<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>

from typing import List

class Solution:
    def findLUSlength(self, strs: List[str]) -> int:
        def is_subsequence(s: str, t: str) -> bool:
            it = iter(t)
            return all(c in it for c in s)

        def is_uncommon(s: str, strs: List[str]) -> bool:
            return sum(is_subsequence(s, other) for other in strs) == 1
        
        # Initialize the length of the longest uncommon subsequence
        longest_length = -1
        
        # Check each string in the list
        for s in strs:
            if is_uncommon(s, strs):
                longest_length = max(longest_length, len(s))
        
        return longest_length

# Example usage
solution = Solution()
print(solution.findLUSlength(["aba", "cdc", "eae"]))  # Output: 3
print(solution.findLUSlength(["aaa", "aaa", "aa"]))   # Output: -1

16500