Table of Contents

**Problem Statement**

the problem “Palindrome Number” states that you are given an integer number. Check if it is a palindrome or not. Solve this problem without converting the given number into a string.

**Example**

12321

true

**Explanation**

12321 is a palindrome number because when we reverse 12321 it gives 12321 which is same as the given number. So 12321 is a palindrome number.

-12321

false

**Explanation**

-12321 is not a palindrome number because when we reverse -12321 it gives 12321- which is not the same as the given number. So -12321 is not a palindrome number.

**Approach for ****Palindrome Number checking**

The first approach that comes in our mind is to convert the given number into a string and check if it is palindrome or not. But we can not do this because this approach is restricted and also we require extra space for the string.

We can also observe one important point that a negative number is never a palindrome number.

So we will try to approach it in a different way. We will reverse the number and store it in a variable and compare if it is equal to the original number. If the reversed number is equal to the original number then the number is a palindrome else it is not a palindrome.

To reverse the given number we will perform( remainder= n%10 ) this gives the last digit of the number. We will generate the reversed number by (reversed=reversed*10+remainder). Now we will divide the number by 10 to get the second last digit. We repeat this process until the value of n is greater than zero.

At last, we will compare if the original number is equal to the reversed number. If yes then the number is a palindrome number else it is not a palindrome number.

**Code**

**C++ code for Palindrome Number**

#include<bits/stdc++.h> using namespace std; bool isPalindrome(int num){ if(num < 0) return false; int reversed = 0, remainder, original = num; while(num != 0) { remainder = num % 10; reversed = reversed * 10 + remainder; num /= 10; } return original == reversed; } int main() { if(isPalindrome(12321)) cout<<"Yes, it is Palindrome"; else cout<<"No, not Palindrome"; }

Yes, it is Palindrome

**Java code for Palindrome Number**

public class check { static boolean isPalindrome(int num){ if(num < 0) return false; int reversed = 0, remainder, original = num; while(num != 0) { remainder = num % 10; reversed = reversed * 10 + remainder; num /= 10; } return original == reversed; } public static void main(String args[]){ if(isPalindrome(12321)) System.out.println("Yes, it is Palindrome"); else System.out.println("No, not Palindrome"); } }

Yes, it is Palindrome

## Complexity Analysis

**Time complexity**

The time complexity to check if a number is a palindrome or not is Â $). Because we are dividing the number by 10 in every iteration. So the time complexity can be said is equal to the number of digits in a number.$

**Space complexity**

**O(1) **because we are using one extra variable to store the reversed number.