890. Find and Replace Pattern

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

Solution 1

Python3JavaC++GoTypeScriptRust

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

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

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;
    }
}

class Solution {
public:
    vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
        vector<string> ans;
        auto match = [](string& s, string& t) {
            int m1[128] = {0};
            int m2[128] = {0};
            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;
        };
        for (auto& word : words)
            if (match(word, pattern)) ans.emplace_back(word);
        return ans;
    }
};

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
}

function findAndReplacePattern(words: string[], pattern: string): string[] {
    return words.filter(word => {
        const map1 = new Map<string, number>();
        const map2 = new Map<string, number>();
        for (let i = 0; i < word.length; i++) {
            if (map1.get(word[i]) !== map2.get(pattern[i])) {
                return false;
            }
            map1.set(word[i], i);
            map2.set(pattern[i], i);
        }
        return true;
    });
}

use std::collections::HashMap;
impl Solution {
    pub fn find_and_replace_pattern(words: Vec<String>, pattern: String) -> Vec<String> {
        let pattern = pattern.as_bytes();
        let n = pattern.len();
        words
            .into_iter()
            .filter(|word| {
                let word = word.as_bytes();
                let mut map1 = HashMap::new();
                let mut map2 = HashMap::new();
                for i in 0..n {
                    if map1.get(&word[i]).unwrap_or(&n) != map2.get(&pattern[i]).unwrap_or(&n) {
                        return false;
                    }
                    map1.insert(word[i], i);
                    map2.insert(pattern[i], i);
                }
                true
            })
            .collect()
    }
}

890. Find and Replace Pattern

Description

Solutions

Solution 1

Comments