890

890. Find and Replace Pattern (Medium)

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 <= 20
• 1 <= words.length <= 50
• words[i].length == pattern.length
• pattern and words[i] are lowercase English letters.

Companies:
Google, Apple

Related Topics:
Array, Hash Table, String

Solution 1.

// OJ: https://leetcode.com/problems/find-and-replace-pattern/
// Author: github.com/lzl124631x
// Time: O(NW) where N is the length of `A` and `W` is word length
// Space: O(1)
class Solution {
public:
    vector<string> findAndReplacePattern(vector<string>& A, string p) {
        vector<string> ans;
        for (auto &s : A) {
            int m[26] = {[0 ... 25] = -1}, r[26] = {[0 ... 25] = -1}, i = 0;
            for (; i < p.size(); ++i) {
                int from = s[i] - 'a', to = p[i] - 'a';
                if ((m[from] != -1 && m[from] != to) || (r[to] != -1 && r[to] != from)) break;
                m[from] = to;
                r[to] = from;
            }
            if (i == p.size()) ans.push_back(s);
        }
        return ans;
    }
};