Check if a number is Palindrome or Not

Problem Statement: Given a number check if it is a palindrome.

An integer is considered a palindrome when it reads the same backward as forward.

Examples:

Example 1:
Input: N = 123
Output: Not Palindrome Number
Explanation: 123 read backwards is 321.Since these are two different numbers 123 is not a palindrome.

Example 2:
Input: N =  121 
Output: Palindrome Number
Explanation: 121 read backwards as 121.Since these are two same numbers 121 is a palindrome.

Solution

Disclaimer: Don’t jump directly to the solution, try it out yourself first.

Approach: We can reverse the original number and compare the original with the reversed number. If both are the same, the number qualifies as a palindrome number.

Say the input number is X. Declare a variable Y to store the reverse and initialize it to 0. Make a copy of X, say dummy that will be used later for comparison.

Let’s understand the procedure to reverse a number.

At every step extract the last digit using the % operator. Suppose X%10 = d.
We will append this last digit, d to Y using a formula 10*Y+d.
The last digit of X has been used. Discard it using X/10.

Repeat these steps for the remaining digits. After every iteration, the size of X will shrink by one digit. Terminate the iteration when X = 0 meaning no new digits are left to be reversed.

The reversed number Y is compared with the dummy variable since X was destroyed while iteration. If Y equals dummy print “Palindrome Number” otherwise “Not Palindrome Number”.

Code:

C++ Code

#include <iostream>

using namespace std;
int reverse(int X) {
   int Y = 0;
   while (X > 0) {
      //Extract the last digit
      int digit = X % 10;
      //Appending last digit
      Y = Y * 10 + digit;
      // Shrinking X by discarding the last digit
      X = X / 10;
   }
   return Y;
}
int main() {
   int X = 121;
   int dummy = X;
   int Y = reverse(X);
   if (dummy == Y) {
      cout << "Palindrome Number" << endl;
   } else {
      cout << "Not Palindrome Number" << endl;
   }
   return 0;
}

Output: Palindrome Number

Time Complexity: O(logN) for reversing N digits of input integer.

Space Complexity: O(1)

Java Code

public class Main {
   static int reverse(int X) {
      int Y = 0;
      while (X > 0) {
         //Extract the last digit
         int digit = X % 10;
         //Appending last digit
         Y = Y * 10 + digit;
         // Shrinking X by discarding the last digit
         X = X / 10;
      }
      return Y;
   }
   public static void main(String[] args) {

      int X = 121;
      int dummy = X;
      int Y = reverse(X);
      if (dummy == Y) {
         System.out.println("Palindrome Number");
      } 
      else {
         System.out.println("Not Palindrome Number");
      }

   }
}

Output: Palindrome Number

Time Complexity: O(logN) for reversing N digits of input integer.

Space Complexity: O(1)

Python Code

def reverse(X):
    Y = 0
    while X > 0:
        # Extract the last digit
        digit = X % 10
        # Appending last digit
        Y = Y * 10 + digit
        # Shrinking X by discarding the last digit
        X = X // 10
    return Y




if __name__ == "__main__":
    X = 121
    dummy = X
    Y = reverse(X)
    if dummy == Y:
        print("Palindrome Number")
    else:
        print("Not Palindrome Number")

Output: Palindrome Number

Time Complexity: O(logN) for reversing N digits of input integer.

Space Complexity: O(1)

Special thanks to Somparna Chakrabarti and Sudip Ghosh for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article