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Find sum of digits in factorial of a number

Sum of Digits of a Number

Last Updated : 07 Feb, 2025

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Given a number n, find the sum of its digits.

Examples :

Input: n = 687
Output: 21
Explanation: The sum of its digits are: 6 + 8 + 7 = 21

Input: n = 12
Output: 3
Explanation: The sum of its digits are: 1 + 2 = 3

Table of Content

Iterative Approach

The idea is to add the digits starting from the rightmost (least significant) digit and moving towards the leftmost (most significant) digit. This can be done by repeatedly extracting the last digit from the number using modulo 10 operation, adding it to the sum, and then removing it using integer division by 10. For eg: n = 1567, then 1567 / 10 = 156.7 = 156(Integer Division).

C++

// Iterative C++ Code to find sum of digits  #include <iostream> using namespace std;  int sumOfDigits(int n) {     int sum = 0;     while (n != 0) {          // Extract the last digit         int last = n % 10;          // Add last digit to sum         sum += last;          // Remove the last digit         n /= 10;     }     return sum; }  int main() {     int n = 12345;     cout << sumOfDigits(n);     return 0; }

C

// Iterative C Code to find sum of digits  int sumOfDigits(int n) {     int sum = 0;     while (n != 0) {          // Extract the last digit         int last = n % 10;          // Add last digit to sum         sum += last;          // Remove the last digit         n /= 10;     }     return sum; }  int main() {     int n = 12345;     printf("%d", sumOfDigits(n));     return 0; }

Java

// Iterative Java Code to find sum of digits  class GfG {     static int sumOfDigits(int n) {         int sum = 0;         while (n != 0) {              // Extract the last digit             int last = n % 10;              // Add last digit to sum             sum += last;              // Remove the last digit             n /= 10;         }         return sum;     }      public static void main(String[] args) {         int n = 12345;         System.out.println(sumOfDigits(n));     } }

Python

# Iterative Python Code to find sum of digits  def sumOfDigits(n):     sum = 0     while n != 0:          # Extract the last digit         last = n % 10          # Add last digit to sum         sum += last          # Remove the last digit         n //= 10     return sum  if __name__ == "__main__": 	n = 12345 	print(sumOfDigits(n))

C#

// Iterative C# Code to find sum of digits  using System; class GfG {     static int sumOfDigits(int n) {         int sum = 0;         while (n != 0) {              // Extract the last digit             int last = n % 10;              // Add last digit to sum             sum += last;              // Remove the last digit             n /= 10;         }         return sum;     }      static void Main() {         int n = 12345;         Console.WriteLine(sumOfDigits(n));     } }

JavaScript

// Iterative JavaScript Code to find sum of digits  function sumOfDigits(n) {     let sum = 0;     while (n !== 0) {          // Extract the last digit         let last = n % 10;          // Add last digit to sum         sum += last;          // Remove the last digit         n = Math.floor(n / 10);     }     return sum; }  // Driver Code let n = 12345; console.log(sumOfDigits(n));

Output

Time Complexity: O(log₁₀n), as we are iterating over all the digits.
Auxiliary Space: O(1)

Recursive Approach

We can also use recursion to find the sum of digits. The idea is to extract the last digit. add it to the sum, and recursively call the function with the remaining number (after removing the last digit).

Base Case: If the number is 0, return 0.
Recursive Case: Extract the last digit and add it to the result of recursive call with n / 10.

C++

// Recursive C++ Code to find sum of digits  #include <iostream> using namespace std;  int sumOfDigits(int n) {          // Base Case     if (n == 0)         return 0;      // Recursive Case     return (n % 10) + sumOfDigits(n / 10); }  int main() {     cout << sumOfDigits(12345);     return 0; }

C

// Recursive C Code to find sum of digits  #include <stdio.h> int sumOfDigits(int n) {          // Base Case      if (n == 0)         return 0;      // Recursive Case     return (n % 10) + sumOfDigits(n / 10); }  int main() {     printf("%d", sumOfDigits(12345));     return 0; }

Java

// Recursive Java Code to find sum of digits  import java.io.*;  class GfG {     static int sumOfDigits(int n) {                  // Base Case         if (n == 0)             return 0;          // Recursive Case         return (n % 10) + sumOfDigits(n / 10);     }      public static void main(String[] args) {         System.out.println(sumOfDigits(12345));     } }

Python

# Recursive Python Code to find sum of digits  def sumOfDigits(n):          # Base Case     if n == 0:         return 0        # Recursive Case     return n % 10 + sumOfDigits(n // 10)   if __name__ == "__main__":     print(sumOfDigits(12345))

C#

// Recursive C# Code to find sum of digits  using System;  class GfG {     static int sumOfDigits(int n) {                  // Base Case         if(n == 0)             return 0;                // Recursive Case         return n % 10 + sumOfDigits(n / 10);     }      static void Main() {         Console.Write(sumOfDigits(12345));     } }

JavaScript

// Recursive JavaScript Code to find sum of digits  function sumOfDigits(n) {          // Base Case     if (n == 0)         return 0;      // Recursive Case     return (n % 10) + sumOfDigits(Math.floor(n / 10)); }  // Driver code console.log(sumOfDigits(12345));

Output

Time Complexity: O(log₁₀n), as we are iterating over all the digits.
Auxiliary Space: O(log₁₀n) because of function call stack.

Taking Input Number as String

The idea is to take the input number as a string and then iterate over all the characters(digits) to find the sum of digits. To find the actual value of a digit from a character, subtract the ASCII value of ‘0’ from the character.

This approach is used when the input number is so large that it cannot be stored in integer data types.

C++

// C++ Code to find the sum of digits by // taking the input number as string  #include <iostream> using namespace std;  int sumOfDigits(string &s) {     int sum = 0;     for (int i = 0; i < s.length(); i++) {                // Extract digit from character         int digit = s[i] - '0';                // Add digit to the sum         sum = sum + digit;     }     return sum; }  int main() {     string s = "123456789123456789123422";      cout << sumOfDigits(s);     return 0; }

C

// C Code to find the sum of digits by // taking the input number as string  #include <string.h> int sumOfDigits(char *s) {     int sum = 0;     for (int i = 0; i < strlen(s); i++) {                // Extract digit from character         int digit = s[i] - '0';                // Add digit to the sum         sum = sum + digit;     }     return sum; }  int main() {     char s[] = "123456789123456789123422";      printf("%d", sumOfDigits(s));     return 0; }

Java

// Java Code to find the sum of digits by // taking the input number as string  class GfG {     static int sumOfDigits(String s) {         int sum = 0;         for (int i = 0; i < s.length(); i++) {                        // Extract digit from character             int digit = s.charAt(i) - '0';                        // Add digit to the sum             sum = sum + digit;         }         return sum;     }      public static void main(String[] args) {         String s = "123456789123456789123422";         System.out.println(sumOfDigits(s));     } }

Python

# Python Code to find the sum of digits by # taking the input number as string  def sumOfDigits(s):     sum = 0     for i in range(len(s)):                # Extract digit from character         digit = ord(s[i]) - ord('0')                 # Add digit to the sum         sum = sum + digit     return sum  if __name__ == "__main__":     s = "123456789123456789123422"     print(sumOfDigits(s))

C#

// C# Code to find the sum of digits by // taking the input number as string  using System;  class GfG {     static int sumOfDigits(string s) {         int sum = 0;         for (int i = 0; i < s.Length; i++) {                        // Extract digit from character             int digit = s[i] - '0';                        // Add digit to the sum             sum = sum + digit;         }         return sum;     }      static void Main() {         string s = "123456789123456789123422";          Console.WriteLine(sumOfDigits(s));     } }

JavaScript

// JavaScript Code to find the sum of digits by // taking the input number as string  function sumOfDigits(s) {     let sum = 0;     for (let i = 0; i < s.length; i++) {                // Extract digit from character         let digit = s[i] - '0';                // Add digit to the sum         sum = sum + digit;     }     return sum; }  // Driver Code let s = "123456789123456789123422"; console.log(sumOfDigits(s));

Output

Time Complexity: O(len(s)), as we are iterating over all the characters(digits).
Auxiliary Space: O(1)

Find sum of digits in factorial of a number

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