<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:
        @cache
        def find(n):
            seen = set()
            sub = ""
            for char in s:
                sub+=char
                if len(sub)==n:
                    st = sub
                    if st in seen:
                        return st
                    seen.add(st)
                    sub = sub[1:]
            return ""
        
        def bin_search(left,right):
            mid = (left + right) // 2
            if left==right:
                return left
            if find(mid)!="":
                return bin_search(mid+1,right)
            else:
                return bin_search(left,mid)
        
        result = bin_search(1,len(s))-1
        return find(result)
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