Day040 - 290. Word Pattern

290. Word Pattern

Given a pattern and a string s, find if s follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in s. Specifically:

• Each letter in pattern maps to exactly one unique word in s.
• Each unique word in s maps to exactly one letter in pattern.
• No two letters map to the same word, and no two words map to the same letter.

Example 1:

Input: pattern = “abba”, s = “dog cat cat dog” Output: true Explanation: The bijection can be established as:

• 'a' maps to "dog".
• 'b' maps to "cat".

Example 2:

Input: pattern = “abba”, s = “dog cat cat fish” Output: false

Example 3:

Input: pattern = “aaaa”, s = “dog cat cat dog” Output: false

Constraints:

• 1 <= pattern.length <= 300
• pattern contains only lower-case English letters.
• 1 <= s.length <= 3000
• s contains only lowercase English letters and spaces ' '.
• s does not contain any leading or trailing spaces.
• All the words in s are separated by a single space.

내 풀이

from collections import defaultdict

class Solution:
    def wordPattern(self, pattern: str, s: str) -> bool:
        
        # s를 list로 변환
        s = list(s.split(" "))

        # pattern의 글자 수와 s의 단어 수가 다르면 return False
        if len(pattern) != len(s): return False

        # pattern과 s의 matching을 기록한 dict와 사용여부를 확인할 set 초기화
        match = defaultdict(str)
        used = set()

        for string, word in zip(pattern, s):
            # string이 match에 없고 word가 사용된 적 없다면
            if string not in match and word not in used:
                # string과 word의 매칭정보 추가 및 사용된 word정보 기록
                match[string] = word
                used.add(word)

            # string과 word가 사용됐던 적이 있는데 짝이 맞지 않는 경우 False
            if match[string] != word:
                return False

        # 모든 iteration을 통과하면 True
        return True

# Time Complexity : \(O(N)\)