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 <= 201 <= words.length <= 50words[i].length == pattern.lengthpattern and words[i] are lowercase English letters.class Solution {
public:
vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
vector<string> res;
map <char, char> m, n;
for(auto e:words){
bool valid = true;
for(int i=0; i<e.size(); i++){
// cout << m[pattern[i]] << "\t" << e[i] << endl;
if(m[pattern[i]]){
if(m[pattern[i]]!=e[i])
valid = false;
}
else if(n[e[i]]){
if(n[e[i]]!=pattern[i])
valid = false;
}
else {
m[pattern[i]] = e[i];
n[e[i]] = pattern[i];
};
}
if (valid){
res.push_back(e);
cout << "true" << endl;
}
else cout << "false" << endl;
m.clear();
n.clear();
}
return res;
}
};