Sum of digit of a number using recursion

Given a number, we need to find sum of its digits using recursion.

Examples: 

Input: 12345
Output: 15
Explanation: Sum of digits → 1 + 2 + 3 + 4 + 5 = 15

Input: 45632
Output: 20

Approach:

To understand the algorithm, consider the number 12345 and refer to the illustration below..

• Extract the last digit: 12345 % 10 = 5, pass 1234 to the next step.
• Extract next digit: 1234 % 10 = 4, pass 123 to the next step.
• Extract next digit: 123 % 10 = 3, pass 12 to the next step.
• Extract next digit: 12 % 10 = 2, pass 1 to the next step.
• Extract last digit: 1 % 10 = 1, pass 0, stopping recursion.

Each step adds the extracted digit to the sum, forming 1 + 2 + 3 + 4 + 5 = 15 

Note: Instead of if (n == 0) return 0;, we can use if (n < 10) return n;, eliminating extra function calls for single-digit numbers without changing the output.

// 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 << result << endl;   return 0;  }  

// 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("%d", result);  return 0; } 

import java.io.*; 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(String args[])  {  int num = 12345;  int result = sum_of_digit(num);  System.out.println(result);  } } 

# 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(result) 

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(result);  } } 

// 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);  console.log(result );  

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 <1 0) 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(log₁₀n), Traverse through all the digits, as there are log₁₀n bits.
Auxiliary Space: O(log₁₀n), due to recursive function calls stored in the call stack.