<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:
void call(string &k,string &ss,int first, set<string>s[],int ind)
{
    if(ind>=ss.size())
    {
        s[first].insert(k);
        return ;
    }
k+=ss[ind];
call(k,ss,first,s,ind+1);
k.pop_back();
call(k,ss,first,s,ind+1);
}
    int findLUSlength(vector<string>& v) {
        int n=v.size(),i,j,count=0,maxi=0;
        set<string>s[n];
        for(i=0;i<n;i++)
        {
            string k;
            call(k,v[i],i,s,0);
        }
        for(i=0;i<n;i++)
        {
            for(auto &it:s[i])
            {
                count=it.size();
              for(j=0;j<n;j++)
              {
                if(i!=j)
                {
                    if(s[j].find(it)!=s[j].end())
                    {
                        count=0;
                        break;
                    }
                }
              }
              if(count!=0)
              {
                maxi=max(maxi,count);
              }  
            }
        }
        if(maxi==0)
        return -1;
        return maxi;
    }
};
37001