<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 {
public:
    string longestDupSubstring(string S)
    {
        std::string_view longest;
        std::unordered_set<std::string_view> set;
        size_t beg = 1;
        size_t end = S.size() - 1;
        while (beg <= end)
        {
            auto len = beg + (end - beg) / 2;
            bool found = false;
            for (size_t i = 0; i != S.size() - len + 1; ++i)
            {
                const auto [it, inserted] = set.emplace(S.data() + i, len);
                if (!inserted)
                {
                    found = true;
                    longest = *it;
                    break;   
                }
            }
            
            if (found)
                beg = len + 1;
            else
                end = len - 1;
            
            set.clear();      
        }
            
        return {longest.begin(), longest.end()};
    }
};
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