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# Find minimum number of currency notes and values that sum to given amount

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

This problem is a simple variation of coin change problem. Here Greedy approach works as the given system is canonical (Please refer this and this for details)
Below is the program implementation to find the 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 % 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``]` `    ``notesCount ``=` `{}` `    `  `    ``for` `note ``in` `notes:` `        ``if` `amount >``=` `note:` `            ``notesCount[note] ``=` `amount``/``/``note` `            ``amount ``=` `amount ``%` `note` `            `  `    ``print` `(``"Currency Count ->"``)` `    ``for` `key, val ``in` `notesCount.items():` `        ``print``(f``"{key} : {val}"``)`   `# Driver code` `amount ``=` `868` `countCurrency(amount)`   `# Code contributed by farzams101`

## 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 % 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 % 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` `% ``\$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)` `?>`

## Javascript

 ``

Output:

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

Time Complexity: O(1), as the algorithm has a fixed number of iterations (9) that does not depend on the size of the input.

Auxiliary Space: O(1), as the algorithm only uses a fixed amount of space to store the notes and note counters, which does not depend on the size of the input.

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