890. Find and Replace Pattern

890. Find and Replace Pattern (Medium)

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

Companies:
Google, Apple

Related Topics:
String

Solution 1.

// OJ: https://leetcode.com/problems/find-and-replace-pattern/
// Author: github.com/lzl124631x
// Time: O(CW) where C is count of words and W is word length
// Space: O(W)
class Solution {
private:
    bool match(string &word, string &pattern) {
        unordered_map<char, char> m;
        unordered_set<char> used;
        for (int i = 0; i < word.size(); ++i) {
            if (m.find(word[i]) == m.end()) {
                if (used.find(pattern[i]) != used.end()) return false;
                m[word[i]] = pattern[i];
                used.insert(pattern[i]);                                            
            } else if (m[word[i]] != pattern[i]) return false;
        }
        return true;
    }
public:
    vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
        vector<string> ans;
        for (auto word : words) {
            if (match(word, pattern)) ans.push_back(word);
        }
        return ans;
    }
};