<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:
    
    bool notPresent(int j, vector<string>&v, int i) {
        vector<int>v1;
        string s;
        int idx=0;
        while(j) {
            if(j&1) {
                v1.push_back(idx);
            }
            j=j>>1;
            idx++;
        }
        for(auto x:v1)
        s=s+v[i][x];
        for(int k=0;k<v.size();k++) {
            int p=0;
            if(k==i) 
            continue;
            for(int j1=0;j1<v[k].length();j1++) {
            
                if (p>=s.length()) {
                    return false;
                }
                if(s[p]==v[k][j1]) { 
                    p++;
                    continue;
                }
            }
            if(p>=s.length()) {
                return false;
            }
        }
        return true;
    }
    int findLUSlength(vector<string>&v) {
        
        int ans=0;
        for(int i=0;i<v.size();i++) {
            int n=v[i].size();
            for(int j=1;j<(1<<n);j++) {
                if(notPresent(j,v,i)) {
                    ans=max(ans, __builtin_popcount(j));
                }
            }
        }
        return ans==0?-1:ans;
    }
};
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