[Leet Code] Find and Replace Pattern

Leet code: https://leetcode.com/problems/find-and-replace-pattern/

Problem:

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:

class Solution(object):
    def findAndReplacePattern(self, words, pattern):
        """
        :type words: List[str]
        :type pattern: str
        :rtype: List[str]
        """
        testing = {}
        testing_2 = []
        incrementer = 1
        for i in pattern:
            if i not in testing:
                testing[i] = incrementer
                testing_2.append(incrementer)
                incrementer += 1
            else:
                testing_2.append(testing[i])
                
        print(testing_2)
                
            
        print testing
        answers = []
        for i in words:
            comparison = {}
            comparison_2 = []
            incrementer = 1
            for j in i:
                if j not in comparison:
                    comparison[j] = incrementer
                    comparison_2.append(incrementer)
                    incrementer += 1
                else:
                    comparison_2.append(comparison[j])
            if testing_2 == comparison_2:
                answers.append(i)
            
                
        return answers