<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:
        from functools import lru_cache
        
        # Memoized recursive function to check if s[i:j+1] can be made a palindrome with up to k deletions
        @lru_cache(None)
        def validPalindrome(s, i, j, k):
            while i < j:
                if s[i] != s[j]:
                    # if we can still remove characters (k > 0), try both cases (skip i or skip j)
                    if k > 0:
                        return validPalindrome(s, i + 1, j, k - 1) or validPalindrome(s, i, j - 1, k - 1)
                    else:
                        return False
                else:
                    i += 1
                    j -= 1
            return True

        return validPalindrome(s, 0, len(s) - 1, 1)
18500