Problem
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 <= 201 <= words.length <= 50words[i].length == pattern.lengthpatternandwords[i]are lowercase English letters.
Solution (Java)
class Solution {
public List<String> findAndReplacePattern(String[] words, String pattern) {
List<String> finalans = new ArrayList<>();
if (pattern.length() == 1) {
Collections.addAll(finalans, words);
return finalans;
}
for (String word : words) {
char[] check = new char[26];
Arrays.fill(check, '1');
HashMap<Character, Character> ans = new HashMap<>();
for (int j = 0; j < word.length(); j++) {
char pat = pattern.charAt(j);
char wor = word.charAt(j);
if (ans.containsKey(pat)) {
if (ans.get(pat) == wor) {
if (j == word.length() - 1) {
finalans.add(word);
}
} else {
break;
}
} else {
if (j == word.length() - 1 && check[wor - 'a'] == '1') {
finalans.add(word);
}
if (check[wor - 'a'] != '1' && check[wor - 'a'] != pat) {
break;
}
if (check[wor - 'a'] == '1') {
ans.put(pat, wor);
check[wor - 'a'] = pat;
}
}
}
}
return finalans;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).