<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:
        def helper(s: str, removed_letter: bool) -> bool:
            f_ptr = 0
            length = len(s)
            s_ptr = length - 1
            half_size = len(s) // 2
            while f_ptr < s_ptr:
                if s[f_ptr] != s[s_ptr] and removed_letter:
                    return False
                if s[f_ptr] != s[s_ptr]:
                    if helper(s[f_ptr:s_ptr], True):
                        return True
                    elif helper(s[f_ptr + 1:s_ptr + 1], True):
                        return True
                    else:
                        return False
                else:
                    f_ptr += 1
                    s_ptr -= 1
            return True
        return helper(s, False) 
17100