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
andwords[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]
Top comments (0)