<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 dup(size):
            table = set()
            tmp = s[:size]
            table.add(tmp)
            for i in range(len(s) - size):
                tmp = tmp[1:] + s[size + i]
                if tmp in table:
                    return [True, tmp]
                table.add(tmp)
            return [False, '']
        
        l, r = 0, len(s) - 1

        res = ''

        while l <= r:
            mid = (l + r) // 2
            flag, st = dup(mid)
            if flag:
                l = mid + 1
                res = st
            else:
                r = mid - 1
        
        

        return res
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