# How To Find The Strings of a Given Pattern

## Problem statement

You are provided with a list of strings named words and a string named pattern. Your task is to find the strings from words that match the given pattern. The order in which you return the answers does not matter.

A word is considered to match the pattern if there is a mapping p of the letters such that, when each letter x in the pattern is replaced with p(x), the word is formed.

Keep in mind that a permutation of letters is a one-to-one correspondence from letters to letters, where each letter is mapped to a distinct letter, and no two letters are mapped 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, "a");
printResult(result);
}

Output:
[mee,aqq,]
[a,b,c,]


### Code explanation

1. The code initializes an empty vector called result, which will store the words that match the given pattern.

2. Two unordered maps (w_to_p and p_to_w) are defined to establish the bijection between characters in words and characters in the pattern. These maps are used to track the mapping relationship between characters in a word and their corresponding characters in the pattern.

3. An integer variable i is declared to be used as an index to iterate through the characters of a word.

4. The code enters a loop that iterates through each word in the words vector using a range-based for loop.

5. Inside the loop, the maps w_to_p and p_to_w are cleared at the beginning of processing each word. This ensures that the maps are empty and ready to establish a new bijection for the current word.

6. The code enters another loop that iterates through the characters of the word.

7. Inside this loop, the code checks whether the character w[i] from the word has already been mapped to a character in the pattern:

• If w[i] is found in the w_to_p map, it means that w[i] was previously mapped to some character (let's call it x) in the pattern. In this case, the code checks if x is not equal to pattern[i]. If x is not equal to pattern[i], it means that w[i] cannot be mapped to pattern[i], so the loop is terminated by break.
• If w[i] is not found in the w_to_p map, the code checks whether pattern[i] has already been mapped to a character in the word:
• If pattern[i] is found in the p_to_w map, it means that pattern[i] was previously mapped to some character in the word. In this case, the loop is terminated by break.
• If neither w[i] nor pattern[i] is found in their respective maps, a bijection between w[i] and pattern[i] is established by adding entries to both maps.
8. After processing all characters of the current word, the code checks if i has reached the length of the word. If i equals w.length(), it means that a bijection has been successfully established for the entire word, and the word matches the pattern. In this case, the word is added to the result vector using result.push_back(w);.

9. After processing all words, the result vector contains all the words that match the pattern in terms of character bijection.

10. The function returns the result vector, which contains the matching words.

### Complexity

This solution efficiently finds and returns words from a vector of strings that match a given pattern in terms of character bijection. It uses two unordered maps to establish and maintain the bijection while iterating through the characters of the words and the pattern.

• Runtime: O(NL), where N = words.length and L = pattern.length.
• Extra space: O(1). The maps w_to_p and p_to_w just map between 26 lowercase English letters.

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