Program for Sum of the digits of a given number

Given a number, find sum of its digits.

Examples : 

Input : n = 687
Output : 21

Input : n = 12
Output : 3

General Algorithm for sum of digits in a given number:

1. Get the number
2. Declare a variable to store the sum and set it to 0
3. Repeat the next two steps till the number is not 0
4. Get the rightmost digit of the number with help of the remainder ‘%’ operator by dividing it by 10 and add it to sum.
5. Divide the number by 10 with help of ‘/’ operator to remove the rightmost digit.
6. Print or return the sum

Below are the solutions to get sum of the digits. 
1. Iterative:

C++

// C program to compute sum of digits in
// number.
#include <iostream>
using namespace std;

/* Function to get sum of digits */
class gfg {
public:
    int getSum(int n)
    {
        int sum = 0;
        while (n != 0) {
            sum = sum + n % 10;
            n = n / 10;
        }
        return sum;
    }
};

// Driver code
int main()
{
    gfg g;
    int n = 687;
    cout << g.getSum(n);
    return 0;
}
// This code is contributed by Soumik

C

// C program to compute sum of digits in
// number.
#include <stdio.h>

/* Function to get sum of digits */
int getSum(int n)
{
    int sum = 0;
    while (n != 0) {
        sum = sum + n % 10;
        n = n / 10;
    }
    return sum;
}

// Driver code
int main()
{
    int n = 687;
    printf(" %d ", getSum(n));
    return 0;
}

Java

// Java program to compute
// sum of digits in number.
import java.io.*;

class GFG {

    /* Function to get sum of digits */
    static int getSum(int n)
    {
        int sum = 0;

        while (n != 0) {
            sum = sum + n % 10;
            n = n / 10;
        }

        return sum;
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 687;

        System.out.println(getSum(n));
    }
}

// This code is contributed by Gitanjali

Python3

# Python 3 program to
# compute sum of digits in
# number.

# Function to get sum of digits


def getSum(n):

    sum = 0
    while (n != 0):

        sum = sum + int(n % 10)
        n = int(n/10)

    return sum


# Driver code
n = 687
print(getSum(n))

C#

// C# program to compute
// sum of digits in number.
using System;

class GFG {
    /* Function to get sum of digits */
    static int getSum(int n)
    {
        int sum = 0;

        while (n != 0) {
            sum = sum + n % 10;
            n = n / 10;
        }

        return sum;
    }

    // Driver code
    public static void Main()
    {
        int n = 687;
        Console.Write(getSum(n));
    }
}

// This code is contributed by Sam007

PHP

<?php
// PHP Code to compute sum
// of digits in number.

// Function to get
// $sum of digits
function getsum($n)
{
    $sum = 0;
    while ($n != 0)
    {
        $sum = $sum + $n % 10;
        $n = $n/10;
    }
    return $sum;
}

// Driver Code
$n = 687;
$res = getsum($n);
echo("$res");

// This code is contributed by
// Smitha Dinesh Semwal.
?>

Javascript

<script>

// Javascript program to compute sum of digits in
// number.

/* Function to get sum of digits */
function getSum(n)
{
    var sum = 0;
    while (n != 0) {
        sum = sum + n % 10;
        n = parseInt(n / 10);
    }
    return sum;
}

// Driver code
var n = 687;
document.write(getSum(n));

</script>

Output

21

Time Complexity : O(logn)

Auxiliary Space: O(1)

How to compute in a single line?
The below function has three lines instead of one line, but it calculates the sum in line. It can be made one-line function if we pass the pointer to sum. 

C++

#include <iostream>
using namespace std;

/* Function to get sum of digits */
class gfg {
public:
    int getSum(int n)
    {
        int sum;

        /* Single line that calculates sum */
        for (sum = 0; n > 0; sum += n % 10, n /= 10)
            ;

        return sum;
    }
};

// Driver code
int main()
{
    gfg g;
    int n = 687;
    cout << g.getSum(n);
    return 0;
}
// This code is contributed by Soumik

C

#include <stdio.h>

/* Function to get sum of digits */
int getSum(int n)
{
    int sum;

    /* Single line that calculates sum */
    for (sum = 0; n > 0; sum += n % 10, n /= 10)
        ;

    return sum;
}

// Driver code
int main()
{
    int n = 687;
    printf(" %d ", getSum(n));
    return 0;
}

Java

// Java program to compute
// sum of digits in number.
import java.io.*;

class GFG {

    /* Function to get sum of digits */
    static int getSum(int n)
    {
        int sum;

        /* Single line that calculates sum */
        for (sum = 0; n > 0; sum += n % 10, n /= 10)
            ;

        return sum;
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 687;

        System.out.println(getSum(n));
    }
}

// This code is contributed by Gitanjali

Python3

# Function to get sum of digits


def getSum(n):

    sum = 0

    # Single line that calculates sum
    while(n > 0):
        sum += int(n % 10)
        n = int(n/10)

    return sum


# Driver code
n = 687
print(getSum(n))

# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# program to compute
// sum of digits in number.
using System;

class GFG {
    static int getSum(int n)
    {
        int sum;

        /* Single line that calculates sum */
        for (sum = 0; n > 0; sum += n % 10, n /= 10)
            ;

        return sum;
    }

    // Driver code
    public static void Main()
    {
        int n = 687;
        Console.Write(getSum(n));
    }
}

// This code is contributed by Sam007

PHP

<?php
// PHP Code for Sum the
// digits of a given number

// Function to get sum of digits
function getsum($n)
{

    // Single line that calculates $sum
    for ($sum = 0; $n > 0; $sum += $n % 10,
                                  $n /= 10);
    return $sum;
}

// Driver Code
$n = 687;
echo(getsum($n));

// This code is contributed by
// Smitha Dinesh Semwal.
?>

Javascript

<script>

// Javascript program to compute
// sum of digits in number.

// Function to get sum of digits
function getSum(n)
{
    let sum;

    // Single line that calculates sum
    for(sum = 0; n > 0;
        sum += n % 10,
        n = parseInt(n / 10))
        ;
    return sum;
}

// Driver code
let n = 687;

document.write(getSum(n));

// This code is contributed by subhammahato348

</script>

Output

21

Time Complexity : O(logn)

Auxiliary Space: O(1)

2. Recursive
Thanks to Ayesha for providing the below recursive solution. 

Algorithm :

1) Get the number
2) Get the remainder and pass the next remaining digits
3) Get the rightmost digit of the number with help of the remainder '%' operator by dividing it by 10 and add it to sum.
   Divide the number by 10 with help of '/' operator to remove the rightmost digit.
4) Check the base case with n = 0
5) Print or return the sum

C++

// C++ program to compute
// sum of digits in number.
#include <iostream>
using namespace std;
class gfg {
public:
    int sumDigits(int no)
    {
        if(no == 0){
          return 0 ;
        }

        return (no % 10) + sumDigits(no / 10) ;
    }
};

// Driver code
int main(void)
{
    gfg g;
    cout << g.sumDigits(687);
    return 0;
}

C

// C program to compute
// sum of digits in number.
#include <stdio.h>

int sumDigits(int no)
{
  if(no == 0){
    return 0 ;
  }

  return (no % 10) + sumDigits(no / 10) ;
}

int main()
{
    printf("%d", sumDigits(687));
    return 0;
}

Java

// Java program to compute
// sum of digits in number.
import java.io.*;

class GFG {

    /* Function to get sum of digits */
    static int sumDigits(int no)
    {
        if(no == 0){
          return 0 ;
        }

        return (no % 10) + sumDigits(no / 10) ;
     }

    // Driver code
    public static void main(String[] args)
    {
        System.out.println(sumDigits(687));
    }
}

// This code is contributed by Gitanjali

Python3

# Python program to compute
# sum of digits in number.


def sumDigits(no):
    return 0 if no == 0 else int(no % 10) + sumDigits(int(no/10))


# Driver code
print(sumDigits(687))

# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# program to compute
// sum of digits in number.
using System;

class GFG {
    /* Function to get sum of digits */
    static int sumDigits(int no)
    {
        return no == 0 ? 0 : no % 10 + sumDigits(no / 10);
    }

    // Driver code
    public static void Main()
    {
        Console.Write(sumDigits(687));
    }
}

// This code is contributed by Sam007

PHP

<?php
// PHP program to compute
// sum of digits in number.
function sumDigits($no)
{
return $no == 0 ? 0 : $no % 10 +
                      sumDigits($no / 10) ;
}

// Driver Code
echo sumDigits(687);

// This code is contributed by aj_36
?>

Javascript

<script>
// Program to compute
// sum of digits in number
  // Function to get sum of digits

     function sumDigits(no)
     {
        if(no == 0){
          return 0 ;
        }

        return (no % 10) + sumDigits(parseInt(no/10)) ;
      }

// Driver code
      document.write(sumDigits(687));

// This is code is contributed by simranarora5sos
</script>

Output

21

Time Complexity : O(logn)

Auxiliary Space: O(logn)

3.Taking input as String

When the number of digits of that number exceeds 10¹⁹ , we can’t take that number as an integer since the range of long long int doesn’t satisfy the given number. So take input as a string, run a loop from start to the length of the string and increase the sum with that character(in this case it is numeric)

Below is the implementation of the above approach

C++14

// C++ implementation of the above approach
#include <iostream>
using namespace std;
int getSum(string str)
{
    int sum = 0;

    // Traversing through the string
    for (int i = 0; i < str.length(); i++) {
        // Since ascii value of
        // numbers starts from 48
        // so we subtract it from sum
        sum = sum + str[i] - 48;
    }
    return sum;
}

// Driver Code
int main()
{
    string st = "123456789123456789123422";
    cout << getSum(st);
    return 0;
}

Java

// Java implementation of the above approach
import java.io.*;
class GFG {

    static int getSum(String str)
    {
        int sum = 0;

        // Traversing through the string
        for (int i = 0; i < str.length(); i++) {

            // Since ascii value of
            // numbers starts from 48
            // so we subtract it from sum
            sum = sum + str.charAt(i) - 48;
        }
        return sum;
    }

    // Driver Code
    public static void main(String[] args)
    {
        String st = "123456789123456789123422";
        System.out.print(getSum(st));
    }
}

// This code is contributed by Dharanendra L V.

Python3

# Python implementation of the above approach
def getSum(n):
    # Initializing sum to 0
    sum = 0
    # Traversing through string
    for i in n:
        # Converting char to int
        sum = sum + int(i)

    return sum


n = "123456789123456789123422"
print(getSum(n))

C#

// C# implementation of the above approach
using System;
public class GFG {
    static int getSum(String str)
    {
        int sum = 0;

        // Traversing through the string
        for (int i = 0; i < str.Length; i++) {

            // Since ascii value of
            // numbers starts from 48
            // so we subtract it from sum
            sum = sum + str[i] - 48;
        }
        return sum;
    }

    // Driver Code
    static public void Main()
    {
        String st = "123456789123456789123422";
        Console.Write(getSum(st));
    }
}

// This code is contributed by Dharanendra L V.

Javascript

<script>
// Javascript implementation of the above approach

function getSum(str)
{
    let sum = 0;

    // Traversing through the string
    for (let i = 0; i < str.length; i++)
    {

        // Since ascii value of
        // numbers starts from 48
        // so we subtract it from sum
        sum = sum + parseInt(str[i]);
    }
    return sum;
}

// Driver Code
let st = "123456789123456789123422";
document.write(getSum(st));

// This code is contributed by subhammahato348.
</script>

PHP

<?php
 // PHP implementation of the above approach
// PHP Code for Sum the
// digits of a given number

// Function to get sum of digits
function getsum($str)
{

      $sum = 0;

  // Traversing through the string
        for ($i = 0; $i<strlen($str); $i++) {

            //Converting char to int
            $sum = $sum + (int)$str[$i];
        }

    return $sum;
}

// Driver Code
$str = "123456789123456789123422";
echo(getsum($str));

// This code is contributed by aadityapburujwale

?>

Output

104

Time Complexity : O(logn)

Auxiliary Space: O(1)

4. Using Tail Recursion 

This problem can also be solved using Tail Recursion. Here is an approach to solving it.

1. Add another variable “Val” to the function and initialize it to ( val = 0 )

2. On every call to the function add the mod value (n%10) to the variable as “(n%10)+val” which is the last digit in n. Along with pass the variable n as n/10. 

3. So on the First call it will have the last digit. As we are passing n/10 as n, It follows until n is reduced to a single digit. 

4. n<10 is the base case so When n < 10, then add the n to the variable as it is the last digit and return the val which will have the sum of digits

C++

// C++ program for the above approach
#include <iostream>
using namespace std;

// Function to check sum of digit using tail recursion
int sum_of_digit(int n, int val)
{
    if (n < 10) {
        val = val + n;
        return val;
    }
    return sum_of_digit(n / 10, (n % 10) + val);
}

// Driver code
int main()
{
    int num = 12345;
    int result = sum_of_digit(num, 0);
    cout << "Sum of digits is " << result;
    return 0;
}

// This code is contributed by subhammahato348

C

// C program for the above approach
#include <stdio.h>

// Function to check sum of digit using tail recursion
int sum_of_digit(int n, int val)
{
    if (n < 10) {
        val = val + n;
        return val;
    }
    return sum_of_digit(n / 10, (n % 10) + val);
}

// Driver code
int main()
{
    int num = 12345;
    int result = sum_of_digit(num, 0);
    printf("Sum of digits is %d", result);
    return 0;
}

// This code is contributed by Sania Kumari Gupta

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class sum_of_digits {

    // Function to check sum
    // of digit using tail recursion
    static int sum_of_digit(int n, int val)
    {
        if (n < 10) {
            val = val + n;
            return val;
        }
        return sum_of_digit(n / 10, (n % 10) + val);
    }

    // Driven Program to check above
    public static void main(String args[])
    {
        int num = 12345;
        int result = sum_of_digit(num, 0);
        System.out.println("Sum of digits is " + result);
    }
}

Python3

# Python3 program for the above approach

# Function to check sum
# of digit using tail recursion
def sum_of_digit(n, val):

    if (n < 10):
        val = val + n
        return val

    return sum_of_digit(n // 10, (n % 10) + val)

# Driver code
num = 12345
result = sum_of_digit(num, 0)

print("Sum of digits is", result)

# This code is contributed by subhammahato348

C#

// C# program for the above approach
using System;

class GFG{

// Function to check sum
// of digit using tail recursion
static int sum_of_digit(int n, int val)
{
    if (n < 10)
    {
        val = val + n;
        return val;
    }
    return sum_of_digit(n / 10, (n % 10) + val);
}

// Driver code
public static void Main()
{
    int num = 12345;
    int result = sum_of_digit(num, 0);

    Console.Write("Sum of digits is " + result);
}
}

// This code is contributed by subhammahato348

Javascript

<script>

// Javascript program for the above approach

// Function to check sum
// of digit using tail recursion
function sum_of_digit(n, val)
{
    if (n < 10)
    {
        val = val + n;
        return val;
    }
    return sum_of_digit(parseInt(n / 10),
    (n % 10) + val);
}

// Driver code
    let num = 12345;
    let result = sum_of_digit(num, 0);

    document.write("Sum of digits is " + result);

// This code is contributed by subhammahato348

</script>

Output

Sum of digits is 15

Time Complexity : O(logn)

Auxiliary Space: O(logn)

Please write comments if you find the above codes/algorithms incorrect, or find better ways to solve the same problem.