Find minimum number of currency notes and values that sum to given amount (original) (raw)

Last Updated : 25 Oct, 2024

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 <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

// 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

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#

// 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 */

JavaScript

PHP

`

**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.