<p>Given an array of strings <code>strs</code>, return <em>the length of the <strong>longest uncommon subsequence</strong> between them</em>. If the longest uncommon subsequence does not exist, return <code>-1</code>.</p>

<p>An <strong>uncommon subsequence</strong> between an array of strings is a string that is a <strong>subsequence of one string but not the others</strong>.</p>

<p>A <strong>subsequence</strong> of a string <code>s</code> is a string that can be obtained after deleting any number of characters from <code>s</code>.</p>

<ul>
	<li>For example, <code>&quot;abc&quot;</code> is a subsequence of <code>&quot;aebdc&quot;</code> because you can delete the underlined characters in <code>&quot;a<u>e</u>b<u>d</u>c&quot;</code> to get <code>&quot;abc&quot;</code>. Other subsequences of <code>&quot;aebdc&quot;</code> include <code>&quot;aebdc&quot;</code>, <code>&quot;aeb&quot;</code>, and <code>&quot;&quot;</code> (empty string).</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> strs = ["aba","cdc","eae"]
<strong>Output:</strong> 3
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> strs = ["aaa","aaa","aa"]
<strong>Output:</strong> -1
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= strs.length &lt;= 50</code></li>
	<li><code>1 &lt;= strs[i].length &lt;= 10</code></li>
	<li><code>strs[i]</code> consists of lowercase English letters.</li>
</ul>

class Solution {
 public:
  int findLUSlength(vector<string>& strs) {
    unordered_set<string> seen;
    unordered_set<string> duplicates;

    for (const string& str : strs)
      if (seen.contains(str))
        duplicates.insert(str);
      else
        seen.insert(str);

    ranges::sort(strs, ranges::greater{},
                 [](const string& s) { return s.length(); });

    for (int i = 0; i < strs.size(); ++i) {
      if (duplicates.contains(strs[i]))
        continue;
      bool isASubsequence = false;
      for (int j = 0; j < i; ++j)
        isASubsequence |= isSubsequence(strs[i], strs[j]);
      if (!isASubsequence)
        return strs[i].length();
    }

    return -1;
  }

 private:
  // Returns true if a is a subsequence of b.
  bool isSubsequence(const string& a, const string& b) {
    int i = 0;
    for (const char c : b)
      if (i < a.length() && c == a[i])
        ++i;
    return i == a.length();
  };
};
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