<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:
        
        q = collections.deque()
        for c in s:
            q.append(c)

        while len(q) >= 2:
            left = q.popleft()
            right = q.pop()
            if left != right:
                return (
                    self.validP(f"{left}{''.join(q)}") or 
                    self.validP(f"{''.join(q)}{right}")
                )                

        return True

    def validP(self, s: str) -> bool:
        q = collections.deque()
        for c in s:
            q.append(c)
        
        while len(q) >= 2:
            if q.popleft() != q.pop():
                return False

        return True
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