<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:
        @lru_cache(2000)
        def dp(i: int, j: int) -> str:
            if i >= j:
                return s[j]
            
            if self.is_palindrome(s[i:j+1]):
                return s[i:j+1]

            left = dp(i+1, j)
            right = dp(i, j-1)
            
            return left if len(left) > len(right) else right
        
#         if len(s) == 1:
#             return s
        
#         if len(s) == 2:
#             if self.is_palindrome(s):
#                 return s
#             return s[0]
        
        return dp(0, len(s) - 1)
        
        
    def is_palindrome(self, s: str) -> bool:
        i, j = 0, len(s) - 1
        while i <= j:
            if s[i] != s[j]:
                return False
            i += 1
            j -= 1
            
        return True
23100