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890. Find and Replace Pattern

Difficulty Topics

Description

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.

Solution

find-and-replace-pattern.py
class Solution:
    def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:

        def h(word):
            s = {}
            res = []

            for x in word:
                if x not in s:
                    s[x] = len(s)

                res.append(s[x])

            return "".join(map(str, res))

        target = h(pattern)
        res = []

        for word in words:
            if h(word) == target:
                res.append(word)

        return res