<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 n = s.size();
        int start = 0;
        int end = n-1;

        while(start <= end){
            if(s[start] == s[end]){
                start++;
                end--;
            }else{
 return is_pal(s.substr(0 , start) + s.substr(start+1)) || is_pal(s.substr(0 , end) + s.substr(end+1)) ;
            }
        }
        return true;
    }
};
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