<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:
        """
        Given a string `s`, return `True` if the string `s` can be a palindrome after
        deleting at most one character from it.  Else, return `False`.

        Notes:
            `s` contains only lowercase English letters (a-z)

        Examples:
            >>> Solution().validPalindrome("aba")
            True
            >>> Solution().validPalindrome("abca")
            True
            >>> Solution().validPalindrome("abc")
            False
        """
        n = len(s)
        midpoint = n // 2
        def bounded_check(left: int, right: int, tries: int) -> bool:
            """
            Check to see if the substring `s[left : right + 1]` is a palindrome.
            """
            while left <= midpoint and left < right:
                if s[left] != s[right]:
                    if not tries:
                        return False
                    tries -= 1
                    # Try removing mismatch on the left first, then try removing the
                    # mismatch on the right.
                    return bounded_check(left + 1, right, tries) or bounded_check(left, right - 1, tries)
                left += 1
                right -= 1
            return True

        return bounded_check(0, n - 1, 1)

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