# Program for Sum of the digits of a given number

• Difficulty Level : Easy
• Last Updated : 04 Aug, 2022

Given a number, find sum of its digits.

Examples : ```Input : n = 687
Output : 21

Input : n = 12
Output : 3```
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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 ` `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 `   `/* 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

 ``

## Javascript

 ``

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 ` `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 `   `/* 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

 ` 0; ``\$sum` `+= ``\$n` `% 10, ` `                                  ``\$n` `/= 10);` `    ``return` `\$sum``;` `}`   `// Driver Code` `\$n` `= 687;` `echo``(getsum(``\$n``));`   `// This code is contributed by ` `// Smitha Dinesh Semwal.` `?>`

## Javascript

 ``

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 ` `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 `   `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

 ``

## Javascript

 ``

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 ` `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

 ``

## PHP

 ``

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 ` `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 `   `// 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

 ``

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.

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