<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 does(self,s,l):
        n = len(s)
        h = {}
        for i in range(n-l+1):
            if s[i:i+l] in h:
                return s[i:i+l]
            h[s[i:i+l]] = 1
        return False

    def longestDupSubstring(self, s: str) -> str:
        
        l = 1
        h = len(s) - 1
        # ans = 0
        out = ""
        while h>=l:
            mid = l + (h-l)//2
            temp = self.does(s,mid)
            print(l,mid,h,temp)
            if temp:
                out = temp
                l = mid + 1
            else:
                h = mid - 1
        return out
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