<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:
        queue: list[tuple[int, int]] = [(0, 1)]
        result = ''
        while queue:
            start, end = queue.pop(0)
            if s[start:end] not in s[start + 1:]:
                if start + 1 < len(s):
                    queue.append((start + 1, start + len(result) + 1))
                continue
            if end + 1 <= len(s):
                queue.append((start, end + 1))
            if len(result) < end - start:
                result = s[start:end]

        return result
        
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