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Formatted question description: https://leetcode.ca/all/890.html

# 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:

Related Topics:
String

## Solution 1.

// OJ: https://leetcode.com/problems/find-and-replace-pattern/
// 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;
}
};

• class Solution {
public List<String> findAndReplacePattern(String[] words, String pattern) {
List<String> matchWords = new ArrayList<String>();
for (String word : words) {
if (matchPattern(word, pattern))
}
return matchWords;
}

public boolean matchPattern(String word, String pattern) {
if (word.length() != pattern.length())
return false;
Map<Character, Character> map1 = new HashMap<Character, Character>();
Map<Character, Character> map2 = new HashMap<Character, Character>();
int length = word.length();
for (int i = 0; i < length; i++) {
char c1 = word.charAt(i), c2 = pattern.charAt(i);
if (map1.containsKey(c1)) {
if (map1.get(c1) != c2)
return false;
} else
map1.put(c1, c2);
if (map2.containsKey(c2)) {
if (map2.get(c2) != c1)
return false;
} else
map2.put(c2, c1);
}
return true;
}
}

############

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

• // OJ: https://leetcode.com/problems/find-and-replace-pattern/
// 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;
}
};

• 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(object):
def findAndReplacePattern(self, words, pattern):
"""
:type words: List[str]
:type pattern: str
:rtype: List[str]
"""
ans = []
set_p = set(pattern)
for word in words:
if len(set(word)) != len(set_p):
continue
fx = dict()
equal = True
for i, w in enumerate(word):
if w in fx:
if fx[w] != pattern[i]:
equal = False
break
fx[w] = pattern[i]
if equal:
ans.append(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()
}
}