<p>You are given a <strong>0-indexed</strong> string <code>s</code> and are tasked with finding two <strong>non-intersecting palindromic </strong>substrings of <strong>odd</strong> length such that the product of their lengths is maximized.</p>

<p>More formally, you want to choose four integers <code>i</code>, <code>j</code>, <code>k</code>, <code>l</code> such that <code>0 &lt;= i &lt;= j &lt; k &lt;= l &lt; s.length</code> and both the substrings <code>s[i...j]</code> and <code>s[k...l]</code> are palindromes and have odd lengths. <code>s[i...j]</code> denotes a substring from index <code>i</code> to index <code>j</code> <strong>inclusive</strong>.</p>

<p>Return <em>the <strong>maximum</strong> possible product of the lengths of the two non-intersecting palindromic substrings.</em></p>

<p>A <strong>palindrome</strong> is a string that is the same forward and backward. A <strong>substring</strong> is a contiguous sequence of characters in a string.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;ababbb&quot;
<strong>Output:</strong> 9
<strong>Explanation:</strong> Substrings &quot;aba&quot; and &quot;bbb&quot; are palindromes with odd length. product = 3 * 3 = 9.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;zaaaxbbby&quot;
<strong>Output:</strong> 9
<strong>Explanation:</strong> Substrings &quot;aaa&quot; and &quot;bbb&quot; are palindromes with odd length. product = 3 * 3 = 9.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution:
    def maxProduct(self, s: str) -> int:
        n = len(s)
        dp, before, after = [0]*n, [0]*n, [0]*n
        c, r = -1, -1

        for i in range(n):
            k = min(dp[2*c-i],r-i) if i<=r else 0
            p, q = i-k, i+k

            while p>=0 and q<n and s[p] == s[q]:
                before[q] = max(before[q],q-p+1)
                after[p] = max(after[p],q-p+1)
                p -= 1
                q += 1

            dp[i] = q-i-1

            if q-1 > r: c, r = i, q-1

        for i in range(1,n):
            before[i] = max(before[i-1],before[i])

        for i in range(n-2,-1,-1):
            after[i] = max(after[i+1],after[i])

        return max([before[i-1]*after[i] for i in range(1,n)])
719