<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:
        def search(L):
            """Search for a duplicate substring of length L using rolling hash."""
            MOD = 2**63 - 1
            p = pow(26, L, MOD)
            current_hash = 0
            for i in range(L):
                current_hash = (current_hash * 26 + ord(s[i]) - ord('a')) % MOD
            
            seen = {current_hash}
            for i in range(1, len(s) - L + 1):
                current_hash = (current_hash * 26 + ord(s[i + L - 1]) - ord('a')) % MOD
                current_hash = (current_hash - (ord(s[i - 1]) - ord('a')) * p) % MOD
                if current_hash in seen:
                    return i
                seen.add(current_hash)
            return -1
        
        left, right = 1, len(s)
        start = -1
        while left <= right:
            mid = (left + right) // 2
            idx = search(mid)
            if idx != -1:
                start = idx
                left = mid + 1
            else:
                right = mid - 1
        
        return s[start:start + left - 1] if start != -1 else ""

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