Valid Palindrome
<p>A phrase is a <strong>palindrome</strong> if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.</p>

<p>Given a string <code>s</code>, return <code>true</code><em> if it is a <strong>palindrome</strong>, or </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;A man, a plan, a canal: Panama&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> &quot;amanaplanacanalpanama&quot; is a palindrome.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;race a car&quot;
<strong>Output:</strong> false
<strong>Explanation:</strong> &quot;raceacar&quot; is not a palindrome.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot; &quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> s is an empty string &quot;&quot; after removing non-alphanumeric characters.
Since an empty string reads the same forward and backward, it is a palindrome.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>s</code> consists only of printable ASCII characters.</li>
</ul>
Two Pointers; String
class Solution:
    def isPalindrome(self, s: str) -> bool:
        i, j = 0, len(s) - 1
        while i < j:
            if not s[i].isalnum():
                i += 1
            elif not s[j].isalnum():
                j -= 1
            elif s[i].lower() != s[j].lower():
                return False
            else:
                i, j = i + 1, j - 1
        return True
