0890-Find-and-Replace-Pattern.py

'''
You have a list of words and a pattern, and you want to know which words in words matches the pattern.

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.)

Return a list of the words in words that match the given pattern. 

You may return the answer in any order.

 

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.
 

Note:

1 <= words.length <= 50
1 <= pattern.length = words[i].length <= 20
'''
class Solution:
    def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
        
        def find_replace(w):
            
            d = {}
            
            return [d.setdefault(c, len(d)) for c in w]
        
        f_pattern = find_replace(pattern)
        
        return [w for w in words if find_replace(w) == f_pattern]