<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)
    def hasRepeating(length):
      seen = set()
      for i in range(N-length+1):
        substr = s[i:i+length]
        if substr in seen: 
          self.res = ''.join(substr)
          return True
        seen.add(substr)

      return False


    # Binary Search
    l, r = 0, N-1
    self.res = ""
    while l <= r:
      mid = (l + r) // 2
      if hasRepeating(mid):
        l = mid + 1
      else:
        r = mid - 1

    return self.res
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