Longest Uncommon Subsequence II
<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>
Array; Hash Table; Two Pointers; String; Sorting
class Solution {
public:
    int findLUSlength(vector<string>& strs) {
        int ans = -1;
        int n = strs.size();
        auto check = [&](const string& s, const string& t) {
            int m = s.size(), n = t.size();
            int i = 0;
            for (int j = 0; i < m && j < n; ++j) {
                if (s[i] == t[j]) {
                    ++i;
                }
            }
            return i == m;
        };
        for (int i = 0, j; i < n; ++i) {
            int x = strs[i].size();
            for (j = 0; j < n; ++j) {
                if (i != j && check(strs[i], strs[j])) {
                    x = -1;
                    break;
                }
            }
            ans = max(ans, x);
        }
        return ans;
    }
};