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[7/29] 890. Find and Replace Pattern

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Given a list of strings words and a string pattern, return a list of words[i] that match pattern. You may return the answer in any order.

A word matches the pattern if there exists a permutation of letters p so that after replacing every letter x in the pattern with p(x), we get the desired word.

Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter.

 

Example 1:

Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb"
Output: ["mee","aqq"]
Explanation: "mee" matches the pattern because there is a permutation {a -> m, b -> e, ...}. 
"ccc" does not match the pattern because {a -> c, b -> c, ...} is not a permutation, since a and b map to the same letter.

Example 2:

Input: words = ["a","b","c"], pattern = "a"
Output: ["a","b","c"]

 

Constraints:

  • 1 <= pattern.length <= 20
  • 1 <= words.length <= 50
  • words[i].length == pattern.length
  • pattern and words[i] are lowercase English letters.
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Submissions
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Runtime: 31 ms, faster than 96.19% of Python3 online submissions for Find and Replace Pattern.
Memory Usage: 13.9 MB, less than 28.72% of Python3 online submissions for Find and Replace Pattern.
class Solution:
    def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
        res = []
        
        for word in words:
            if len(word) != len(pattern): continue
            conv_dict = {}
            temp = ""
            used = set()
            for idx, char in enumerate(word):
                if char not in conv_dict:
                    if pattern[idx] in used: break
                    conv_dict[char] = pattern[idx]
                    used.add(pattern[idx])
                temp += conv_dict[char]
            
            if temp == pattern: res.append(word)
        
        return res
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