Skip to content
Related Articles

Related Articles

Find out the minimum number of coins required to pay total amount
  • Last Updated : 25 Apr, 2019
GeeksforGeeks - Summer Carnival Banner

Given a total amount of N and unlimited number of coins worth 1,  10 and 25 currency coins. Find out the minimum number of coins you need to use to pay exactly amount N.

Examples:

Input : N = 14
Output : 5
You will use one coing of value 10 and 
four coins of value 1.

Input :  N = 88
Output : 7

Approach:
There are three different cases:

  1. If value of N < 10, then coins that have value 1 can only be used for payment.
  2. When N > 9 and < 25, then coins that have value 1 and 10 will be used for payment. Here, to minimize the number of coins used, coins with value 10 will be preferred mostly.
  3. When N > 24. Then all coins of value 1, 10 and 25 will be used for payment. To minimize the number of coins, the primary aim will be to use coin with value 25 first as much as possible then coin with value 10 and then with value 1.

Below is the implementation of the above approach:

C++




// C++ program to find the minimum number
// of coins required
  
#include <iostream>
using namespace std;
  
// Function to find the minimum number
// of coins required
int countCoins(int n)
{
    int c = 0;
  
    if (n < 10) {
        // counts coins which have value 1
        // we will need n coins of value 1
        return n;
    }
    if (n > 9 && n < 25) {
        // counts coins which have value 1 and 10
        c = n / 10 + n % 10;
        return c;
    }
    if (n > 24) {
        // counts coins which have value 25
        c = n / 25;
  
        if (n % 25 < 10) {
  
            // counts coins which have value 1 and 25
            c = c + n % 25;
            return c;
        }
  
        if (n % 25 > 9) {
            // counts coins which have value 1, 10 and 25
            c = c + (n % 25) / 10 + (n % 25) % 10;
            return c;
        }
    }
}
  
// Driver Code
int main()
{
    int n = 14;
  
    cout << countCoins(n);
  
    return 0;
}

Java




// Java program to find the minimum number
// of coins required
  
class GFG {
  
    // Function to find the minimum number
    // of coins required
    static int countCoins(int n)
    {
        int c = 0;
        if (n < 10) {
            // counts coins which have value 1
            // we will need n coins of value 1
            return n;
        }
        if (n > 9 && n < 25) {
  
            // counts coins which have value 1 and 10
            c = n / 10 + n % 10;
  
            return c;
        }
        if (n > 24) {
            // counts coins which have value 25
            c = n / 25;
  
            if (n % 25 < 10) {
                // counts coins which have value 1 and 25
                c = c + n % 25;
  
                return c;
            }
            if (n % 25 > 9) {
                // counts coins which have value 1, 10 and 25
                c = c + (n % 25) / 10 + (n % 25) % 10;
  
                return c;
            }
        }
  
        return c;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int n = 14;
        System.out.println(countCoins(n));
    }
}

Python3




# Python3 program to find the minimum number
# of coins required
  
# Function to find the minimum number
# of coins required
def countCoins(n):
  
    c = 0
  
    if (n < 10):
          
        # counts coins which have value 1
        # we will need n coins of value 1
        return n
      
    if (n > 9 and n < 25):
          
        # counts coins which have value 1 and 10
        c = n // 10 + n % 10
        return c
  
    if (n > 24):
          
        # counts coins which have value 25
        c = n // 25
  
        if (n % 25 < 10):
  
            # counts coins which have value
            # 1 and 25
            c = c + n % 25
            return c
  
        if (n % 25 > 9):
              
            # counts coins which have value
            # 1, 10 and 25
            c = (c + (n % 25) // 10 +
                     (n % 25) % 10)
            return c
  
# Driver Code
n = 14
  
print(countCoins(n))
  
# This code is contributed by mohit kumar

C#




// C# program to find the minimum number 
// of coins required 
using System;
  
class GFG
  
    // Function to find the minimum number 
    // of coins required 
    static int countCoins(int n) 
    
        int c = 0; 
        if (n < 10) 
        
            // counts coins which have value 1 
            // we will need n coins of value 1 
            return n; 
        
        if (n > 9 && n < 25)
        
  
            // counts coins which have value 1 and 10 
            c = n / 10 + n % 10; 
  
            return c; 
        
        if (n > 24)
        
            // counts coins which have value 25 
            c = n / 25; 
  
            if (n % 25 < 10) 
            
                // counts coins which have value 1 and 25 
                c = c + n % 25; 
  
                return c; 
            
            if (n % 25 > 9)
            
                // counts coins which have value 1, 10 and 25 
                c = c + (n % 25) / 10 + (n % 25) % 10; 
  
                return c; 
            
        
  
        return c; 
    
  
    // Driver Code 
    public static void Main() 
    
        int n = 14; 
        Console.WriteLine(countCoins(n)); 
    }
  
// This code is contributed by Ryuga

PHP




<?php
// PHP program to find the minimum number
// of coins required
  
// Function to find the minimum number
// of coins required
function countCoins($n)
{
    $c = 0;
    if ($n < 10)
    {
        // counts coins which have value 1
        // we will need n coins of value 1
        return $n;
    }
      
    if ($n > 9 && $n < 25) 
    {
  
        // counts coins which have value 1 and 10
        $c = (int)($n / 10 + $n % 10);
  
        return $c;
    }
      
    if ($n > 24) 
    {
        // counts coins which have value 25
        $c = (int)($n / 25);
  
        if ($n % 25 < 10) 
        {
              
            // counts coins which have value 1 and 25
            $c = $c + $n % 25;
  
            return $c;
        }
        if ($n % 25 > 9)
        {
            // counts coins which have value 1, 10 and 25
            $c = $c + ($n % 25) / 10 + ($n % 25) % 10;
  
            return $c;
        }
    }
  
    return $c;
}
  
// Driver Code
$n = 14;
echo(countCoins($n));
  
// This code is contributed Code_Mech
Output:
5

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up
Recommended Articles
Page :