Leetcode - Find and Replace Pattern Solution
Poor Man
- Published
- Updated
- Reading time
- 1 min read
- Length
- 232 words
You have a list of words and a pattern, and you want to know which words in words matches the pattern.
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.)
Return a list of the words in words that match the given pattern.
You may return the answer in any order.
Example 1:
Text
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.Note:
1 <= words.length <= 501 <= pattern.length = words[i].length <= 20
Solution in Python#
Python
class Solution:
@staticmethod
def recreate(word):
word_map = defaultdict(int)
recreated_word = 0
index = 0
for j,i in enumerate(word):
key_index = word_map.get(i)
if not key_index:
index+=1
word_map[i] = index
recreated_word +=((10**j)*word_map[i])
return recreated_word
def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
recreated_pattern = self.recreate(pattern)
return [word for word in words if self.recreate(word) == recreated_pattern]