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:
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 <= 501 <= pattern.length = words[i].length <= 20
class Solution {
public List<String> findAndReplacePattern(String[] words, String pattern) {
List<String> ans = new ArrayList<>();
if (words == null || words.length == 0 || pattern == null || pattern.length() == 0) {
return ans;
}
for (String word : words) {
if (match(word, pattern)) {
ans.add(word);
}
}
return ans;
}
private boolean match(String s, String p) {
if (s.length() != p.length()) {
return false;
}
Map<Character, Character> map = new HashMap<>();
Map<Character, Character> rMap = new HashMap<>();
for (int i = 0; i < s.length(); i++) {
char cs = s.charAt(i);
char cp = p.charAt(i);
if (!map.containsKey(cs)) {
if (rMap.containsKey(cp)) {
return false;
}
map.put(cs, cp);
rMap.put(cp, cs);
} else {
if (map.get(cs) != cp || rMap.get(cp) != cs) {
return false;
}
}
}
return true;
}
}
Code (Java):
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