<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)
        dp = [[0] * (n + 1) for _ in range(n + 1)]
        ans = [0, 0]        # [start, end] of longest palindrome substring

        # for i in range(n):
        #     dp[i][i] = 1    # middle diagnal is always true, every 1 character is a palindrome by itself
        #     ans = [i, i]    # save longest palindrome (every charater is one, so we just save the last one)
        
        # for i in range(n - 1):
        #     if s[i] == s[i + 1]:    # check if the diagnal character is the same as the character to its right
        #         dp[i][i + 1] = 1    # if the neighbor character is the same, set it to 1 
        #         ans = [i, i + 1]    # save longest palindrome (last one if more than one pair is a palidrome)

        for expand in range(n):
            for i in range(n):  
                j = i + expand
                if 0 <= i < n and 0 <= j < n:
                    if s[i] == s[j] and (i == j or i + 1 == j or dp[i + 1][j - 1] == 1):
                        ans = [i, j]
                        dp[i][j] = 1

        return s[ans[0]: ans[1] + 1] 
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