<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 calc(n):
            nonlocal ans
            if not n:
                return True
            visited = set()
            for i in range(len(s)-n+1):
                if s[i:i+n] in visited:
                    ans = s[i:i+n]
                    return True
                visited.add(s[i:i+n])
            return False

        l = 0
        h = len(s)-1
        ans = ''
        while l <= h:
            mid = (l+h)//2
            if calc(mid):
                l = mid+1
            else:
                h = mid-1
        return ans
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