Strings: Checking for Palindromes

In this post, we’ll walk through a common interview question—checking if a given string is a palindrome. This problem is a great exercise for understanding pointers, loops, and conditional logic in Java.

Problem Statement

Write a Java method that checks if a given string is a palindrome. A palindrome is a word or phrase that reads the same forwards and backwards (e.g., “noon” or “madam”).

Solution Overview

The solution leverages a two-pointer technique to check characters from both ends of the string, moving towards the center. By comparing characters at corresponding positions, we can determine if the string is a palindrome without having to reverse it.

Key Points of the Approach:

1. 
   Two-Pointer Technique: Check characters from both directions.


2. 
   Early Exit: Stop as soon as a mismatch is found.


3. 
   Optimization: Only traverse up to half the string length for efficiency.

Code Solution

Here’s the code for the solution:

public class StringPalindromeQuestion {



    // Method to check if a given string is a palindrome

    private boolean isPalindrome(String string) {

        if (string != null) {

            for (int i = 0, j = string.length() - 1; i < string.length()

            / 2; i++, j--) {

                if (string.charAt(i) != string.charAt(j)) {

                    return false;

                }

            }

        }

        return true;

    }



    public static void main(String[] args) {

        StringPalindromeQuestion palindrome = new StringPalindromeQuestion();



        String oddString = "abcdcba";    // Palindrome with odd length

        String evenString = "abccba";    // Palindrome with even length

        String nonPalindrome = "asfgsa"; // Not a palindrome



        // Result: true

        System.out.println(palindrome.isPalindrome(oddString));



        // Result: true

        System.out.println(palindrome.isPalindrome(evenString));



        // Result: false

        System.out.println(palindrome.isPalindrome(nonPalindrome));



        // Testing with null

        // Result: true

        System.out.println(palindrome.isPalindrome(null));

    }

}

Explanation

1. Two-Pointer Approach:

• We initialize two pointers: one at the start (i) and one at the end (j).




• We compare characters at these positions (string.charAt(i) and string.charAt(j)) and increment i and decrement j after each comparison.




• The loop runs only up to string.length() / 2, ensuring efficient traversal regardless of whether the length is odd or even.

2. Odd vs. Even Length:

• For even-length strings (e.g., "abccba"), the method checks up to the midpoint, so no middle character remains unchecked.




• For odd-length strings (e.g., "abcdcba"), the middle character naturally does not affect palindrome status.

3. Null Handling:

The method checks if the string is null at the beginning to avoid NullPointerException.

Example Output

• Odd-length Palindrome: "abcdcba" returns true.




• Even-length Palindrome: "abccba" returns true.




• Non-Palindrome: "asfgsa" returns false.




• Null String: returns true (a null input is considered a palindrome by this implementation).

Interview Tip 💡

Understanding two-pointer techniques is valuable for solving many string-based problems efficiently. This technique avoids extra space complexity and makes code execution faster by limiting unnecessary comparisons.

Conclusion

This solution provides a clean and efficient way to check for palindromes in Java. Try using this approach with different string inputs to further solidify your understanding of pointer manipulation and string traversal.

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Happy Coding!