<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:
        ss = [x for x in s]

        left = 0
        right = len(ss) - 1
        status = False

        l_status = True
        while left < right:
            if ss[left] != ss[right]:
                if status == True:
                    l_status = False
                    break
                else:
                    left += 1
                    status = True
                    continue
            left += 1
            right -= 1
        
        left = 0
        right = len(ss) - 1
        status = False

        r_status = True
        while left < right:
            if ss[left] != ss[right]:
                if status == True:
                    r_status = False
                    break
                else:
                    right -= 1
                    status = True
                    continue
            left += 1
            right -= 1
            
        return l_status or r_status
17600