// Efficient : Time Complexity : O(n), Space Complexity : Θ(1)

    static boolean isPalindrome(String str)
    {
 
        // Pointers pointing to the beginning
        // and the end of the string
        int begin = 0, end = str.length() - 1;
 
        // While there are characters to compare
        while (begin < end) {
 
            // If there is a mismatch
            if (str.charAt(begin) != str.charAt(end))
                return false;
 
            // Increment first pointer and
            // decrement the other
            begin++;
            end--;
        }
 
        // Given string is a palindrome
        return true;
    }


// Naive : Time Complexity : Θ(n), Space Complexity : Θ(n)

    static boolean isPalindrome(String str)
    {
      StringBuilder rev = new StringBuilder(str);
      rev.reverse();  // StringBuilder is mutable & has a function called reverse
      
      return str.equals(rev.toString());
    }

// Time Complexity : O(n), Space Complexity : O(n)

import java.io.*;
import java.util.*;

class GFG 
{
	static boolean isPalindrome(String str, int start, int end)
	{
		if(start >= end)
			return true;

		return ((str.charAt(start)==str.charAt(end)) && 
			     isPalindrome(str, start + 1, end - 1));
	}
    public static void main(String [] args) 
    {
    	String s = "GeekskeeG";

    	System.out.println(isPalindrome(s, 0, s.length() -1));
    }
}