Two pointers: valid palindrome

A palindrome reads the same forwards and backwards. The catch in this classic (LeetCode #125) is that we only care about alphanumeric characters, and we ignore case. So "A man, a plan, a canal: Panama" counts as a palindrome — strip the punctuation and spaces, lowercase it, and you get amanaplanacanalpanama, which is a mirror of itself.

Brute force would build a cleaned string and compare it to its reverse. The slicker move is two pointers: one starting at the far left, one at the far right, walking toward the middle. Skip anything that isn't a letter or digit, then compare the characters you land on.

left, right = 0, len(s) - 1
while left < right:
    # skip non-alphanumerics on each side, then compare
    ...

If the two pointers ever disagree, it's not a palindrome. If they cross without a single mismatch, it is.

Your task

Write is_palindrome(s) returning True if s is a palindrome when you consider only alphanumeric characters and treat upper/lowercase as equal, otherwise False.

is_palindrome("A man, a plan, a canal: Panama")   # True
is_palindrome("race a car")                        # False
is_palindrome("")                                  # True   (nothing to fail)
is_palindrome(".,")                                # True   (no alphanumerics)

Hint: c.isalnum() tells you whether c is a letter or digit, and c.lower() normalizes case. An empty string — or one made entirely of punctuation — is trivially a palindrome.