<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 longestPalindrome(string s) {
        string ans = "";
        auto check = [&](int l, int r, string str)
        {
            string left,right;
            while(l >= 0 && r < s.size())
            {
                if(s[l] == s[r])
                {
                    left.push_back(s[l]);
                    right.push_back(s[r]);
                    l--;
                    r++;
                }
                else
                {
                    break;
                }
            }
            reverse(left.begin(),left.end());
            string tmp = left + str + right;
            return tmp;
        };
        for(int i = 0; i < s.size(); i++)
        {
            string str;
            str += s[i];
            string odd = check(i-1,i+1,str);
            string even;
            if(i+1 < s.size() && s[i] == s[i+1])
            {
                str += s[i+1];
                even = check(i-1,i+2,str);
            }
            // cout << i << " " << odd << " " << even << endl; 
            if(odd.size() > ans.size()) ans = odd;
            if(even.size() > ans.size()) ans = even;
        }
        return ans;
    }
};
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