## Description

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

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.

## Explanation

Group word characters by positions and check which word group matches pattern.

## Python Solution

```
class Solution:
def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
pattern_group = defaultdict(list)
for i, c in enumerate(pattern):
pattern_group[c].append(i)
results = []
for word in words:
group = defaultdict(list)
for i, c in enumerate(word):
group[c].append(i)
if list(group.values()) == list(pattern_group.values()):
results.append(word)
return results
```

- Time Complexity: O(N).
- Space Complexity: O(N).