Sunday, March 10, 2019

Leetcode 890. Find and Replace Pattern

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 <= 50
  • 1 <= 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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