<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:
        l = len(s)
        if l <=0: return ''

        result = []
        result += [(i, s[i]+ s[i+1], 0) for i in range(0, len(s)-1) if s[i]==s[i+1]]
        result += [(i, v, 1) for i,v in enumerate(s)]
        
        new_result = result

        #start = 1

        while new_result != [] : # or start:
            #start = 0
            
            result = new_result
            new_result = []

            for i, v, odd in result:
                
                half = len(v)//2
                
                if odd:
                    if i-half-1 >=0 and i+half+1 < l and s[i-half-1] == s[i+half+1]:
                        new_result.append((i, s[i-half-1]+v+s[i+half+1], odd))
                else:
                    #print(i-half, i+half+1)
                    if i-half >=0 and i+half+1 < l and s[i-half] == s[i+half+1]:
                        new_result.append((i, s[i-half]+v+s[i+half+1], odd))
                #print(result, new_result)                   
                    
        return  result[0][1]

                



19300