<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>

from functools import lru_cache
class Solution:
    def validPalindrome(self, s: str) -> bool:
        return self.recursiveIsValidPalindrome(s, 1)

    @lru_cache()
    def recursiveIsValidPalindrome(self, s, removals_remaining):
        if removals_remaining < 0:
            return False

        l, r = 0, len(s) - 1
        while l < r:
            if s[l] != s[r]:
                # remove the left character
                if self.recursiveIsValidPalindrome(s[l+1:r+1], removals_remaining - 1):
                    return True

                # remove the right character
                if self.recursiveIsValidPalindrome(s[l:r], removals_remaining - 1):
                    return True

                return False
            
            l += 1
            r -= 1

        return True
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