<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:

    def longestPalindrome(self, s: str) -> str:
        
        longest = ""
        for i in range(len(s)):
            left = i
            right = i

            while left>=0 and right<len(s) and s[left] == s[right]:
                left -= 1
                right += 1
            
            odd_palindrome = s[left +1: right]
            print(odd_palindrome, "odd")
            if len(odd_palindrome) > len(longest):
                longest = odd_palindrome

            left = i
            right = i+1

            while left>=0 and right<len(s) and s[left] == s[right]:
                left -= 1
                right += 1

            even_palindrome = s[left +1: right]
            print(even_palindrome, "even")
            if len(even_palindrome) > len(longest):
                longest = even_palindrome

        return longest 





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