Hello Everyone,
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:
 Get the number
 Declare a variable to store the sum and set it to 0
 Repeat the next two steps till the number is not 0
 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.
 Print or return the sum
Below are the solutions to get sum of the digits.
1. Iterative:
// 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;
}
Output
21
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 oneline function if we pass the pointer to sum.
#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;
}
Output
21
2. Recursive
// C program to compute
// sum of digits in number.
#include <stdio.h>
using
namespace
std;
int
sumDigits(
int
no)
{
return
no == 0 ? 0 : no % 10 + sumDigits(no / 10);
}
int
main(
void
)
{
printf
(
"%d"
, sumDigits(687));
return
0;
}
Output
21
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++ 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;
}
Output
104
 Using Tail Recursion
This problem can also be solved using Tail Recursion. Here is an approach to solving it.

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

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.

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.

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++ 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;
}
Output
Sum of digits is 15