<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 right = s.length()-1;
        int diff = 0;
        string reversedSR, reversedSL;
        for (int i = 0; i < s.length()/2; i++) {
            if (s[i] != s[right-i]) {
                string temp = s;            
                if (s[i] == s[right-i-1]) {
                    s.erase(right-i, 1);
                    reversedSR = s;
                    reverse(reversedSR.begin(), reversedSR.end());
                }
                if (temp[i+1] == temp[right-i]) {
                    temp.erase(i, 1);
                    reversedSL = temp;
                    reverse(reversedSL.begin(), reversedSL.end());
                }
                return (reversedSR == s || reversedSL == temp);
            }
        }
        return true;
    }
};
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