<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 solve(int ind,string &temp,string &s,set<string>&st)
    {
        if(ind>=s.size())
        {
            st.insert(temp);
            return ;
        }
        temp+=s[ind];
        solve(ind+1,temp,s,st);
        temp.pop_back();

        solve(ind+1,temp,s,st);
    }
    int findLUSlength(vector<string>& strs) {
        map<string,int>mp;
        for(auto it:strs)
        {
            set<string>st;
            string temp="";
            solve(0,temp,it,st);
            for(auto t:st)
            {
                mp[t]++;
            }
        }
        int ans=-1;
        for(auto it:mp)
        {
            if(it.second==1)
            {

                ans=max(ans,(int)it.first.size());
            }
        }
        return ans;

    }
};
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