<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 {
    private:
        bool solve(string s){
            for(int i=0, j=s.size()-1; i<j; ++i, --j) if(s[i] != s[j]) return false;

            return true;
        }

    public:
        bool validPalindrome(string &s){
            cin.tie(0);
            cout.sync_with_stdio(0);

            for(int i=0, j=s.size()-1; i<j; ++i, --j){
                if(s[i] != s[j]){
                    return solve(s.substr(i, j-i)) || solve(s.substr(i+1, j-i));
                }
            }

            return true;
        }
};
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