21. Find and Replace Pattern.cpp

/*
Find and Replace Pattern
========================

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.
*/

class Solution
{
public:
  string standardString(string s)
  {
    unordered_map<char, char> m;
    char ch = 'a';
    for (int i = 0; i < s.size(); ++i)
    {
      if (m.find(s[i]) == m.end())
        m[s[i]] = ch++;
      s[i] = m[s[i]];
    }
    return s;
  }

  vector<string> findAndReplacePattern(vector<string> &words, string pattern)
  {
    vector<string> ans;
    for (auto &word : words)
    {
      if (word.size() == pattern.size() && standardString(word) == standardString(pattern))
        ans.push_back(word);
    }
    return ans;
  }
};