Given a number, find sum of its digits.
Examples :
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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++
#include <iostream>
using namespace std;
class gfg {
public:
int getSum(int n)
{
int sum = 0;
while (n != 0) {
sum = sum + n % 10;
n = n / 10;
}
return sum;
}
};
int main()
{
gfg g;
int n = 687;
cout << g.getSum(n);
return 0;
}
|
C
#include <stdio.h>
int getSum(int n)
{
int sum = 0;
while (n != 0) {
sum = sum + n % 10;
n = n / 10;
}
return sum;
}
int main()
{
int n = 687;
printf(" %d ", getSum(n));
return 0;
}
|
Java
import java.io.*;
class GFG {
static int getSum(int n)
{
int sum = 0;
while (n != 0) {
sum = sum + n % 10;
n = n / 10;
}
return sum;
}
public static void main(String[] args)
{
int n = 687;
System.out.println(getSum(n));
}
}
|
Python3
def getSum(n):
sum = 0
while (n != 0):
sum = sum + int(n % 10)
n = int(n/10)
return sum
n = 687
print(getSum(n))
|
C#
using System;
class GFG {
static int getSum(int n)
{
int sum = 0;
while (n != 0) {
sum = sum + n % 10;
n = n / 10;
}
return sum;
}
public static void Main()
{
int n = 687;
Console.Write(getSum(n));
}
}
|
PHP
<?php
function getsum($n)
{
$sum = 0;
while ($n != 0)
{
$sum = $sum + $n % 10;
$n = $n/10;
}
return $sum;
}
$n = 687;
$res = getsum($n);
echo("$res");
?>
|
Javascript
<script>
function getSum(n)
{
var sum = 0;
while (n != 0) {
sum = sum + n % 10;
n = parseInt(n / 10);
}
return sum;
}
var n = 687;
document.write(getSum(n));
</script>
|
How to compute in a single line?
The below function has three lines instead of one line, but it calculates the sum in line. It can be made one-line function if we pass the pointer to sum.
C++
#include <iostream>
using namespace std;
class gfg {
public:
int getSum(int n)
{
int sum;
for (sum = 0; n > 0; sum += n % 10, n /= 10)
;
return sum;
}
};
int main()
{
gfg g;
int n = 687;
cout << g.getSum(n);
return 0;
}
|
C
#include <stdio.h>
int getSum(int n)
{
int sum;
for (sum = 0; n > 0; sum += n % 10, n /= 10)
;
return sum;
}
int main()
{
int n = 687;
printf(" %d ", getSum(n));
return 0;
}
|
Java
import java.io.*;
class GFG {
static int getSum(int n)
{
int sum;
for (sum = 0; n > 0; sum += n % 10, n /= 10)
;
return sum;
}
public static void main(String[] args)
{
int n = 687;
System.out.println(getSum(n));
}
}
|
Python3
def getSum(n):
sum = 0
while(n > 0):
sum += int(n % 10)
n = int(n/10)
return sum
n = 687
print(getSum(n))
|
C#
using System;
class GFG {
static int getSum(int n)
{
int sum;
for (sum = 0; n > 0; sum += n % 10, n /= 10)
;
return sum;
}
public static void Main()
{
int n = 687;
Console.Write(getSum(n));
}
}
|
PHP
<?php
function getsum($n)
{
for ($sum = 0; $n > 0; $sum += $n % 10,
$n /= 10);
return $sum;
}
$n = 687;
echo(getsum($n));
?>
|
Javascript
<script>
function getSum(n)
{
let sum;
for(sum = 0; n > 0;
sum += n % 10,
n = parseInt(n / 10))
;
return sum;
}
let n = 687;
document.write(getSum(n));
</script>
|
2. Recursive
Thanks to Ayesha for providing the below recursive solution.
C++
#include <iostream>
using namespace std;
class gfg {
public:
int sumDigits(int no)
{
return no == 0 ? 0 : no % 10 + sumDigits(no / 10);
}
};
int main(void)
{
gfg g;
cout << g.sumDigits(687);
return 0;
}
|
C
#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;
}
|
Java
import java.io.*;
class GFG {
static int sumDigits(int no)
{
return no == 0 ? 0 : no % 10 + sumDigits(no / 10);
}
public static void main(String[] args)
{
System.out.println(sumDigits(687));
}
}
|
Python3
def sumDigits(no):
return 0 if no == 0 else int(no % 10) + sumDigits(int(no/10))
print(sumDigits(687))
|
C#
using System;
class GFG {
static int sumDigits(int no)
{
return no == 0 ? 0 : no % 10 + sumDigits(no / 10);
}
public static void Main()
{
Console.Write(sumDigits(687));
}
}
|
PHP
<?php
function sumDigits($no)
{
return $no == 0 ? 0 : $no % 10 +
sumDigits($no / 10) ;
}
echo sumDigits(687);
?>
|
Javascript
<script>
function sumDigits(no)
{
return no == 0 ? 0 : no % 10+ sumDigits(parseInt(no/10));
}
document.write(sumDigits(687));
</script>
|
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
#include <iostream>
using namespace std;
int getSum(string str)
{
int sum = 0;
for (int i = 0; i < str.length(); i++) {
sum = sum + str[i] - 48;
}
return sum;
}
int main()
{
string st = "123456789123456789123422";
cout << getSum(st);
return 0;
}
|
Java
import java.io.*;
class GFG {
static int getSum(String str)
{
int sum = 0;
for (int i = 0; i < str.length(); i++) {
sum = sum + str.charAt(i) - 48;
}
return sum;
}
public static void main(String[] args)
{
String st = "123456789123456789123422";
System.out.print(getSum(st));
}
}
|
Python3
def getSum(n):
sum = 0
for i in n:
sum = sum + int(i)
return sum
n = "123456789123456789123422"
print(getSum(n))
|
C#
using System;
public class GFG {
static int getSum(String str)
{
int sum = 0;
for (int i = 0; i < str.Length; i++) {
sum = sum + str[i] - 48;
}
return sum;
}
static public void Main()
{
String st = "123456789123456789123422";
Console.Write(getSum(st));
}
}
|
Javascript
<script>
function getSum(str)
{
let sum = 0;
for (let i = 0; i < str.length; i++)
{
sum = sum + parseInt(str[i]);
}
return sum;
}
let st = "123456789123456789123422";
document.write(getSum(st));
</script>
|
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++
#include <iostream>
using namespace std;
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);
}
int main()
{
int num = 12345;
int result = sum_of_digit(num, 0);
cout << "Sum of digits is " << result;
return 0;
}
|
Java
import java.io.*;
import java.lang.*;
import java.util.*;
class sum_of_digits {
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);
}
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
def sum_of_digit(n, val):
if (n < 10):
val = val + n
return val
return sum_of_digit(n // 10, (n % 10) + val)
num = 12345
result = sum_of_digit(num, 0)
print("Sum of digits is", result)
|
C#
using System;
class GFG{
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);
}
public static void Main()
{
int num = 12345;
int result = sum_of_digit(num, 0);
Console.Write("Sum of digits is " + result);
}
}
|
Javascript
<script>
function sum_of_digit(n, val)
{
if (n < 10)
{
val = val + n;
return val;
}
return sum_of_digit(parseInt(n / 10),
(n % 10) + val);
}
let num = 12345;
let result = sum_of_digit(num, 0);
document.write("Sum of digits is " + result);
</script>
|
OutputSum of digits is 15
Please write comments if you find the above codes/algorithms incorrect, or find better ways to solve the same problem.
