<p>Given a string <code>s</code>, return <code>true</code> <em>if the </em><code>s</code><em> can be palindrome after deleting <strong>at most one</strong> character from it</em>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;aba&quot;
<strong>Output:</strong> true
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;abca&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> You could delete the character &#39;c&#39;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;abc&quot;
<strong>Output:</strong> false
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution:
    def validPalindrome(self, s: str) -> bool:
        s_m = [l.lower() for l in s if l.isalnum()]
        lptr, rptr = 0, len(s_m) - 1

        def vp1(s_s) -> bool:
            lptr, rptr = 0, len(s_s) - 1 
            while lptr < rptr:
                if s_s[lptr] != s_s[rptr]:
                    return False
                lptr, rptr = lptr + 1, rptr - 1
            return True            

        while lptr < rptr:
            if s_m[lptr] == s_m[rptr]:
                lptr += 1
                rptr -= 1
                continue

            return vp1(s_m[lptr: rptr]) or vp1(s_m[lptr + 1: rptr + 1])

        return True
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