## DEV Community

Abhishek Chaudhary

Posted on

# 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.

SOLUTION:

``````class Solution:
def keygen(self, s):
smap = {}
slist = []
i = 0
for c in s:
if c not in smap:
smap[c] = i
i += 1
slist.append(smap[c])
return slist

def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
plist = self.keygen(pattern)
return [w for w in words if self.keygen(w) == plist]
``````