Frequently Asked Java Program 01: Java Program to Check If a Given Number is Palindrome

Palindrome program is a frequently asked programming interview question to both freshers or experienced in interviews. We will going to learn that in this post.

Let’s start with basics first. We must be aware what is Palindrome.

What is palindrome?

We read a word or group of words from left to right. If we can read them from right to left same as we read left to right , is called a palindrome.

Examples: “Refer” ,  “level” , “Madam” , “Nurses run” , 1991 are palindrome.

A number can also be a palindrome number as shown above in example. In this post, we will be learn about a palindrome number. In next post, we will learn about string palindrome.

A palindromic number or numeral palindrome is a number that remains the same when its digits are reversed.

For example: 16461.

Logic to find out a numeral palindrome:

1. 1. We need to extract last digit from input number and keep it at first position for a new number.
   2. Extract the second last digit and put in the next position of new number.
   3. We can achieve that using multiple ways like reverse traversing or charAt methods etc. Here we will see include some number concepts.
   4. We can extract last digit of a number(remainder) if we divide number by 10.

For examples:
a. 13/10 : Remainder= 3 which is last digit in number 13.
b. 100/10 : Remainder= 0 which is last digit in number 10.

5. New number which is actually reverse number, will be formed by formula “(reverseNumber*10) + remainder”.
6. Now divide the original number by 10 to remove last digit of it.
7. Repeat above steps till original number is less than zero.

Java Program:

package NumberSeries;

import java.util.Scanner;

public class PalindromeNumber {

        public static void main(String[] args) {

                // Taking input from user
                Scanner sc = new Scanner(System.in);
                System.out.println("Please input the number to find if it is  palindrome or not:");
                int inputByUser = sc.nextInt();
                System.out.println("Input Number to be checked for palindrome: " + inputByUser);
                
                // Closing input stream
                sc.close();

                // Copy input number to a temporary variable to keep original value intact
                int temp = inputByUser;
                int revNumber = 0;

                // checking if number is negative
                if (inputByUser < 0) {
                        System.out.println("Negative number.Enter positive number.");
                        System.exit(1);

                } // checking if number is single digit only
                else if (inputByUser >= 0 && inputByUser <= 9) {
                        System.out.println(inputByUser + " is palindrome as it is single digit number.");
                        System.exit(1);
                } else {
                        while (temp > 0)

                        {
                                // extracting last digit of number
                                int rem = temp % 10;
                                // forming number
                                revNumber = revNumber * 10 + rem;
                                // removing last digit from number
                                temp = temp / 10;
                        }
                        
                        System.out.println("Input By User:" + inputByUser);
                        System.out.println("Reverse number:" + revNumber);
                        
                        // Comparing if input number and reversed number are same
                        if (inputByUser == revNumber) {
                                System.out.println(inputByUser + " is a Palindrome Number");
                        } else {
                                System.out.println(inputByUser + " is not a Palindrome Number");
                        }

                }

        }
}

Output:

You can run above program for multiple inputs and if it fails for any condition, let me know.

#HappyCoding