<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:
    
    unordered_map<string,int> mp;
    void func(string &str, int n, int i, string &s)
    {
        if(i==n)
        {
            mp[s]++;
            return;
        }

        s+= str[i];
        func(str, n, i+1, s);
        s.pop_back();
        func(str, n, i+1, s);
    }
    int findLUSlength(vector<string>& strs) 
    {
        for(auto &str : strs)
        {
            string s="";
            func(str, str.size(), 0, s);
        }

        int ans=0;
        for(auto &it : mp)
            if(it.second==1)
                ans= max(ans, (int)(it.first.size()));
        return ans==0?-1:ans;
    }
};
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