521. Longest Uncommon Subsequence I

Description

Given two strings a and b, return the length of the longest uncommon subsequence between a and b. If no such uncommon subsequence exists, return -1.

An uncommon subsequence between two strings is a string that is a subsequence of exactly one of them.

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. So the answer would be -1.

Constraints:

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

💡 Hint 1:
Think very simple.

💡 Hint 2:
If a == b, the answer is -1.

💡 Hint 3:
Otherwise, the answer is the string a or the string b.

Submitted Code

class Solution(object):
    def findLUSlength(self, a, b):
        """
        :type a: str
        :type b: str
        :rtype: int
        """
        if a == b:
            return -1
        elif len(a) == len(b):
            return len(a)
        else:
            return max(len(a), len(b)) 

Runtime: ** ms | Beats **100.00%
Memory: 12.36 MB | Beats 80.08%

‘가장 긴’ uncommon subsequence를 찾아야 하기 때문에, a와 b가 서로 다르다면 둘 중 길이가 더 긴 문자열 자체가 longest uncommon subsequence이 된다.

Other Solutions

1st

class Solution(object):
    def findLUSlength(self, a, b):
        return -1 if a==b else max(len(a), len(b))

time complexity: 𝑂(1)
space complexity: 𝑂(1)

파이썬의 경우 한 줄 코드로 작성 가능하다.