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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 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

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 <bits/stdc++.h>
using namespace std;
  
// function to count and 
// print currency notes
void countCurrency(int amount)
{
    int notes[9] = { 2000, 500, 200, 100,
                     50, 20, 10, 5, 1 };
    int noteCounter[9] = { 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[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
        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

<?php
// PHP program to accept an amount
// and count number of notes
  
// function to count and 
// prcurrency notes
function countCurrency($amount)
{
    $notes = array(2000, 500, 200, 100,
                   50, 20, 10, 5, 1);
    $noteCounter = array(0, 0, 0, 0, 0, 
                         0, 0, 0, 0, 0);
      
    // count notes using
    // Greedy approach
    for ($i = 0; $i < 9; $i++) 
    {
        if ($amount >= $notes[$i])
        {
            $noteCounter[$i] = intval($amount
                                      $notes[$i]);
            $amount = $amount
                      $noteCounter[$i] * 
                      $notes[$i];
        }
    }     
    // Print notes
    echo ("Currency Count ->"." ");
    for ($i = 0; $i < 9; $i++) 
    {
        if ($noteCounter[$i] != 0) 
        {
            echo ($notes[$i] . " : " .
                  $noteCounter[$i] . " ");
        }
    }
}
  
// 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


This article is attributed to GeeksforGeeks.org

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