How to Explain Valid Palindrome in a Coding Interview (Step-by-Step)

Valid Palindrome looks trivial — until the interviewer asks you to handle spaces, punctuation, and mixed case in one clean pass. The two-pointer approach with in-place filtering is what separates a polished answer from a naïve reverse-and-compare.

In this post, you’ll learn how to walk an interviewer through LeetCode 125 with confidence: the cleaning step, the two-pointer technique, the alphanumeric check, and a tight complexity argument.

Problem Statement

Examples

Constraints

• 1 <= s.length <= 2 * 105
• s consists only of printable ASCII characters

What Is the Problem?

“I’m given a string that may contain spaces, punctuation, and mixed case. I need to strip everything except letters and digits, lowercase whatever remains, and then check if the result reads the same forwards and backwards. An empty string after cleaning counts as a palindrome.”

Restating the cleaning rules clearly — alphanumeric only, case-insensitive — tells the interviewer you’ve understood the full scope of the problem before touching the keyboard.

Step 1: Compare Approaches Before Jumping to Code

“The clean-and-reverse approach works and is fine if the interviewer is happy with O(n) space. But with two pointers we can do the filtering and the comparison in a single pass without allocating any new strings, bringing space down to O(1).”

The Key Insight — Skip, Don’t Strip

“The insight is that I don’t need to materialise a cleaned version of the string. I can treat the two-pointer comparison and the filtering as the same step. Each character is visited at most once by either pointer, so the whole thing is O(n) time and O(1) space.”

Step 2: Explain the Algorithm in Plain English

Two-Pointer Visualised — “A man, a plan, a canal: Panama”

After cleaning, the effective string is amanaplanacanalpanama. Here’s how the pointers move through the original string, skipping non-alphanumeric characters on the fly.

Step 3: Talk Through the Code

class Solution:
    def isPalindrome(self, s: str) -> bool:
        left, right = 0, len(s) - 1

        while left < right:
            # Advance left past any non-alphanumeric character
            while left < right and not s[left].isalnum():
                left += 1

            # Advance right past any non-alphanumeric character
            while left < right and not s[right].isalnum():
                right -= 1

            # Both pointers are now on alphanumeric characters
            if s[left].lower() != s[right].lower():
                return False   # Mismatch — not a palindrome

            left  += 1
            right -= 1

        return True  # All characters matched — valid palindrome

Point out the use of Python's built-in .isalnum() — it returns True for both letters and digits, which is exactly the definition given in the problem. Using .lower() at comparison time (rather than pre-processing) keeps the space usage at O(1).

Step 4: Highlight Edge Cases

Step 5: State Time and Space Complexity

"Time is O(n) because across the entire run, the left pointer only ever moves right and the right pointer only ever moves left — each character is touched at most once. Space is O(1) because we compare in-place using two indices. The clean-and-reverse approach is also O(n) time but O(n) space — the two-pointer approach is strictly better on memory."

Interview Checklist for Valid Palindrome

• Restate — "alphanumeric characters only, case-insensitive, check if it reads the same forwards and backwards"
• Clarify that digits are alphanumeric — confirm with the interviewer if unsure
• Mention the naïve clean-and-reverse approach first, then explain why two pointers is better
• State the key insight — skip non-alphanumeric characters in-place rather than stripping them first
• Use .isalnum() — covers both letters and digits cleanly
• Use .lower() at comparison time, not as a preprocessing step
• Guard both inner while loops with left < right to prevent pointer crossing
• Call out: empty effective string → true
• Call out: single character → always true
• State O(n) time and O(1) space — and contrast with the O(n) space alternative

Related Problems to Practice

If you can solve Valid Palindrome cleanly, try these next: