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522. Longest Uncommon Subsequence II

Difficulty Topics

Description

Given an array of strings strs, return the length of the longest uncommon subsequence between them. If the longest uncommon subsequence does not exist, return -1.

An uncommon subsequence between an array of strings is a string that is a subsequence of one string but not the others.

A subsequence of a string s is a string that can be obtained after deleting any number of characters from s.

  • For example, "abc" is a subsequence of "aebdc" because you can delete the underlined characters in "aebdc" to get "abc". Other subsequences of "aebdc" include "aebdc", "aeb", and "" (empty string).

 

Example 1:

Input: strs = ["aba","cdc","eae"]
Output: 3

Example 2:

Input: strs = ["aaa","aaa","aa"]
Output: -1

 

Constraints:

  • 2 <= strs.length <= 50
  • 1 <= strs[i].length <= 10
  • strs[i] consists of lowercase English letters.

Solution

longest-uncommon-subsequence-ii.py
class Solution:
    def findLUSlength(self, words: List[str]) -> int:

        def getDuplicates():
            sset = set()
            duplicates = set()

            for word in words:
                if word in sset:
                    duplicates.add(word)
                sset.add(word)

            return duplicates

        def isSubsequence(a, b):
            i = j = 0

            while i < len(a) and j < len(b):
                if a[i] == b[j]: j += 1
                i += 1

            return j == len(b)

        words.sort(key = len, reverse = 1)
        duplicates = getDuplicates()
        if words[0] not in duplicates: return len(words[0])

        for i, word in enumerate(words):
            if word not in duplicates:
                for j in range(i):
                    if isSubsequence(words[j], words[i]): break
                    if j == i - 1: return len(word)

        return -1