<p>Given a string <code>s</code>, consider all <em>duplicated substrings</em>: (contiguous) substrings of s that occur 2 or more times.&nbsp;The occurrences&nbsp;may overlap.</p>

<p>Return <strong>any</strong> duplicated&nbsp;substring that has the longest possible length.&nbsp;If <code>s</code> does not have a duplicated substring, the answer is <code>&quot;&quot;</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "banana"
<strong>Output:</strong> "ana"
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "abcd"
<strong>Output:</strong> ""
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= s.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution:
    def longestDupSubstring(self, s: str) -> str:
        self.longest_duplicated_substring = ""
        def search(L):
            """Search for a duplicate substring of length L using rolling hash."""
            seen = set()
            cur = s[:L]
            seen.add(cur)
            for i in range(L, len(s)):
                cur = cur[1:] + s[i]
                if cur in seen:
                    self.longest_duplicated_substring = cur
                    return True
                seen.add(cur)
            return False
        
        left, right = 0, len(s) - 1
        while left <= right:
            mid = left + (right - left) // 2
            if search(mid):
                left = mid + 1
            else:
                right = mid - 1
        return self.longest_duplicated_substring
        

        

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