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Java program to check whether a string is a Palindrome

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A string is said to be a palindrome if it is the same if we start reading it from left to right or right to left. In this article, we will learn how to check if a string is a palindrome in Java.

So let us consider a string “str”, now the task is just to find out with its reverse string is the same as it is. 

Example of Palindrome

Input : str = “abba” 

Output: Yes

Input : str = “geeks”

Output: No 

Methods for Palindrome String in Java

There are three major methods to check string palindrome in Java as mentioned below:

  1. Naive Method
  2. Two Pointer Method
  3. Recursive Method

1. Naive Approach to Check Palindrome String  in Java

By Reversing the given string and Comparing. We can check if the given string is a palindrome by comparing the original string with its reversed version. 

Below is the implementation of the above approach:

Java

// Java Program to implement
// Basic Approach to check if
// string is a Palindrome
import java.io.*;
 
// Driver Class
class GFG {
    // main function
    public static boolean isPalindrome(String str)
    {
        // Initializing an empty string to store the reverse
        // of the original str
        String rev = "";
 
        // Initializing a new boolean variable for the
        // answer
        boolean ans = false;
 
        for (int i = str.length() - 1; i >= 0; i--) {
            rev = rev + str.charAt(i);
        }
 
        // Checking if both the strings are equal
        if (str.equals(rev)) {
            ans = true;
        }
        return ans;
    }
    public static void main(String[] args)
    {
        // Input string
        String str = "geeks";
 
        // Convert the string to lowercase
        str = str.toLowerCase();
        boolean A = isPalindrome(str);
        System.out.println(A);
    }
}

                    

Output
false

The complexity of the above method

Time Complexity:-The time complexity of the given code is O(n), where n is the length of the input string. This is because the for loop iterates through each character in the string once to create the reverse string.

Space Complexity:- The space complexity of the code is O(n), where n is the length of the input string. This is because the reverse string is created and stored in a separate string variable, which takes up space in memory proportional to the length of the input string. In addition, the other variables used in the code (i, str, and ans) take up a constant amount of space that is independent of the input size.

In the above example, if we write ABba in place of abba, then also we should get the output as yes.  So, we must change the case of the string to either lowercase or uppercase before we check it for a palindrome. If we do not do this, we will get unexpected results. This is because the compiler checks the characters based on their ASCII value and the ASCII value of A is not the same as a

2. Two Pointer Approach for Palindrome Program in Java

Our approach will be that we will first convert the string to lowercase. Then, we will take two pointers i pointing to the start of the string and j pointing to the end of the string. Keep incrementing i and decrementing j while i < j and at every step check whether the characters at these pointers are the same or not. If not then the string is not a palindrome else it is.

Example 1:

Java

// Java program to check whether a
// string is a Palindrome
// Using two pointing variables
 
// Main class
public class GFG {
 
    // Method
    // Returning true if string is palindrome
    static boolean isPalindrome(String str)
    {
 
        // Pointers pointing to the beginning
        // and the end of the string
        int i = 0, j = str.length() - 1;
 
        // While there are characters to compare
        while (i < j) {
 
            // If there is a mismatch
            if (str.charAt(i) != str.charAt(j))
                return false;
 
            // Increment first pointer and
            // decrement the other
            i++;
            j--;
        }
 
        // Given string is a palindrome
        return true;
    }
 
    // Method 2
    // main driver method
    public static void main(String[] args)
    {
        // Input string
        String str = "geeks";
 
        // Convert the string to lowercase
        str = str.toLowerCase();
        // passing bool function till holding true
        if (isPalindrome(str))
 
            // It is a palindrome
            System.out.print("Yes");
        else
 
            // Not a palindrome
            System.out.print("No");
    }
}

                    

Output
No

The complexity of the above method

Time Complexity:- The time complexity of the given code is O(n), where n is the length of the input string. This is because the while loop iterates through half of the string to check if it is a palindrome. Since we only check half of the string, the number of iterations is proportional to n/2, which is O(n).

Space Complexity:- The space complexity of the code is O(1), because the code only uses a constant amount of additional space that is independent of the input size. The only variables used in the code are i, j, and str, which each take up a constant amount of space. No additional space is created in the code.

Example 2:

Java

// Java Program to check Whether
// the String is Palindrome
// or Not
 
// Main class
class GFG {
 
    // Method 1
    // Returns true if string is a palindrome
    static boolean isPalindrome(String str)
    {
 
        // Pointers pointing to the beginning
        // and the end of the string
        int i = 0, j = str.length() - 1;
 
        // While there are characters to compare
        while (i < j) {
 
            // If there is a mismatch
            if (str.charAt(i) != str.charAt(j))
                return false;
 
            // Increment first pointer and
            // decrement the other
            i++;
            j--;
        }
 
        // Given string is a palindrome
        return true;
    }
 
    // Main driver method
    public static void main(String[] args)
    {
        String str = "geeks";
        String str2 = "RACEcar";
 
        // Change strings to lowercase
        str = str.toLowerCase();
        str2 = str2.toLowerCase();
 
        // For string 1
        System.out.print("String 1 :");
 
        if (isPalindrome(str))
            System.out.print("It is a palindrome");
        else
            System.out.print("It is not a palindrome");
 
        // new line for better readability
        System.out.println();
 
        // For string 2
        System.out.print("String 2 :");
        if (isPalindrome(str2))
            System.out.print("It is a palindrome");
        else
            System.out.print("It is not a palindrome");
    }
}

                    

Output
String 1 :It is not a palindrome
String 2 :It is a palindrome

The complexity of the above method

Time Complexity:- The time complexity of the given code is O(n), where n is the length of the input string. This is because the while loop in the `isPalindrome` method iterates through half of the string to check if it is a palindrome. Since we only check half of the string, the number of iterations is proportional to n/2, which is O(n).

Space Complexity:- The space complexity of the code is O(1), because the code only uses a constant amount of additional space that is independent of the input size. The only variables used in the code are i, j, str, and str2, which each take up a constant amount of space. No additional space is created in the code.

3. Recursive Approach For Palindrome Program in Java

The approach is very simple. Just like the two-pointer approach, we will check the first and the last value of the string but this time it will be through recursion.

  1. We will take two pointers i pointing to the start of the string and j pointing to the end of the string. 
  2. Keep incrementing i and decrementing j while i < j and at every step 
  3. Check whether the characters at these pointers are the same or not. We are doing this through recursion – (i+1, j-1
  4. If all the characters are the same on the ith and jth index till i>=j condition satisfies, print true else false

Below is the implementation of the above approach:

Java

// Java program to check whether a
// string is a Palindrome using recursion
import java.io.*;
 
// Driver Class
class GFG {
    public static boolean isPalindrome(int i, int j,
                                       String A)
    {
        // comparing the two pointers
        if (i >= j) {
            return true;
        }
 
        // comparing the characters on those pointers
        if (A.charAt(i) != A.charAt(j)) {
            return false;
        }
 
        // checking everything again recursively
        return isPalindrome(i + 1, j - 1, A);
    }
 
    public static boolean isPalindrome(String A)
    {
        return isPalindrome(0, A.length() - 1, A);
    }
 
    // main function
    public static void main(String[] args)
    {
        // Input string
        String A = "geeks";
 
        // Convert the string to lowercase
        A = A.toLowerCase();
        boolean str = isPalindrome(A);
        System.out.println(str);
    }
}

                    

Output
false

The complexity of the above method

Time Complexity:- The time complexity of the given code is O(n), where n is the length of the input string. This is because the `isPalindrome` function recursively calls itself for the characters at positions (i+1, j-1) until the pointers i and j cross each other or the characters at the pointers are not equal. Since we compare each character in the string exactly once, the time complexity is O(n).

Space Complexity:- The space complexity of the code is O(n), where n is the length of the input string. This is because each recursive call creates a new stack frame that stores the current values of the function parameters and local variables. In the worst case, the function call stack may grow as large as n/2 (when the input string is a palindrome), so the space complexity is O(n).

Reference

To know more about Palindrome refer Program for String Palindrome.



Last Updated : 17 Jul, 2023
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