<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:
        l = 0
        r = len(s) - 1

        Q = [[l, r, 0]]
        i = 0
        while i < len(Q):
            l, r, count = Q[i]
            i += 1
            if count > 1:
                continue
                
            if l >= r:
                return True
            
            
            if s[l] == s[r]:
                Q.append([l + 1, r - 1, count])
            else:
                if s[l + 1] == s[r]:
                    Q.append([l + 1, r, count + 1])
                if s[l] == s[r - 1]:
                    Q.append([l, r - 1, count + 1])
        
        return False
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