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Leetcode Problem #890 (Medium): Find and Replace Pattern
Description:
(Jump to: Solution Idea || Code: JavaScript | Python | Java | C++)
Given a list of strings
wordsand a stringpattern, return a list ofwords[i]that matchpattern. You may return the answer in any order.A word matches the pattern if there exists a permutation of letters
pso that after replacing every letterxin the pattern withp(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.
Examples:
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 <= 201 <= words.length <= 50words[i].length == pattern.lengthpatternandwords[i]are lowercase English letters.
Idea:
(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)
Right away, we can realize that if we can remap characters in an attempt to match the pattern, it doesn't actually matter which characters map to other characters, just that the locations are consistent.
At this point, then, the goal is to make the comparison as easy as possible. To do that, we can reimagine the words as an alphabetic sequence, where the first new character we come across is always masked to "a", the second to "b", and so on. If we apply this same process to the pattern first, then it should be much easier to compare the words to the pattern.
First, we can define a helper function to translate characters for us. We'll have to create a map or array structure (codex) to keep track of the character mapping for a given word. The translate function will then check to see if the character has already been mapped, and if so, return its mapped value. If not, it assigns it the next unused alphabetic value.
We can then easily translate the pattern into a cipher mask which we can then compare to each word in words using another helper function. The compare function will clear the codex for each word, then we can compare each character of word to the corresponding character in cipher. If at any time we fail to match, we can quickly return out of the comparison and continue with the next word. If the translated word fully matches cipher, it can be added to our answer array (ans).
Then we can return ans once we're done.
- Time Complexity: O(N * M) where N is the length of words and M is the length of each word/pattern
-
Space Complexity: O(M) for the codex
- or O(N + M) if you count the space of the output (ans)
Javascript Code:
(Jump to: Problem Description || Solution Idea)
var findAndReplacePattern = function(words, pattern) {
let ans = [], codex = new Map()
const translate = char => {
if (!codex.has(char))
codex.set(char, String.fromCharCode(97 + codex.size))
return codex.get(char)
}
const compare = word => {
codex.clear()
for (let i = 0; i < word.length; i++)
if (translate(word[i]) !== cipher[i])
return
ans.push(word)
}
let cipher = new Array(pattern.length)
for (let i = 0; i < pattern.length; i++)
cipher[i] = translate(pattern.charAt(i))
words.forEach(compare)
return ans
};
Python Code:
(Jump to: Problem Description || Solution Idea)
class Solution:
def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
ans, codex = [], defaultdict()
def translate(c: str) -> str:
if c not in codex:
codex[c] = chr(97 + len(codex))
return codex[c]
def compare(word: str) -> None:
codex.clear()
for i in range(len(word)):
if translate(word[i]) != cipher[i]:
return
ans.append(word)
cipher = [translate(c) for c in pattern]
for word in words:
compare(word)
return ans
Java Code:
(Jump to: Problem Description || Solution Idea)
class Solution {
List<String> ans;
Map<Character, Character> codex;
char[] cipher;
public List<String> findAndReplacePattern(String[] words, String pattern) {
ans = new ArrayList<>();
codex = new HashMap<>();
cipher = pattern.toCharArray();
for (int i = 0; i < pattern.length(); i++)
cipher[i] = translate(cipher[i]);
for (String word : words) compare(word);
return ans;
}
private char translate(char c) {
if (!codex.containsKey(c))
codex.put(c, (char)(97 + codex.size()));
return codex.get(c);
}
private void compare(String word) {
codex.clear();
for (int i = 0; i < word.length(); i++)
if (translate(word.charAt(i)) != cipher[i]) return;
ans.add(word);
}
}
C++ Code:
(Jump to: Problem Description || Solution Idea)
class Solution {
public:
vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
ans.resize(0);
codex.clear();
cipher = pattern;
for (int i = 0; i < pattern.size(); i++)
cipher[i] = translate(cipher[i]);
for (auto& word : words) compare(word);
return ans;
}
private:
vector<string> ans;
unordered_map<char, char> codex;
string cipher;
char translate(char& c) {
if (codex.find(c) == codex.end())
codex[c] = (char)(97 + codex.size());
return codex[c];
}
void compare(string& word) {
codex.clear();
for (int i = 0; i < word.length(); i++)
if (translate(word[i]) != cipher[i]) return;
ans.push_back(word);
}
};
Top comments (1)
Version in ruby.
I decided to use the incoming pattern as a cipher, not sure how does the middleware cipher add up to performance