Maximum Product of the Length of Two Palindromic Substrings
<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>
String; Hash Function; Rolling Hash
class Solution {
public:
    long long maxProduct(string s) {
        long long res = 0, l = 0, n = s.size();
        vector<int> m(n), r(n);

        for (int i = 0, l = 0, r = -1; i < n; ++i) {
            int k = (i > r) ? 1 : min(m[l + r - i], r - i + 1);
            while (0 <= i - k && i + k < n && s[i - k] == s[i + k])
                k++;
            m[i] = k--;
            if (i + k > r) {
                l = i - k;
                r = i + k;
            }
        }

        queue<array<int, 2>> q, q1;

        for (int i = n - 1; i >= 0; --i) {
            while (!q.empty() && q.front()[0] - q.front()[1] > i - 1)
                q.pop();
            r[i] = 1 + (q.empty() ? 0 : (q.front()[0] - i) * 2);
            q.push({i, m[i]});
        }

        for (int i = 0; i < n - 1; i++) {
            while (!q1.empty() && q1.front()[0] + q1.front()[1] < i + 1)
                q1.pop();
            l = max(l, 1ll + (q1.empty() ? 0 : (i - q1.front()[0]) * 2));
            res = max(res, l * r[i + 1]);
            q1.push({i, m[i]});
        }

        return res;
    }
};
