Leetcode 890. Find and Replace Pattern 找同樣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.

常見作法是create 兩個字典

這裡有一個辦法，只用一個字典

最後檢查domain 跟range是不是一樣大小

如果是一樣，保證onto就是答案了

歡迎參考下方解法:

class Solution:
    def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
        
        
        def check(words):
            m={}
            for i,x in enumerate(pattern):
                if x not in m:
                    m[x]=words[i]
                else:
                    if m[x]!=words[i]:
                        return False
            return len(set(m.keys()))==len(set(m.values()))
        
        ans=[]
        for i in words:
            if check(i):
                ans.append(i)
        
        return ans