human
What is a the Python program you requested given the following:
Solve in Python:
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
assistant
#Here is the Python program you requested.

class Solution:
    def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
        def match(list1 , pattern):
            for i in range(len(pattern)):
                if list1[i] != pattern[i]:
                    return False
            return True
        
        def number_arr(list1):
            dict = {}
            arr = []
            last_given_value = 0
            for i in list1:
                if i not in dict:
                    last_given_value +=1
                    dict[i] = last_given_value
                    arr.append(dict[i])
                else:
                    arr.append(dict[i])
            return arr
        
        pat = number_arr(pattern)
        ans = []
        for word in words:
            word_list = number_arr(word)
            if match(pat , word_list):
                ans.append(word)
        return ans