<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 check_palindrome(self, s_list) -> bool:
        s_length = len(s_list)
        if s_length % 2 == 0:
            val = s_length//2
        else:
            val = s_length//2 + 1

        for i in range(val):
            if s_list[i] != s_list[-i-1]:
                return False, i
        
        return True, 0

    def validPalindrome(self, s: str) -> bool:
        s_list = list(s)
        s_length = len(s_list)

        check, i = self.check_palindrome(s_list)

        if check:
            return True
        else:
            
            check1, _ = self.check_palindrome(s_list[i+1:s_length-i])

            if check1:
                return True

            check2, _ = self.check_palindrome(s_list[i:s_length-(i+1)])

            return check2


            
        
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