Sum of Digit is Pallindrome or not

 Sum of Digit is Pallindrome or not

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Given a number N.Find if the digit sum(or sum of digits) of N is a Palindrome number or not.
Note:A Palindrome number is a number which stays the same when reversed.Example- 121,131,7 etc.

Example 1:

Input:
N=56
Output:
1
Explanation:
The digit sum of 56 is 5+6=11.
Since, 11 is a palindrome number.Thus,
answer is 1.

Example 2:

Input:
N=98
Output:
0
Explanation:
The digit sum of 98 is 9+8=17.
Since 17 is not a palindrome,thus, answer
is 0.

Your Task:
You don't need to read input or print aything.Your Task is to complete the function isDigitSumPalindrome() which takes a number N as input parameter and returns 1 if the Digit sum of N is a palindrome.Otherwise it returns 0.

Expected Time Complexity:O(LogN)
Expected Auxillary Space:O(1)

Constraints:
1<=N<=10⁹

 

^{********  SOLUTION ********************************}

 

<sup>// { Driver Code Starts
// Initial Template for C++

#include <bits/stdc++.h>
using namespace std;

 // } Driver Code Ends


// User function Template for C++

class Solution {
  public:
    int isDigitSumPalindrome(int N) {
        int isDigitSum=0;
        int m=N;
        int digit;
        int remainder,reverseNum=0,originalNum;

        while(m){
            digit=m%10;
            isDigitSum=isDigitSum+digit;
            m=m/10;
        }

        originalNum=isDigitSum;

        while(isDigitSum){
            remainder=isDigitSum%10;
            reverseNum=reverseNum*10+remainder;
            isDigitSum/=10;
        }


        if(originalNum==reverseNum){
          return 1;
        }else{
          return 0;
        }

        
    }
};

// { Driver Code Starts.
int main() {
    int t;
    cin >> t;
    while (t--) {
        int N;
        cin >> N;
        Solution ob;
        cout << ob.isDigitSumPalindrome(N) << "\n";
    }
}
  // } Driver Code Ends</sup>