# Find And Replace Pattern Problem

## Description

LeetCode Problem 890.

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:

``````1
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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:

``````1
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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.

## Sample C++ Code

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class Solution {
public:
vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
int n = words.size();
vector<map<char,char>> m(n);

for (int i = 0; i < n; ++i) {
set<int> s;
for (int j = 0; j < words[i].size(); ++j) {
if (s.find(pattern[j]) == s.end()) {
s.insert(pattern[j]);
m[i][words[i][j]] = pattern[j];
}
}
}
vector<string> ans;
for (int i = 0; i < n; ++i) {
string k = words[i];
for (int j = 0; j < words[i].size(); ++j) {
words[i][j] = m[i][words[i][j]];
}
if (words[i] == pattern) {
ans.push_back(k);
}
}
return ans;
}
};
``````