#   # How to solve Leetcode 890. Find and Replace Pattern

## How to create a bijection map

### Problem statement

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.

### Solution: Construct the bijection and check the condition

#### Code

``````#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
vector<string> result;
// need two maps for the bijection
unordered_map<char,char> w_to_p, p_to_w;
int i;
for (string& w : words) {
w_to_p.clear();
p_to_w.clear();
i = 0;
while (i < w.length()) {
if (w_to_p.find(w[i]) != w_to_p.end()) {
// w[i] was mapped to some letter x
// but x != pattern[i]
if (w_to_p[w[i]] != pattern[i]) {
break;
}
} else {
if (p_to_w.find(pattern[i]) != p_to_w.end()) {
// w[i] was not mapped to any letter yet
// but pattern[i] was already mapped to some letter
break;
}
// build the bijection w[i] <-> pattern[i]
w_to_p[w[i]] = pattern[i];
p_to_w[pattern[i]] = w[i];
}
i++;
}
if (i == w.length()) {
result.push_back(w);
}
}
return result;
}
void printResult(const vector<string>& result) {
cout << "[";
for (const string& s : result) {
cout << s << ",";
}
cout << "]\n";
}
int main() {
vector<string> words{"abc","deq","mee","aqq","dkd","ccc"};
auto result = findAndReplacePattern(words, "abb");
printResult(result);
words = {"a", "b", "c"};
result = findAndReplacePattern(words, "abb");
printResult(result);
}
``````
``````Output:
[mee,aqq,]
[a,b,c,]
``````

#### Complexity

• Runtime: `O(N*L)`, where `N = words.length` and `L = pattern.length`.

• Extra space: `O(1)` if `N` or `L` is very larger than 26. The maps `w_to_p` and `p_to_w` just map between 26 lowercase English letters.