Title: Find and Replace Pattern Source: leetcode.com
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:
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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 <= 50
- 1 <= pattern.length = words[i].length <= 20
Python solution
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''' https://leetcode.com/problems/find-and-replace-pattern/ ''' class Solution(object): def findAndReplacePattern(self, words, pattern): """ :type words: List[str] :type pattern: str :rtype: List[str] """ res = [] for word in words: pd = {} wd = {} i = 0 flag = True for c in pattern: wc = word[i] if c in pd and wc != pd[c]: flag = False break if wc in wd and c != wd[wc]: flag = False break pd[c] = wc wd[wc] = c i += 1 if flag: res.append(word) return res |