<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 isPalindrome(string s,int i,int j){
        while(i<=j){
            if(s[i]==s[j]){
                i++;
                j--;
            }
            else{
                return false;
            }
        }

        return true;
    }

    bool validPalindrome(string s) {
        int n=s.size();

        int i=0,j=n-1;
        int x=0,y=0;
        
        while(i<=j){
            if(s[i]==s[j]){
                i++;
                j--;
            }
            else if(s[i]!=s[j]){
                if(s[i+1]==s[j]){
                    x=isPalindrome(s,i+1,j);
                }
                if(s[i]==s[j-1]){
                    y=isPalindrome(s,i,j-1);
                }
                return x|y;
            }
        }

        return true;
    }
};
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