LeetCode 솔루션 분류
[7/29] 890. Find and Replace Pattern
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Medium
1928121Add to ListShareGiven 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 <= 201 <= words.length <= 50words[i].length == pattern.lengthpatternandwords[i]are lowercase English letters.
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101,354
Submissions
134,082
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Runtime: 31 ms, faster than 96.19% of Python3 online submissions for Find and Replace Pattern.
Memory Usage: 13.9 MB, less than 28.72% of Python3 online submissions for Find and Replace Pattern.
Memory Usage: 13.9 MB, less than 28.72% of Python3 online submissions for Find and Replace Pattern.
class Solution:
def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
res = []
for word in words:
if len(word) != len(pattern): continue
conv_dict = {}
temp = ""
used = set()
for idx, char in enumerate(word):
if char not in conv_dict:
if pattern[idx] in used: break
conv_dict[char] = pattern[idx]
used.add(pattern[idx])
temp += conv_dict[char]
if temp == pattern: res.append(word)
return res
