<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 isPalindrome(const std::string& s) {
        int first = 0;
        int last = s.size()-1;

        if(s[first] != s[last]){
            return false;
        }

        if(s.empty()){
            return false;
        }

        if(s.size() == 1){
            return true;
        }


        while(first < last)
        {
            if(s[first] != s[last]){
                return false;
            }

            first++;
            last--;
        }

        return true;
    }

    bool validPalindrome(const std::string& s) {
        int left = 0;
        int right = s.size() - 1;

        while (left < right) {
            if (s[left] != s[right]) {
                // If mismatch, try skipping either left or right character
                // std::string skipLeft = s.substr(0, left) + s.substr(left + 1);
                // std::string skipRight = s.substr(0, right) + s.substr(right + 1);
                // return isPalindrome(skipLeft) || isPalindrome(skipRight);
                return isPalindrome(s.substr(left + 1, right - left)) || 
                       isPalindrome(s.substr(left, right - left));
            }
            ++left;
            --right;
        }

        return true;
    }
};
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