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++ 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 % 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;
}
// 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 */
}
# 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
// 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 % 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 */
<script>
// Javascript program to accept an amount
// and count number of notes
// function to count and
// print currency notes
function countCurrency(amount)
{
let notes = [ 2000, 500, 200, 100, 50, 20, 10, 5, 1 ];
let noteCounter = Array(9).fill(0);
// count notes using Greedy approach
for (let i = 0; i < 9; i++) {
if (amount >= notes[i]) {
noteCounter[i] = Math.floor(amount / notes[i]);
amount = amount % notes[i];
}
}
// Print notes
document.write("Currency Count ->" + "<br/>");
for (let i = 0; i < 9; i++) {
if (noteCounter[i] != 0) {
document.write(notes[i] + " : "
+ noteCounter[i] + "<br/>");
}
}
}
// driver code
let amount = 868;
countCurrency(amount);
</script>
<?php
// PHP program to accept an amount
// and count number of notes
// function to count and
// print currency 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 % $notes[$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
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.