7/26 每日一題(返回最長不同的子序列)

 Given two strings a and b, return the length of the longest uncommon subsequence between a and b. If the longest uncommon subsequence does not exist, return -1.

An uncommon subsequence between two strings is a string that is a subsequence of one but not the other.

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: a = "aba", b = "cdc"
Output: 3
Explanation: One longest uncommon subsequence is "aba" because "aba" is a subsequence of "aba" but not "cdc".
Note that "cdc" is also a longest uncommon subsequence.

Example 2:

Input: a = "aaa", b = "bbb"
Output: 3
Explanation: The longest uncommon subsequences are "aaa" and "bbb".

Example 3:

Input: a = "aaa", b = "aaa"
Output: -1
Explanation: Every subsequence of string a is also a subsequence of string b. Similarly, every subsequence of string b is also a subsequence of string a.

 

Constraints:

• 1 <= a.length, b.length <= 100
• a and b consist of lower-case English letters.

class Solution:

    def findLUSlength(self, a: str, b: str) -> int:

        # i,j=0,0

        # while i<len(a) and i <len(b):

        #     if a[i] != b[i]:

        #         j +=1 #如果最後j==0,表示元素可能相同

        #     i +=1

        # if j==0 and len(a)==len(b): #元素相同且長度相同

        #     return -1

        # else:

        #      return max(len(a),len(b))

       

        if a == b:

            return -1

        return max(len(a),len(b))