890-find-and-replace-pattern

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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(Python)

class Solution:
    def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
        
        def is_pattern(s, t):
            s_map = {}
            t_map = {}
            
            n= len(s)
            if n != len(t):
                return False
            
            for i in range(n):
                if s[i] not in s_map:
                    s_map[s[i]] = t[i]
                elif s_map[s[i]] != t[i]:
                        return False
                    
                if t[i] not in t_map:
                    t_map[t[i]] = s[i]
                elif t_map[t[i]] != s[i]:
                    return False
            return True
        
        res = []
        for word in words:
            if is_pattern(word, pattern):
                res.append(word)
        return res