<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:
    def findLUSlength(self, strs: List[str]) -> int:
        
        mval = len(max(strs, key = lambda x : len(x)))
        co = 0

        for s in strs:
            if len(s) == mval:
                co += 1
                if co > 1:
                    break
        
        if co == 1:
            return mval
        

        def generateSubSeq(i, j, k):
            nonlocal res, pick

            if k == 0:
                subseq = ''.join(pick)
                res[subseq].add(i)
                return

            if j >= len(strs[i]):
                return
            
            pick.append(strs[i][j])
            generateSubSeq(i, j + 1, k - 1)
            pick.pop()

            generateSubSeq(i, j + 1, k)
        
        for solution in range(mval, 0, -1):
            
            res = defaultdict(set)
            pick = []

            for idx, word in enumerate(strs):
                generateSubSeq(idx, 0, solution)

            for k, v in res.items():
                if len(v) == 1:
                    return solution
        
        return -1

            

            

                

        
18900