README_EN.md

890. Find and Replace Pattern

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Description

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.

Solutions

Python3

class Solution:
    def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
        def match(s, t):
            m1, m2 = [0] * 128, [0] * 128
            for i in range(n):
                if m1[ord(s[i])] != m2[ord(t[i])]:
                    return False
                m1[ord(s[i])] = m2[ord(t[i])] = i + 1
            return True

        n = len(pattern)
        return [word for word in words if match(word, pattern)]

Java

class Solution {
    public List<String> findAndReplacePattern(String[] words, String pattern) {
        List<String> ans = new ArrayList<>();
        for (String word : words) {
            if (match(word, pattern)) {
                ans.add(word);
            }
        }
        return ans;
    }

    private boolean match(String s, String t) {
        int[] m1 = new int[128];
        int[] m2 = new int[128];
        for (int i = 0; i < s.length(); ++i) {
            char c1 = s.charAt(i);
            char c2 = t.charAt(i);
            if (m1[c1] != m2[c2]) {
                return false;
            }
            m1[c1] = i + 1;
            m2[c2] = i + 1;
        }
        return true;
    }
}

C++

class Solution {
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;
    }

    bool match(string s, string t) {
        vector<int> m1(128);
        vector<int> m2(128);
        for (int i = 0; i < s.size(); ++i)
        {
            if (m1[s[i]] != m2[t[i]]) return 0;
            m1[s[i]] = i + 1;
            m2[t[i]] = i + 1;
        }
        return 1;
    }
};

Go

func findAndReplacePattern(words []string, pattern string) []string {
	match := func(s, t string) bool {
		m1, m2 := make([]int, 128), make([]int, 128)
		for i := 0; i < len(s); i++ {
			if m1[s[i]] != m2[t[i]] {
				return false
			}
			m1[s[i]] = i + 1
			m2[t[i]] = i + 1
		}
		return true
	}
	var ans []string
	for _, word := range words {
		if match(word, pattern) {
			ans = append(ans, word)
		}
	}
	return ans
}

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