# LeetCode 890. Find and Replace Pattern

## 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).