# Coding Sample Questions | Part 4 | Codewindow.in

Problem Statement :

Find the minimum number of coins required to form any value between 1 to N, both inclusive. Cumulative value of coins
should not exceed N. Coin denominations are 1 Rupee, 2
Rupee and 5 Rupee.
Let’s understand the problem using the following example.
Consider the value of N is 13, then the minimum number of coins required to formulate any value between 1 and 13, is 6.
One 5 Rupee, three 2 Rupee and two 1 Rupee coins are required to realize any value between 1 and 13. Hence this is
However, if one takes two 5 Rupee coins, one 2 rupee coins and two 1 rupee coins, then to all values between 1 and 13 are achieved. But since the cumulative value of all coins equals 14, i.e., exceeds 13, this is not the answer.

Input Format:
A single integer value.

Output Format:
Four Space separated Integer Values
1st – Total Number of coins
2nd – number of 5 Rupee coins.
3rd – number of 2 Rupee coins.
4th – number of 1 Rupee coins.

Constraints:
0<n<1000

Sample Input 1 :
13
Sample Output 1 :
6 1 3 2

Solution:

C

```//www.codewindow.in
//start
#include<stdio.h>
int main(){
int number,one,two,five;
scanf("%d",&number);

five = ((number-4)/5);

if(((number-5*five) % 2) == 0)
one=2;
else
one=1;

two=(number-5*five-one)/2 ;
printf("%d %d %d %d",one+two+five,five,two,one);

return 0;
}

//end
```

JAVA

```//www.codewindow.in
//start
import java.util.*;
public class Main
{
public static void main(String []args)
{
int number,one,two,five;
Scanner ip=new Scanner(System.in);

number=ip.nextInt();

five = ((number-4)/5);

if(((number-5*five) % 2) == 0)
one=2;
else
one=1;
two=(number-5*five-one)/2 ;

System.out.printf("%d %d %d %d",one+two+five,five,two,one);
}
}

//end
```

Python

```#www.codewindow.in

number = int(input())
five = int((number-4)/5)

if((number-5*five) % 2) == 0:
one=2
else:
one=1
two=(number-5*five-one)//2
print(one+two+five,five,two,one)

#end
```