public class PalindromeNumber { public static void main(String[] args) { // Simple Palindrome Check (Using Integer Arithmetic) int number = 121; boolean isPalindrome = isPalindromeNumber(number); System.out.println("Is " + number + " a palindrome? " + isPalindrome); // Using String Conversion int number1 = 12321; boolean isPalindrome1 = isPalindromeNumber1(number1); System.out.println("Is " + number1 + " a palindrome? " + isPalindrome1); // Recursive Palindrome Check int number2 = 1221; boolean isPalindrome2 = isPalindromeNumber2(number2); System.out.println("Is " + number2 + " a palindrome? " + isPalindrome2); } public static boolean isPalindromeNumber(int number) { // Negative numbers are not palindromes if (number < 0) { return false; } int original = number; // Save the original number int reversed = 0; while (number != 0) { int digit = number % 10; // Extract the last digit reversed = reversed * 10 + digit; // Append the digit to reversed number /= 10; // Remove the last digit } return original == reversed; } // Using String Conversion public static boolean isPalindromeNumber1(int number1) { String numStr = Integer.toString(number1); String reversedStr = new StringBuilder(numStr).reverse().toString(); return numStr.equals(reversedStr); } // Recursive Palindrome Check public static boolean isPalindromeNumber2(int number2) { return number2 == reverseNumber(number2, 0); } private static int reverseNumber(int number2, int reversed) { if (number2 == 0) { return reversed; } return reverseNumber(number2 / 10, reversed * 10 + number2 % 10); } } /* * 1. Arithmetic Approach: * • Time Complexity: O(log₁₀(n)) (Number of digits in the number) * • Space Complexity: O(1) * 2. String Conversion: * • Time Complexity: O(n) (n is the number of digits) * • Space Complexity: O(n) (for string conversion and reversal) * 3. Recursive Approach: * • Time Complexity: O(log₁₀(n)) * • Space Complexity: O(log₁₀(n)) (recursion stack) */