<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:
        # Edge Case
        if not s:
            return ""

        # Dynamic Programming
        # dp[start, end] -> if s[start: end + 1] if palindromic or not
        longest_palindrome = s[0]
        dp = []
        # initialization
        for i in range(len(s)):
            current = [False] * len(s)
            current[i] = True
            dp.append(current)

        for end in range(1, len(s)):
            for start in range(end-1, -1, -1):
                if s[start] == s[end]:
                    if end - start == 1 or dp[start + 1][end - 1] is True:
                        dp[start][end] = True
                        if end - start + 1 > len(longest_palindrome):
                            longest_palindrome = s[start: end + 1]

        return longest_palindrome
24200