LeetCode 890. Find and Replace Pattern

Description

https://leetcode.com/problems/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.

Explanation

Group word characters by positions and check which word group matches pattern.

Python Solution

class Solution:
    def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
        
        pattern_group = defaultdict(list)
        
        for i, c in enumerate(pattern):
            pattern_group[c].append(i)
            
        results = []
        
        for word in words:
            group = defaultdict(list)
            
            for i, c in enumerate(word):
                group[c].append(i)
            
            if list(group.values()) == list(pattern_group.values()):
                results.append(word)
        
        return results
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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