<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 isPalindrome(self, s: str) -> bool:
        new_s_arr = []
        
        for c in s:
            if c.isalnum():
                new_s_arr.append(c.lower())
        
        s = "".join(new_s_arr)
        
        l, r = 0, len(s) - 1
        while l <= r:
            l_char, r_char = s[l], s[r]
            if l_char != r_char:
                return False
            l += 1
            r -= 1
        
        return True
    
    def validPalindrome(self, s: str) -> bool:
        '''
        for each character in the string, I want to explore whether or not deleting will lead to a palindrome. 
        '''
        
        if len(s) <= 1 or self.isPalindrome(s):
            return True
            
            
        l, r = 0, len(s) - 1
        while l <= r:
            if s[l] == s[r]:
                l += 1
                r -= 1
                continue
                
            s_withoutl = s[:l] + s[l + 1:]
            if self.isPalindrome(s_withoutl):
                return True
            
            s_withoutr = s[:r] + s[r + 1:]
            if self.isPalindrome(s_withoutr):
                return True
            return False
            
        return True
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