<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 isPal(string s) {
        int n = s.size();
        for(int i = 0; i < n / 2; i++) {
            if(s[i] != s[n - i - 1]) return false;
        }
        return true;
    }

    bool validPalindrome(string s) {
        if(isPal(s)) return true;

        int l = 0, r = s.size() - 1, skip = 1;
        while(l <= r) {
            if(s[l] != s[r]) {
                return isPal(s.substr(l, r - l)) || isPal(s.substr(l + 1, r - l));
            }
            ++l, --r;
        }


        return true;
    }
};
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