<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(self, s:str):
        p = 0
        q = len(s) - 1
        flag = True
        while p < q:
            if s[p] != s[q]:
                flag = False
            p += 1
            q -= 1
        return flag
    def validPalindrome(self, s: str) -> bool:
        p = 0
        q = len(s) - 1
        s = list(s)

        while p < q:
            if s[p] != s[q]:
                return self.check(s[:p] + s[p+1:]) or self.check(s[:q] + s[q+1:])
            p += 1
            q -= 1
        return True

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