# [Valid Palindrome](https://leetcode.com/problems/valid-palindrome/)
###### tags: `Leetcode`, `Easy`, `Two Pointers`

## Approach
* Initialize two pointers `left` and `right` to the `0` and `len(nums) - 1` respectively
* Check if char at left and char at right is alphanumeric
* if char at left is equal to char at right then increment left and decrement right by 1 each
* else return False
* If char at left is not alphanumeric then increment left
* If char at right is not alphanumeric then decrement right
* Return True

## Asymptotic Analysis
### Time Complexity: **O(N)**
### Space Complexity: **O(1)**

## Code

class ValidPalindrome:

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


s = "A man, a plan, a canal: Panama"
print(ValidPalindrome.is_palindrome(s))