Check if the sum of digits of N is palindrome

Given an integer N, the task is to check whether the sum of digits of N is palindrome or not.
Example:

Input: N = 56 
Output: Yes 
Explanation: Digit sum is (5 + 6) = 11, which is a palindrome.

Input: N = 51241 
Output: No 

Approach: Find the sum of digits of N and store it in a variable sum. Now check whether sum is palindrome or not using the approach discussed in this article.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the
// sum of digits of n
int digitSum(int n)
{
    int sum = 0;
    while (n > 0) {
        sum += (n % 10);
        n /= 10;
    }
    return sum;
}

// Function that returns true
// if n is palindrome
bool isPalindrome(int n)
{
    // Find the appropriate divisor
    // to extract the leading digit
    int divisor = 1;
    while (n / divisor >= 10)
        divisor *= 10;

    while (n != 0) {
        int leading = n / divisor;
        int trailing = n % 10;

        // If first and last digit
        // not same return false
        if (leading != trailing)
            return false;

        // Removing the leading and trailing
        // digit from number
        n = (n % divisor) / 10;

        // Reducing divisor by a factor
        // of 2 as 2 digits are dropped
        divisor = divisor / 100;
    }
    return true;
}

// Function that returns true if
// the digit sum of n is palindrome
bool isDigitSumPalindrome(int n)
{

    // Sum of the digits of n
    int sum = digitSum(n);

    // If the digit sum is palindrome
    if (isPalindrome(sum))
        return true;
    return false;
}

// Driver code
int main()
{
    int n = 56;

    if (isDigitSumPalindrome(n))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

// Function to return the
// sum of digits of n
static int digitSum(int n)
{
    int sum = 0;
    while (n > 0)
    {
        sum += (n % 10);
        n /= 10;
    }
    return sum;
}

// Function that returns true
// if n is palindrome
static boolean isPalindrome(int n)
{
    // Find the appropriate divisor
    // to extract the leading digit
    int divisor = 1;
    while (n / divisor >= 10)
        divisor *= 10;

    while (n != 0)
    {
        int leading = n / divisor;
        int trailing = n % 10;

        // If first and last digit
        // not same return false
        if (leading != trailing)
            return false;

        // Removing the leading and trailing
        // digit from number
        n = (n % divisor) / 10;

        // Reducing divisor by a factor
        // of 2 as 2 digits are dropped
        divisor = divisor / 100;
    }
    return true;
}

// Function that returns true if
// the digit sum of n is palindrome
static boolean isDigitSumPalindrome(int n)
{

    // Sum of the digits of n
    int sum = digitSum(n);

    // If the digit sum is palindrome
    if (isPalindrome(sum))
        return true;
    return false;
}

// Driver code
public static void main(String []args)
{
    int n = 56;

    if (isDigitSumPalindrome(n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}

// This code is contributed by Surendra_Gangwar

Python3

# Python3 implementation of the approach

# Function to return the
# sum of digits of n
def digitSum(n) :

    sum = 0;
    while (n > 0) :
        sum += (n % 10);
        n //= 10;

    return sum;

# Function that returns true
# if n is palindrome
def isPalindrome(n) :

    # Find the appropriate divisor
    # to extract the leading digit
    divisor = 1;
    while (n // divisor >= 10) :
        divisor *= 10;

    while (n != 0) :
        leading = n // divisor;
        trailing = n % 10;

        # If first and last digit
        # not same return false
        if (leading != trailing) :
            return False;

        # Removing the leading and trailing
        # digit from number
        n = (n % divisor) // 10;

        # Reducing divisor by a factor
        # of 2 as 2 digits are dropped
        divisor = divisor // 100;

    return True;

# Function that returns true if
# the digit sum of n is palindrome
def isDigitSumPalindrome(n) :

    # Sum of the digits of n
    sum = digitSum(n);

    # If the digit sum is palindrome
    if (isPalindrome(sum)) :
        return True;
    return False;

# Driver code
if __name__ == "__main__" :

    n = 56;

    if (isDigitSumPalindrome(n)) :
        print("Yes");
    else :
        print("No");

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the
// sum of digits of n
static int digitSum(int n)
{
    int sum = 0;
    while (n > 0)
    {
        sum += (n % 10);
        n /= 10;
    }
    return sum;
}

// Function that returns true
// if n is palindrome
static bool isPalindrome(int n)
{
    // Find the appropriate divisor
    // to extract the leading digit
    int divisor = 1;
    while (n / divisor >= 10)
        divisor *= 10;

    while (n != 0)
    {
        int leading = n / divisor;
        int trailing = n % 10;

        // If first and last digit
        // not same return false
        if (leading != trailing)
            return false;

        // Removing the leading and trailing
        // digit from number
        n = (n % divisor) / 10;

        // Reducing divisor by a factor
        // of 2 as 2 digits are dropped
        divisor = divisor / 100;
    }
    return true;
}

// Function that returns true if
// the digit sum of n is palindrome
static bool isDigitSumPalindrome(int n)
{

    // Sum of the digits of n
    int sum = digitSum(n);

    // If the digit sum is palindrome
    if (isPalindrome(sum))
        return true;
    return false;
}

// Driver code
static public void Main ()
{
    int n = 56;

    if (isDigitSumPalindrome(n))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}

// This code is contributed by ajit

Javascript

<script>
    // Javascript implementation of the approach

    // Function to return the
    // sum of digits of n
    function digitSum(n)
    {
        let sum = 0;
        while (n > 0)
        {
            sum += (n % 10);
            n = parseInt(n / 10, 10);
        }
        return sum;
    }

    // Function that returns true
    // if n is palindrome
    function isPalindrome(n)
    {
        // Find the appropriate divisor
        // to extract the leading digit
        let divisor = 1;
        while (parseInt(n / divisor, 10) >= 10)
            divisor *= 10;

        while (n != 0)
        {
            let leading = parseInt(n / divisor, 10);
            let trailing = n % 10;

            // If first and last digit
            // not same return false
            if (leading != trailing)
                return false;

            // Removing the leading and trailing
            // digit from number
            n = parseInt((n % divisor) / 10, 10);

            // Reducing divisor by a factor
            // of 2 as 2 digits are dropped
            divisor = parseInt(divisor / 100, 10);
        }
        return true;
    }

    // Function that returns true if
    // the digit sum of n is palindrome
    function isDigitSumPalindrome(n)
    {

        // Sum of the digits of n
        let sum = digitSum(n);

        // If the digit sum is palindrome
        if (isPalindrome(sum))
            return true;
        return false;
    }

    let n = 56;

    if (isDigitSumPalindrome(n))
        document.write("Yes");
    else
        document.write("No");

// This code is contributed by suresh07.
</script>

Output

Yes

Time Complexity: O(logN)
Auxiliary Space: O(1)

Another Approach: The idea is to find the sum of digits of N and store it in a variable sum and then convert the sum into a string say s1 and check whether reversing the string s1 is equal to s1, If yes, then this sum of the digit of the given number N is palindrome otherwise not.

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

int main()
{
    int num = 56;
    int sum = 0;

    while (num != 0) {
        int temp = (num % 10);
        sum = sum + temp;
        num /= 10;
    }

    // Driver Code

    // convert sum to string
    string str = to_string(sum);

    // reverse the string
    string string_rev = "" + str;
    reverse(string_rev.begin(), string_rev.end());

    // check the palindrome
    cout << ((str == string_rev) ? "Yes" : "No");
}

// This code is contributed by phasing17

Java

/*package whatever //do not write package name here */

import java.io.*;

class GFG {
    public static void main(String[] args)
    {
        int n = 56;
        int sum = 0;
        while (n != 0) {
            int temp = n % 10;

            // Addition of digits
            sum = sum + temp;
            n = n / 10;
        }

        // Converting int sum to string
        String str = String.valueOf(sum);

        // Reverse the string
        String rev
            = new StringBuilder(str).reverse().toString();
        if (str.equals(rev)) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }
    }
}
// This code is contributed by keerthikarathan123

Python3

# code
num = int(56)
sum = int(0)

while num != 0:
    temp = int(num % 10)
    sum = sum+temp
    num = num/10

# convert sum to string
string = str(sum)

# reverse the string
string_rev = string[:: -1]

# check the palindrome
if string == string_rev:
    print("Yes")
else:
    print("No")

# This code is contributed by keerthikarathan123

C#

// C# code to implement the approach
using System;
using System.Text;
using System.Collections.Generic;

class GFG {

  public static string Reverse( string s )
  {
    char[] charArray = s.ToCharArray();
    Array.Reverse( charArray );
    return new string( charArray );
  }
  public static void Main(string[] args)
  {
    int n = 56;
    int sum = 0;
    while (n != 0) {
      int temp = n % 10;

      // Addition of digits
      sum = sum + temp;
      n = n / 10;
    }

    // Converting int sum to string
    string str = Convert.ToString(sum);

    // Reverse the string
    string rev  = Reverse(str);
    if (str.Equals(rev))
      Console.WriteLine("Yes");

    else {
      Console.WriteLine("No");
    }
  }
}

// This code is contributed by phasing17

Javascript

// JS code to implement the approach
let num = 56
let sum = 0

while(num != 0)
{
    let temp = (num % 10)
    sum = sum+temp
    num = Math.floor(num/10)
}

// Driver Code

// convert sum to string
let string = "" + sum

// reverse the string
let string_rev = string.split("").reverse().join("")

// check the palindrome
console.log( (string == string_rev) ? "Yes" : "No")

// This code is contributed by phasing17

Output

Yes

Time Complexity: O(N)
Auxiliary Space: O(1)