<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 valid(string s){
        int l = 0, r = s.size()-1;
        while(l<r){
            if(s[l]!=s[r]){
                return false;
            }
            l++; r--;
        }
        return true;
    }
    bool validPalindrome(string s) {
        int l=0,r=s.size()-1;
        while(l<r){
            if(s[l]!=s[r]){
                if(r-l==1) return true;
                if(s[l]==s[r-1]){
                    r-=2; l++;
                    if(valid(s.substr(l,r-l+1))){
                        return true;
                    }
                    l--; r+=2;
                }
                if(s[l+1]==s[r]){
                    l+=2; r--;
                    if(valid(s.substr(l,r-l+1))){
                        return true;
                    }
                    l-=2; r++;
                }
                return false;
            }
            else{
                l++; r--;
            }
        }
        return true;
    }
};
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