Given a number, we need to find sum of its digits using recursion.
Examples:
Input : 12345 Output : 15 Input : 45632 Output :20
The step-by-step process for a better understanding of how the algorithm works.
Let the number be 12345.
Step 1-> 12345 % 10 which is equal-too 5 + ( send 12345/10 to next step )
Step 2-> 1234 % 10 which is equal-too 4 + ( send 1234/10 to next step )
Step 3-> 123 % 10 which is equal-too 3 + ( send 123/10 to next step )
Step 4-> 12 % 10 which is equal-too 2 + ( send 12/10 to next step )
Step 5-> 1 % 10 which is equal-too 1 + ( send 1/10 to next step )
Step 6-> 0 algorithm stops
following diagram will illustrate the process of recursion
// Recursive C++ program to find sum of digits // of a number #include <bits/stdc++.h> using namespace std;
// Function to check sum of digit using recursion int sum_of_digit( int n)
{ if (n == 0)
return 0;
return (n % 10 + sum_of_digit(n / 10));
} // Driven code int main()
{ int num = 12345;
int result = sum_of_digit(num);
cout << "Sum of digits in " << num
<< " is " <<result << endl;
return 0;
} // THis code is contributed by // SHUBHAMSINGH10 |
// Recursive C program to find sum of digits // of a number #include <stdio.h> // Function to check sum of digit using recursion int sum_of_digit( int n)
{ if (n == 0)
return 0;
return (n % 10 + sum_of_digit(n / 10));
} // Driven Program to check above int main()
{ int num = 12345;
int result = sum_of_digit(num);
printf ( "Sum of digits in %d is %d\n" , num, result);
return 0;
} |
// Recursive java program to // find sum of digits of a number import java.io.*;
class sum_of_digits
{ // Function to check sum
// of digit using recursion
static int sum_of_digit( int n)
{
if (n == 0 )
return 0 ;
return (n % 10 + sum_of_digit(n / 10 ));
}
// Driven Program to check above
public static void main(String args[])
{
int num = 12345 ;
int result = sum_of_digit(num);
System.out.println( "Sum of digits in " +
num + " is " + result);
}
} // This code is contributed by Anshika Goyal. |
# Recursive Python3 program to # find sum of digits of a number # Function to check sum of # digit using recursion def sum_of_digit( n ):
if n = = 0 :
return 0
return (n % 10 + sum_of_digit( int (n / 10 )))
# Driven code to check above num = 12345
result = sum_of_digit(num)
print ( "Sum of digits in" ,num, "is" , result)
# This code is contributed by "Sharad_Bhardwaj". |
// Recursive C# program to // find sum of digits of a number using System;
class GFG {
// Function to check sum
// of digit using recursion
static int sum_of_digit( int n)
{
if (n == 0)
return 0;
return (n % 10 + sum_of_digit(n / 10));
}
// Driven Program to check above
public static void Main()
{
int num = 12345;
int result = sum_of_digit(num);
Console.WriteLine( "Sum of digits in " +
num + " is " + result);
}
} // This code is contributed by Anant Agarwal. |
<?php // Recursive PHP program // to find sum of digits // of a number // Function to check sum of // digit using recursion function sum_of_digit( $n )
{ if ( $n == 0)
return 0;
return ( $n % 10 +
sum_of_digit( $n / 10));
} // Driven Code $num = 12345;
$result = sum_of_digit( $num );
echo ( "Sum of digits in " . $num . " is " . $result );
// This code is contributed by Ajit. ?> |
<script> // Recursive Javascript program to find sum of digits // of a number // Function to check sum of digit using recursion function sum_of_digit(n)
{ if (n == 0)
return 0;
return (n % 10 + sum_of_digit(parseInt(n / 10)));
} // Driven code var num = 12345;
var result = sum_of_digit(num);
document.write( "Sum of digits in " + num
+ " is " +result );
</script> |
Output:
Sum of digits in 12345 is 15
Besides writing (n==0 , then return 0) in the code given above we can also write it in this manner , there will be no change in the output .
if(n<10) return n; By writing this there will be no need to call the function for the numbers which are less than 10
Time complexity : O(logn) where n is the given number.
Auxiliary space : O(logn) due to recursive stack space.