<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 validPalindrome(string s) {
        string bruh=s;
        int i=0,j=s.length()-1,a=0;
        while(i<j){
            if(s[i]!=s[j] && a==0){
                a=1;
                s.erase(j,1);
                bruh.erase(i,1);
                break;
            }
            i++,j--;
        }
        i=0,j=s.length()-1;
        int bhaifinal=0;
        while(i<j){
            if(s[i]!=s[j]) bhaifinal=1;
            i++,j--;
        }
        i=0,j=bruh.length()-1;
        int suarfinal=0;
        while(i<j){
            if(bruh[i]!=bruh[j]) suarfinal=1;
            i++,j--;
        }
        // for(int i=0;i<s.length();i++) cout<<bruh[i]<<" ";
        if(suarfinal==0 || bhaifinal==0) return true;
        else return false;

    }
};
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