動機

就是硬幹 但isSubsequence很神，原來iterator可以這麼用

  • 用iterator比對時，沒比對到就會iterator往下，比對到就停

Problem

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:

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

Sol

用iterator比對時，沒比對到就會iterator往下，比對到就停

class Solution:
    def findLUSlength(self, words):
        def isSubsequence(s, t):
            t = iter(t)
            return all(c in t for c in s)
 
        words.sort(key = lambda x:-len(x))
        for i, word in enumerate(words):
            if all(not isSubsequence(word, words[j]) for j in range(len(words)) if j != i): 
                return len(word)
        
        return -1