<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:
        n = len(s)
        if n == 0:
            return ""

        # dp[i][j] will be True if s[i:j+1] is a palindrome
        dp = [[False] * n for _ in range(n)]
        start = 0
        max_len = 1

        # Every single character is a palindrome
        for i in range(n):
            dp[i][i] = True

        # Check for two-character palindromes
        for i in range(n - 1):
            if s[i] == s[i + 1]:
                dp[i][i + 1] = True
                start = i
                max_len = 2

        # Fill the table for substrings of length 3 and greater
        for length in range(3, n + 1):  # length is the length of the substring
            for i in range(n - length + 1):
                j = i + length - 1  # end index of the current substring
                if s[i] == s[j] and dp[i + 1][j - 1]:
                    dp[i][j] = True
                    start = i
                    max_len = length

        return s[start:start + max_len]
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