884. Uncommon Words from Two Sentences

Description

A sentence is a string of single-space separated words where each word consists only of lowercase letters.

A word is uncommon if it appears exactly once in one of the sentences, and does not appear in the other sentence.

Given two sentences s1 and s2, return a list of all the uncommon words. You may return the answer in any order.

Example 1:

• Input: s1 = “this apple is sweet”, s2 = “this apple is sour”
• Output: [“sweet”,”sour”]
• Explanation: The word “sweet” appears only in s1, while the word “sour” appears only in s2.

Example 2:

• Input: s1 = “apple apple”, s2 = “banana”
• Output: [“banana”]

Constraints:

• 1 <= s1.length, s2.length <= 200
• s1 and s2 consist of lowercase English letters and spaces.
• s1 and s2 do not have leading or trailing spaces.
• All the words in s1 and s2 are separated by a single space.

Submitted Code

class Solution(object):
    def uncommonFromSentences(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: List[str]
        """
        counting = {}
        result = []

        for word in s1.split():
            counting[word] = counting.get(word, 0) + 1
        
        for word in s2.split():
            counting[word] = counting.get(word, 0) + 1

        for k, v in counting.items():
            if v == 1: result.append(k)

        return result

Runtime: 0 ms | Beats 100.00%
Memory: 12.59 MB | Beats 15.36%

s1과 s2의 모든 단어를 합쳐 카운팅하면 한 번만 등장하는 단어가 uncommon words가 된다.

Other Solutions

1st

class Solution(object):
    def uncommonFromSentences(self, A, B):
        c = collections.Counter((A + " " + B).split())
        return [w for w in c if c[w] == 1]

time complexity: 𝑂(𝑚+𝑛)
space complexity: 𝑂(𝑚+𝑛)

collections 모듈로 단어 등장 빈도를 카운팅하는 딕셔너리를 빠르게 생성할 수 있다.