Solution for interview question: Palindrome String

1. Understanding the Problem

The task is to determine if a given string is a palindrome.
Ignore non-alphanumeric characters, and treat upper- and lower-case letters as the same.
The goal is to avoid extra storage, so we'll use character-by-character comparison instead of storing a reversed string.

2. Use Two Pointers to Compare Characters

Use a two-pointer technique, one starting from the beginning and one from the end of the string.
Increment the left pointer and decrement the right pointer after each comparison.

def is_palindrome(s: str) -> bool:
    left, right = 0, len(s) - 1
    return left, right

3. Skip Non-Alphanumeric Characters

As you move the two pointers, skip any characters that are not alphanumeric (like spaces, punctuation).
Use Python's isalnum() method to check if a character is alphanumeric.

def is_palindrome(s: str) -> bool:
    left, right = 0, len(s) - 1
    while left < right:
        # Skip non-alphanumeric characters
        while left < right and not s[left].isalnum():
            left += 1
        while left < right and not s[right].isalnum():
            right -= 1
    return left, right

4. Compare Characters (Ignore Case)

Compare the characters at the left and right pointers, ignoring case.
If the characters are not equal, the string is not a palindrome and you can return False immediately.

def is_palindrome(s: str) -> bool:
    left, right = 0, len(s) - 1
    while left < right:
        # Skip non-alphanumeric characters
        while left < right and not s[left].isalnum():
            left += 1
        while left < right and not s[right].isalnum():
            right -= 1
        # Compare characters (ignoring case)
        if s[left].lower() != s[right].lower():
            return False
        left += 1
        right -= 1
    return True

5. Return the Result

If all characters match as you move the pointers, the string is a palindrome.
Once the two pointers meet or cross, return True to indicate the string is a palindrome.

def is_palindrome(s: str) -> bool:
    left, right = 0, len(s) - 1
    while left < right:
        while left < right and not s[left].isalnum():
            left += 1
        while left < right and not s[right].isalnum():
            right -= 1
        if s[left].lower() != s[right].lower():
            return False
        left += 1
        right -= 1
    return True

6. Handle Edge Cases

Consider edge cases like an empty string or a string containing only non-alphanumeric characters.
These should return True since they can be considered palindromes.

def is_palindrome(s: str) -> bool:
    left, right = 0, len(s) - 1
    while left < right:
        while left < right and not s[left].isalnum():
            left += 1
        while left < right and not s[right].isalnum():
            right -= 1
        if s[left].lower() != s[right].lower():
            return False
        left += 1
        right -= 1
    return True  # Empty strings or strings with only non-alphanumeric characters are palindromes