LeetCode - 890. Find and Replace Pattern

890. Find and Replace Pattern - esay

문제

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.

제한사항

• 1 <= words.length <= 50
• 1 <= pattern.length = words[i].length <= 20

입출력 예

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.

풀이

• Hash

class Solution {
public:
    vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
        std::vector<string> answer;

        for (const auto& str : words) {
            bool flag = true;
            std::map<char, char> m1;
            std::map<char, char> m2;

            // 두 map에 순차적으로 word, pattern의 문자를 서로 교차 저장하며,
            // key가 이미 저장되어 있지 않다면 새로 해당 문자를 key로 하여 상대 문자를 저장하고,
            // 지금 저장할 key와 매칭되는 상대의 문자가 같은지 검사
            for (auto i = 0 ; i < str.size() ; ++i) {
                char s = str[i];
                char p = pattern[i];
                
                if (m1.find(s) == m1.end()) {
                    m1[s] = p;
                }
                
                if (m2.find(p) == m2.end()) {
                    m2[p] = s;
                }
                
                if (m1[s] != p || m2[p] != s) {
                    flag = false;
                    break;
                }
            }

            if (flag) {
                answer.push_back(str);                
            }
        }
        
        return answer;
    }
};