<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 palindrome_s(string s, int start, int end, bool b) {
            s.erase(start,1);
            end--;
            while(start<end)
            {
                if(s[start]==s[end])
                {
                    start++;
                    end--;
                }
                else
                {
                    b=false;
                    break;
                }
            }
            return b;
        }

    bool palindrome_e(string s, int start, int end, bool b)
    {
        s.erase(end,1);
        end--;

        while(start<end)
        {
            if(s[start]==s[end])
            {
                start++;
                end--;
            }
            else
            {
                b=false;
                break;
            }
        }
        return b;
    }

    bool validPalindrome(string s) {

        int start=0, end = s.length()-1;
        int count=0;
        bool b=true,b1,b2;

        while(s[start]==s[end] && start<end)
        {
            start++;
            end--;
        }

        if(count<2)
        {
            b1 = palindrome_s(s,start,end,b);
            b2 = palindrome_e(s,start,end,b);
            b= b1 || b2;
            count++;
        }
        else{b=false;}
        return b; 
    }
};
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