCounter[i]);``            ``}``        ``}``    ``}``     ` `    ``// driver function ``    ``public` `static` `void` `main(String argc[]){``        ``int` `amount = ``868``;``        ``countCurrency(amount);``    ``}``     ` `    ``/* This code is contributed by Sagar Shukla */``}`

## C#

 `// C# program to accept an amount``// and count number of notes``using` `System;`` ` `public` `class` `GfG{`` ` `    ``// function to count and ``    ``// print currency notes``    ``public` `static` `void` `countCurrency(``int` `amount)``    ``{``        ``int``[] notes = ``new` `int``[]{ 2000, 500, 200, 100, 50, 20, 10, 5, 1 };``        ``int``[] noteCounter = ``new` `int``;``     ` `        ``// count notes using Greedy approach``        ``for` `(``int` `i = 0; i < 9; i++) {``            ``if` `(amount >= notes[i]) {``                ``noteCounter[i] = amount / notes[i];``                ``amount = amount - noteCounter[i] * notes[i];``            ``}``        ``}``     ` `        ``// Print notes``        ``Console.WriteLine(``"Currency Count ->"``);``        ``for` `(``int` `i = 0; i < 9; i++) {``            ``if` `(noteCounter[i] != 0) {``                ``Console.WriteLine(notes[i] + ``" : "``                    ``+ noteCounter[i]);``            ``}``        ``}``    ``}``     ` `    ``// Driver function ``    ``public` `static` `void` `Main(){``        ``int` `amount = 868;``        ``countCurrency(amount);``    ``}``     ` ` ` `}`` ` `/* This code is contributed by vt_m */`

## PHP

 `= ``\$notes``[``\$i``])``        ``{``            ``\$noteCounter``[``\$i``] = ``intval``(``\$amount` `/ ``                                      ``\$notes``[``\$i``]);``            ``\$amount` `= ``\$amount` `- ``                      ``\$noteCounter``[``\$i``] * ``                      ``\$notes``[``\$i``];``        ``}``    ``}     ``    ``// Print notes``    ``echo` `(``"Currency Count ->"``.``"\n"``);``    ``for` `(``\$i` `= 0; ``\$i` `< 9; ``\$i``++) ``    ``{``        ``if` `(``\$noteCounter``[``\$i``] != 0) ``        ``{``            ``echo` `(``\$notes``[``\$i``] . ``" : "` `.``                  ``\$noteCounter``[``\$i``] . ``"\n"``);``        ``}``    ``}``}`` ` `// Driver Code``\$amount` `= 868;``countCurrency(``\$amount``);`` ` `// This code is contributed by ``// Manish Shaw(manishshaw1)``?>`

Output:
```Currency  Count ->
500 : 1
200 : 1
100 : 1
50 : 1
10 : 1
5 : 1
1 : 3
```

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