Valid Palindrome in Java: Solution, Explanation & Practice

Check if string is palindrome considering only alphanumeric

Problem summary

Given a string, determine if it's a palindrome considering only alphanumeric characters and ignoring case. Use two pointers from both ends.

Starter code

public class Main {
    public static void main(String[] args) {
        String s = "A man, a plan, a canal: Panama";
        
        // Two pointers: left and right
        // Skip non-alphanumeric characters
        // Compare characters (case-insensitive)
        // Print: Is palindrome: true
    }
}

Expected output and test cases

A man, a plan, a canal: Panama
```
Is palindrome: true
```
race a car
```
Is palindrome: false
```
```
Is palindrome: true
```

Hints

Use Character.isLetterOrDigit() to check valid characters
Use Character.toLowerCase() for case-insensitive comparison
Move pointers inward, skipping invalid characters
If characters don't match, it's not a palindrome

Validated solution

Reveal Java solution

public class Main {
    static boolean validPalindrome(String text) {
        int left = 0, right = text.length() - 1;
        while (left < right) {
            while (left < right && !Character.isLetterOrDigit(text.charAt(left))) left++;
            while (left < right && !Character.isLetterOrDigit(text.charAt(right))) right--;
            if (Character.toLowerCase(text.charAt(left)) != Character.toLowerCase(text.charAt(right))) return false;
            left++; right--;
        }
        return true;
    }
    public static void main(String[] args) {
        System.out.println("Is palindrome: " + validPalindrome("A man, a plan, a canal: Panama"));
    }
}

How to approach the problem

Use two pointers and skip characters outside the stated alphanumeric definition before comparing. This avoids allocating a cleaned copy while still applying the same normalization at both ends.

Approach

Move inward past punctuation and spaces.
Compare lowercased alphanumeric characters.
Stop immediately on a mismatch.

Time and space complexity

Time: O(n). Space: O(1).

Edge cases to test

A blank or punctuation-only string is valid under this definition.
Character.isLetterOrDigit handles more than ASCII.

Common mistakes

Removing only spaces but not other punctuation.
Advancing a pointer after a mismatch instead of returning false.

Follow-up challenge

Make the scan code-point-aware for full Unicode text.

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