**Problem Statement: **Given an integer, find the sum of digits of that integer.

**Examples:**

Example 1:Input:N = 472Output:13Explanation:The digits in the number are 4,7 and 2. Summing them up gives us a value of 13Example 2:Input:N = 102Output: 3Explanation:The digits in the number are 0, 1, and 2. Summing them up gives us a value of 3

**Solution:**

** Disclaimer**:

*Don’t jump directly to the solution, try it out yourself first.*

**Intuition: **We can see that we somehow need to extract the individual digits of a number and sum them up. Is there any operator that we know of that can be used to extract digits from a number?

**Modulus operation(%) **: 10%3 = 1(Modulus operation gives us the remainder when we divide a fraction)

**Divisor operation(/) **: 10/3 = 3(Divisor operation gives us the quotient when we divide a fraction)

**Note:** Did you notice we get the last digit of a number when we perform a modulus operation with a value of 10

**Approach: **

- Get the last digit of the number
- Add the last digit to a sum variable
- Divide the number by 10 and repeat the process until the number is not zero

**Code:**

## C++ Code

```
#include<bits/stdc++.h>
using namespace std;
int getSum(int n)
{
int d,sum=0;
while(n!=0)
{
d=n%10;
sum+=d;
n=n/10;
}
return sum;
}
int main()
{
int n=472;
cout<<"Sum of digits of the given number is "<<getSum(n);
return 0;
}
```

**Output:** Sum of digits of the given number is 13

**Time Complexity: O(N), N is the no. of digits**

**Space Complexity: O(1)**

## Java Code

```
import java.io.*;
class Test
{
static private int getSum(int n)
{
int d;
int sum = 0;
while (n != 0)
{
d = n % 10;
sum += d;
n = n / 10;
}
return sum;
}
public static void main(String[] args)
{
int n = 472;
System.out.print("Sum of digits of the given number is "+getSum(n));
}
}
```

**Output:** Sum of digits of the given number is 13

**Time Complexity: O(N), N is the no. of digits**

**Space Complexity: O(1)**

Special thanks toplease check out this articleNaman Dagafor contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam,