How to Solve: Valid Palindrome – Complete Guide for FAANG Interviews

How to Solve: Valid Palindrome – Complete Guide for FAANG Interviews

? Introduction

"This problem appears in 1 out of every 5 string-manipulation interviews—mastering it proves you can handle edge cases, optimize for time/space, and write clean, bug-free code under pressure."

? WHAT YOU NEED TO KNOW FIRST

Before tackling "Valid Palindrome," ensure you understand: 1. Two-pointer technique – Efficiently traverse strings/arrays from both ends. 2. ASCII character manipulation – Check if a character is alphanumeric (letters/digits). 3. String preprocessing – Convert to lowercase and filter non-alphanumeric characters.

? KEY TECHNIQUES & PATTERNS

? STEP-BY-STEP FRAMEWORK (Mental Checklist)

Follow these steps every time you solve a "Valid Palindrome" problem:

1. Clarify the problem
2. Ask: "Should we consider case sensitivity? Are spaces/punctuation allowed?"

3. Example: "A man, a plan, a canal: Panama" → True (case-insensitive, ignore non-alphanumeric).

4. Preprocess the string

5. Convert to lowercase.

6. Remove all non-alphanumeric characters (use isalnum() or ASCII checks).

7. Initialize two pointers

8. left = 0, right = len(s) - 1.

9. Compare characters

10. While left < right:

    • If s[left] != s[right], return False.
    • Increment left, decrement right.

11. Return True if loop completes

12. If all comparisons pass, the string is a palindrome.

13. Optimize for edge cases

14. Empty string → True.
15. Single character → True.

16. All non-alphanumeric → True (after preprocessing).

17. Test with examples

18. "racecar" → True.
19. "race a car" → False.
20. " " → True.

? WORKED EXAMPLES

Example 1 – Basic (LeetCode 125)

Problem: Given a string s, return True if it is a palindrome (case-insensitive, ignore non-alphanumeric).

Solution:

def isPalindrome(s: str) -> bool:

    # Step 1: Preprocess (lowercase + filter non-alphanumeric)

    s = [c.lower() for c in s if c.isalnum()]

    s = ''.join(s)


    # Step 2: Two-pointer comparison

    left, right = 0, len(s) - 1

    while left < right:

        if s[left] != s[right]:

            return False

        left += 1

        right -= 1

    return True

Time Complexity: O(n) (two passes: preprocessing + two-pointer). Space Complexity: O(n) (new string for filtered characters).

Why this works: - Preprocessing ensures we only compare relevant characters. - Two-pointer efficiently checks symmetry in O(n/2) time.

Example 2 – Medium (Follow-up: No Extra Space)

Problem: Solve the same problem without using extra space (modify the string in-place).

Solution:

def isPalindrome(s: str) -> bool:

    left, right = 0, len(s) - 1

    while left < right:

        # Skip non-alphanumeric from left

        while left < right and not s[left].isalnum():

            left += 1

        # Skip non-alphanumeric from right

        while left < right and not s[right].isalnum():

            right -= 1

        # Compare (case-insensitive)

        if s[left].lower() != s[right].lower():

            return False

        left += 1

        right -= 1

    return True

Time Complexity: O(n) (single pass with skips). Space Complexity: O(1) (no extra space).

Why this works: - Two-pointer skips non-alphanumeric characters on the fly, avoiding preprocessing. - Case-insensitive comparison is done during traversal.

Example 3 – Hard (Follow-up: Almost Palindrome)

Problem: Given a string s, return True if it can be a palindrome by removing at most one character.

Solution:

def validPalindrome(s: str) -> bool:

    def is_palindrome(s, left, right):

        while left < right:

            if s[left] != s[right]:

                return False

            left += 1

            right -= 1

        return True


    left, right = 0, len(s) - 1

    while left < right:

        if s[left] != s[right]:

            # Try skipping left or right character

            return is_palindrome(s, left + 1, right) or is_palindrome(s, left, right - 1)

        left += 1

        right -= 1

    return True

Time Complexity: O(n) (worst case: two passes). Space Complexity: O(1) (no extra space).

Why this works: - If a mismatch is found, check both possibilities (skip left or right character). - Recursively verify if the remaining substring is a palindrome.

❌ Common Mistakes

? INTERVIEW TrapS

? 1-Minute Recap (Closing Script)

"Alright, let’s lock this in. For Valid Palindrome: 1. Preprocess first – lowercase and filter non-alphanumeric. If space is tight, skip during traversal. 2. Two-pointer – start at both ends, compare, and move inward. 3. Edge cases – empty string, single character, all spaces. 4. Follow-ups – if they ask for one deletion, use a helper function to check both skip-left and skip-right. 5. Optimize – always ask if you can avoid extra space.

You’ve got this. Now go write that clean, bug-free code!

? Final Notes

• Practice variations: Try solving with Unicode characters or custom alphabets.
• Time yourself: Aim for <10 minutes on the basic problem, <15 minutes for follow-ups.
• Whiteboard first: Sketch the two-pointer approach before coding.

This framework ensures you never panic when "Valid Palindrome" appears in an interview. Good luck! ?