<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 {
public:
    bool validPalindrome(string s) {
        int counter = 0;
        int low  = 0, high = s.size() - 1;
        
        while (low < high) {
            if (s[low] != s[high]) {
                // If there's a mismatch, try both options: skipping low or skipping high
                return isPalindrome(s, low + 1, high) || isPalindrome(s, low, high - 1);
            } else {
                low++;
                high--;
            }
        }
        
        return true;
    }

private:
    // Helper function to check if a string is a palindrome between two indices
    bool isPalindrome(const string &s, int low, int high) {
        while (low < high) {
            if (s[low] != s[high]) {
                return false;
            }
            low++;
            high--;
        }
        return true;
    }
};

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