<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:
        n = len(s)
        ans = ""
        
        # Use binary search to find the longest duplicated substring
        def search(len_sub: int) -> str:
            seen = set()
            for i in range(n - len_sub + 1):
                sub = s[i:i + len_sub]
                if sub in seen:
                    return sub
                seen.add(sub)
            return ""
        
        low, high = 1, n - 1
        while low <= high:
            mid = (low + high) // 2
            dup_sub = search(mid)
            if dup_sub:
                ans = dup_sub
                low = mid + 1  # Try for a longer substring
            else:
                high = mid - 1  # Try for a shorter substring
        
        return ans

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