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]:
        
        ans = []
        
        for i in words:
            temp = i
            
            if len(pattern) == len(temp):
                as1 = {}
                got = []
                found = 1
                for i1 in range(len(temp)):
                    if pattern[i1] not in as1:
                        if temp[i1] in got:
                            found = 0 
                            break
                        else:
                            as1[pattern[i1]] = temp[i1]
                            got.append(temp[i1])
                    else:
                        if temp[i1] != as1.get(pattern[i1]):
                            found = 0
                            break
                            
                if found == 1:
                    ans.append(temp)
                    
                    
        return ans
                        
                        
                        
                    
                    
                
            

class Solution:
    def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
        
        ans = []
        
        for i in words:
            temp = i
            
            if len(pattern) == len(temp):
                as1 = {}
                got = []
                found = 1
                for i1 in range(len(temp)):
                    if pattern[i1] not in as1:
                        if temp[i1] in got:
                            found = 0 
                            break
                        else:
                            as1[pattern[i1]] = temp[i1]
                            got.append(temp[i1])
                    else:
                        if temp[i1] != as1.get(pattern[i1]):
                            found = 0
                            break
                            
                if found == 1:
                    ans.append(temp)
                    
                    
        return ans
                        
                        
                        
                    
                    
                
            
