<p>Given a string <code>s</code>, return <em>the longest</em> <span data-keyword="palindromic-string"><em>palindromic</em></span> <span data-keyword="substring-nonempty"><em>substring</em></span> in <code>s</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;babad&quot;
<strong>Output:</strong> &quot;bab&quot;
<strong>Explanation:</strong> &quot;aba&quot; is also a valid answer.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;cbbd&quot;
<strong>Output:</strong> &quot;bb&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s</code> consist of only digits and English letters.</li>
</ul>

class Solution {
public:
    string result;
    int get_len(int itr1,int itr2,string s)
    {
        int len=0;
        while(itr1>-1 and itr2<s.size())
        {
            if(s[itr1]==s[itr2])
            {
                len++;
                itr1--;
                itr2++;
            }
            else
            {
                break;
            }
        }
        return len;
    }

    string longestPalindrome(string s) 
    {
        int max_ans=0;
        for(int itr=0;itr<s.size();itr++)
        {
            int ans=0;
            cout<<itr<<endl;
            
            if(itr>0 and s[itr-1]==s[itr])
            {
                ans=get_len(itr-1,itr,s);
                cout<<"ans="<<ans<<endl;
                if(max_ans<ans*2)
                {
                    max_ans=ans*2;
                    result=s.substr(itr-ans,ans*2);
                }
                
            }
            if(itr>0 and itr<s.size())
            {
                ans=get_len(itr-1,itr+1,s);
                cout<<"ans="<<ans<<endl;
                if(max_ans<ans*2+1)
                {
                    max_ans=ans*2+1;
                    result=s.substr(itr-ans,(ans*2)+1);
                }
                ans++;
            }
            if(max_ans==0)
            {
                    max_ans=1;
                    result=s[itr];
                
            }
            max_ans=max(max_ans,ans);


        }

    return result;

    }
};
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