Given a number, find the sum of its digits.
Examples :
Input: n = 687
Output: 21
Input: n = 12
Output: 3
Follow the below steps to solve the problem:
- 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 adding it to the sum.
- Divide the number by 10 with help of ‘/’ operator to remove the rightmost digit.
- Print or return the sum
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
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
if __name__ == "__main__":
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("$res");
?>
|
Javascript
<script>
function getSum(n)
{
var sum = 0;
while (n != 0) {
sum = sum + n % 10;
n = parseInt(n / 10);
}
return sum;
}
var n = 687;
document.write(getSum(n));
</script>
|
Time Complexity: O(log N)
Auxiliary Space: O(1)
How to compute in a single line?
The below function has three lines instead of one line, but it calculates the sum in one line using for loop. It can be made one-line function if we pass the pointer to the sum.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
class gfg {
public:
int getSum(int n)
{
int sum;
for (sum = 0; n > 0; sum += n % 10, n /= 10)
;
return sum;
}
};
int main()
{
gfg g;
int n = 687;
cout << g.getSum(n);
return 0;
}
|
C
#include <stdio.h>
int getSum(int n)
{
int sum;
for (sum = 0; n > 0; sum += n % 10, n /= 10)
;
return sum;
}
int main()
{
int n = 687;
printf(" %d ", getSum(n));
return 0;
}
|
Java
import java.io.*;
class GFG {
static int getSum(int n)
{
int sum;
for (sum = 0; n > 0; sum += n % 10, n /= 10)
;
return sum;
}
public static void main(String[] args)
{
int n = 687;
System.out.println(getSum(n));
}
}
|
Python3
def getSum(n):
sum = 0
while(n > 0):
sum += int(n % 10)
n = int(n/10)
return sum
if __name__ == "__main__":
n = 687
print(getSum(n))
|
C#
using System;
class GFG {
static int getSum(int n)
{
int sum;
for (sum = 0; n > 0; sum += n % 10, n /= 10)
;
return sum;
}
public static void Main()
{
int n = 687;
Console.Write(getSum(n));
}
}
|
PHP
<?php
function getsum($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>
|
Time Complexity: O(log N)
Auxiliary Space: O(1)
Sum of the digits of a given number using recursion:
Follow the below steps to solve the problem:
- Get the number
- Get the remainder and pass the next remaining digits
- Get the rightmost digit of the number with help of the remainder ‘%’ operator by dividing it by 10 and adding it to the sum.
- Divide the number by 10 with help of the ‘/’ operator to remove the rightmost digit.
- Check the base case with n = 0
- Print or return the sum
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
class gfg {
public:
int sumDigits(int no)
{
if (no == 0) {
return 0;
}
return (no % 10) + sumDigits(no / 10);
}
};
int main(void)
{
gfg g;
cout << g.sumDigits(687);
return 0;
}
|
C
#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
import java.io.*;
class GFG {
static int sumDigits(int no)
{
if (no == 0) {
return 0;
}
return (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))
if __name__ == "__main__":
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)
{
if(no == 0){
return 0 ;
}
return (no % 10) + sumDigits(parseInt(no/10)) ;
}
document.write(sumDigits(687));
</script>
|
Time Complexity: O(log N)
Auxiliary Space: O(log N)
Sum of the digits of a given number with 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)
Follow the below steps to solve the problem:
- Declare a variable sum equal to zero
- Run a loop from zero to the length of the input string
- Add the value of each character into the sum, by converting the character into it’s integer value
- Return sum
Below is the implementation of the above approach:
C++14
#include <bits/stdc++.h>
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
if __name__ == "__main__":
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));
}
}
|
PHP
<?php
function getsum($str)
{
$sum = 0;
for ($i = 0; $i<strlen($str); $i++) {
$sum = $sum + (int)$str[$i];
}
return $sum;
}
$str = "123456789123456789123422";
echo(getsum($str));
?>
|
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>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Sum of the digits of a given number using tail recursion:
Follow the below steps to solve the problem:
- 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 passing 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
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
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;
}
|
C
#include <stdio.h>
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);
printf("Sum of digits is %d", 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)
if __name__ == "__main__":
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
Time Complexity: O(log N)
Auxiliary Space: O(log N)
Please write comments if you find the above codes/algorithms incorrect, or find better ways to solve the same problem.