"""
So, if we start traversing inwards, from both ends of the input string, we can expect to see the same characters, in the same order.

The resulting algorithm is simple:

- Set two pointers, one at each end of the input string-
- If the input is palindromic, both the pointers should point to equivalent characters, at all times. 1
- If this condition is not met at any point of time, we break and return early.
- We can simply ignore non-alphanumeric characters by continuing to traverse further.
- Continue traversing inwards until the pointers meet in the middle.
"""

class Solution:
    def isPalindrome(self, s: str) -> bool:

        def is_alphabet(c):
            return ord(c) - ord("a") < 26 or ord(c) - ord("A") < 26

        if len(s) == 1:
            return is_alphabet(s[0])

        if len(s) == 0:
            return False

        start = 0
        end = len(s) - 1

        while (start<end):
            if not is_alphabet(s[start]):
                start = start+1
                continue

            if not is_alphabet(s[end]):
                end = end - 1
                continue

            if s[start].lower() == s[end].lower():
                start = start + 1
                end = end - 1
                continue
            else:
                return False

        return True