<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:
   int is_pal(string s){
    int i = 0 ; int j = s.length() - 1 ;
    while(i < j){
        if(s[i] != s[j]){
            return false ;
        }
        i++;j-- ;
    }
    return true ;
}

bool validPalindrome(string s) {
    int i = 0 ; 
    int j = s.length()- 1 ;
    while(i <= j){
        if(s[i] == s[j]){
            i++;j--;
        }
        else{
            
            return is_pal(s.substr(0 , i) + s.substr(i+1)) || is_pal(s.substr(0 , j) + s.substr(j+1)) ;
        }
    }
    return true ;
}
};
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