<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) {
        int res = 0;

        for (int i = 0; i < s.size(); i++) {
            res = max(res, lengthPalindrome(s, i).second);
        }
        for (int i = 0; i < s.size(); i++) {
            pair<int, int> length = lengthPalindrome(s, i);
            if (length.second == res)
                return s.substr(length.first, length.second);
        }
        return "";
    }

    pair<int, int> lengthPalindrome(string s, int index) {
        int expand = 0;

        while (index - expand >= 0 && index + expand < s.size() && s[index - expand] == s[index + expand])
            expand++;

        int expandP = 0;
        while (index - expandP >= 0 && index + 1 + expandP < s.size() && s[index - expandP] == s[index + 1 + expandP])
            expandP++;
        if (expand <= expandP)
            return {index - expandP + 1, expandP * 2};
        else
            return {index - expand + 1, expand * 2 - 1};
    }
};
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