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

        return s[left + 1 : right]

    def longestPalindrome(self, s: str) -> str:
        """
        Approach
        - We need to take into consideration the 2 types of palindromes, odds and evens
        - We can iterate over the s and try both cases using to ptrs
        - In this case, our starting point will be the iteration index, and the prev and
          next position will be our ptrs.
        """
        ans = ""

        for i in range(len(s)):
            # Check for odd palindrome
            odd_largest = self.iterate_palindrome(i - 1, i + 1, s)
            ans = odd_largest if len(odd_largest) > len(ans) else ans

            # Check for even palindrome
            even_largest = self.iterate_palindrome(i, i + 1, s)
            ans = even_largest if len(even_largest) > len(ans) else ans


        return ans 


16500