Sum of Digits of a Number

Last Updated : 18 Jun, 2025

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

[Approach 1] Digit Extraction - O(log₁₀n) Time and O(1) Space

We can sum the digits of a number by repeatedly extracting the last digit using n % 10, adding it to the sum, and then removing it by dividing n by 10 using integer division.

C++

#include <iostream>
using namespace std;

int sumOfDig(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 << sumOfDig(n);
    return 0;
}

#include<stdio.h>

int sumOfDig(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", sumOfDig(n));
    return 0;
}

Java

class GfG {
    static int sumOfDig(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(sumOfDig(n));
    }
}

Python

def sumOfDig(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(sumOfDig(n))

using System;
class GfG {
    static int sumOfDig(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(sumOfDig(n));
    }
}

JavaScript

function sumOfDig(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;
}

let n = 12345;
console.log(sumOfDig(n));

Output

[Approach 2] Using Recursion - O(log₁₀n) Time and O(log₁₀n) Space

We can use recursion to find the sum of digits. The idea is to extract the last digit, add it to the sum of digits of the remaining number, and repeat.
Base Case: If the number is 0, return 0.
Recursive Case: Return (n % 10) + sumOfDig(n / 10)

C++

#include <iostream>
using namespace std;

int sumOfDig(int n) {
    
    // Base Case
    if (n == 0)
        return 0;

    // Recursive Case
    return (n % 10) + sumOfDig(n / 10);
}

int main() {
    cout << sumOfDig(12345);
    return 0;
}

#include <stdio.h>
int sumOfDig(int n) {
    
    // Base Case 
    if (n == 0)
        return 0;

    // Recursive Case
    return (n % 10) + sumOfDig(n / 10);
}

int main() {
    printf("%d", sumOfDig(12345));
    return 0;
}

Java

import java.io.*;

class GfG {
    static int sumOfDig(int n) {
        
        // Base Case
        if (n == 0)
            return 0;

        // Recursive Case
        return (n % 10) + sumOfDig(n / 10);
    }

    public static void main(String[] args) {
        System.out.println(sumOfDig(12345));
    }
}

Python

def sumOfDig(n):
    # Base Case
    if n == 0:
        return 0
  
    # Recursive Case
    return n % 10 + sumOfDig(n // 10)

if __name__ == "__main__":
    print(sumOfDig(12345))

using System;

class GfG {
    static int sumOfDig(int n) {
        
        // Base Case
        if(n == 0)
            return 0;
      
        // Recursive Case
        return n % 10 + sumOfDig(n / 10);
    }

    static void Main() {
        Console.Write(sumOfDig(12345));
    }
}

JavaScript

function sumOfDig(n) {
    
    // Base Case
    if (n == 0)
        return 0;

    // Recursive Case
    return (n % 10) + sumOfDig(Math.floor(n / 10));
}

console.log(sumOfDig(12345));

Output

[Approach 3] String Conversion

Convert the number to a string and iterate through each character (digit). For each character, subtract the ASCII value of '0' to get the actual digit, then add it to the sum.

Note: This method is especially useful when the number is too large to fit in standard integer types.

C++

#include <iostream>
#include <string>
using namespace std;

// Function to calculate sum of digits using string conversion

int sumOfDig(int n) {
    // Convert number to string
    string s = to_string(n);  
    int sum = 0;

    // Loop through each character, convert to digit, and add to sum
    for (char ch : s) {
        sum += ch - '0';
    }

    return sum;
}

int main() {
    int n = 12345;
    cout << sumOfDig(n) << endl;
    return 0;
}

#include <stdio.h>
#include <string.h>

// Function to calculate sum of digits using string conversion
int sumOfDig(int n) {
    // Convert number to string
    char s[20];
    sprintf(s, "%d", n);
    int sum = 0;

    // Loop through each character, convert to digit, and add to sum
    for (int i = 0; i < strlen(s); i++) {
        sum += s[i] - '0';
    }

    return sum;
}

int main() {
    int n = 12345;
    printf("%d\n", sumOfDig(n));
    return 0;
}

Java

class GfG {

    // Function to calculate sum of digits using string conversion
    static int sumOfDig(int n) {
        // Convert number to string
        String s = Integer.toString(n);
        int sum = 0;

        // Loop through each character, convert to digit, and add to sum
        for (char ch : s.toCharArray()) {
            sum += ch - '0';
        }

        return sum;
    }

    public static void main(String[] args) {
        int n = 12345;
        System.out.println(sumOfDig(n));
    }
}

Python

# Function to calculate sum of digits using string conversion
def sumOfDig(n):
    # Convert number to string
    s = str(n)
    sum = 0

    # Loop through each character, convert to digit, and add to sum
    for ch in s:
        sum += int(ch)

    return sum

n = 12345
print(sumOfDig(n))

using System;

class GfG
{
    // Function to calculate sum of digits using string conversion
    public static int sumOfDigits(int n)
    {
        // Convert number to string
        string s = n.ToString();
        int sum = 0;

        // Loop through each character, convert to digit, and add to sum
        foreach (char ch in s)
        {
            sum += ch - '0';
        }

        return sum;
    }

    public static void Main()
    {
        int n = 12345;
        Console.WriteLine(sumOfDigits(n));
    }
}

JavaScript

function sumOfDig(n) {
    // Convert number to string
    let s = n.toString();
    let sum = 0;

    // Loop through each character, convert to digit, and add to sum
    for (let ch of s) {
        sum += parseInt(ch);
    }

    return sum;
}


let n = 12345;
console.log(sumOfDig(n));

Output

Time Complexity: O(d) – we iterate over each of the d digits, where d ≈ log₁₀(n) (count of digits)
Auxiliary Space: O(d) - to store all d digits as characters.

Sum of Digits of a Number

kartik

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Practice Tags :

Sum of Digits of a Number

[Approach 1] Digit Extraction - O(log10n) Time and O(1) Space

[Approach 2] Using Recursion - O(log10n) Time and O(log10n) Space

[Approach 3] String Conversion

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[Approach 1] Digit Extraction - O(log₁₀n) Time and O(1) Space

[Approach 2] Using Recursion - O(log₁₀n) Time and O(log₁₀n) Space