Formatted question description:

2227. Encrypt and Decrypt Strings (Hard)

You are given a character array keys containing unique characters and a string array values containing strings of length 2. You are also given another string array dictionary that contains all permitted original strings after decryption. You should implement a data structure that can encrypt or decrypt a 0-indexed string.

A string is encrypted with the following process:

  1. For each character c in the string, we find the index i satisfying keys[i] == c in keys.
  2. Replace c with values[i] in the string.

A string is decrypted with the following process:

  1. For each substring s of length 2 occurring at an even index in the string, we find an i such that values[i] == s. If there are multiple valid i, we choose any one of them. This means a string could have multiple possible strings it can decrypt to.
  2. Replace s with keys[i] in the string.

Implement the Encrypter class:

  • Encrypter(char[] keys, String[] values, String[] dictionary) Initializes the Encrypter class with keys, values, and dictionary.
  • String encrypt(String word1) Encrypts word1 with the encryption process described above and returns the encrypted string.
  • int decrypt(String word2) Returns the number of possible strings word2 could decrypt to that also appear in dictionary.


Example 1:

["Encrypter", "encrypt", "decrypt"]
[[['a', 'b', 'c', 'd'], ["ei", "zf", "ei", "am"], ["abcd", "acbd", "adbc", "badc", "dacb", "cadb", "cbda", "abad"]], ["abcd"], ["eizfeiam"]]
[null, "eizfeiam", 2]

Encrypter encrypter = new Encrypter([['a', 'b', 'c', 'd'], ["ei", "zf", "ei", "am"], ["abcd", "acbd", "adbc", "badc", "dacb", "cadb", "cbda", "abad"]);
encrypter.encrypt("abcd"); // return "eizfeiam". 
                           // 'a' maps to "ei", 'b' maps to "zf", 'c' maps to "ei", and 'd' maps to "am".
encrypter.decrypt("eizfeiam"); // return 2. 
                              // "ei" can map to 'a' or 'c', "zf" maps to 'b', and "am" maps to 'd'. 
                              // Thus, the possible strings after decryption are "abad", "cbad", "abcd", and "cbcd". 
                              // 2 of those strings, "abad" and "abcd", appear in dictionary, so the answer is 2.



  • 1 <= keys.length == values.length <= 26
  • values[i].length == 2
  • 1 <= dictionary.length <= 100
  • 1 <= dictionary[i].length <= 100
  • All keys[i] and dictionary[i] are unique.
  • 1 <= word1.length <= 2000
  • 1 <= word2.length <= 200
  • All word1[i] appear in keys.
  • word2.length is even.
  • keys, values[i], dictionary[i], word1, and word2 only contain lowercase English letters.
  • At most 200 calls will be made to encrypt and decrypt in total.


Related Topics:
Hash Table, String

Similar Questions:

Solution 1. Hash Map

encrypt is straightforward. For decrypt, we can precompute ans store the frequencies of encrypted permitted original strings in a map freq, and return freq[word] in decrypt.

// OJ:
// Time:
//      Encrypter: O(K + V + D) where `K` is the length of `keys`, `V` and `D` are the sizes of all the contents in `values` and `dict`, respectively.
//      encrypt: O(W) where `W` is the length of the input word
//      decrypt: O(W)
// Space: O(K + V + D)
class Encrypter {
    unordered_map<char, string> m; // mapping from character keys to corresponding value strings
    unordered_map<string, int> freq; // frequency map of encrypted permitted original strings
    Encrypter(vector<char>& keys, vector<string>& values, vector<string>& dictionary) {
        for (int i = 0; i < keys.size(); ++i) {
            m[keys[i]] = values[i];
        for (auto &s : dictionary) {
            auto e = encrypt(s);
            if (e.size()) freq[e]++;
    string encrypt(string s) {
        string ans;
        for (char c : s) {
            if (m.count(c) == 0) return ""; // invalid key character met, return empty string.
            ans += m[c];
        return ans;
    int decrypt(string s) {
        return freq.count(s) ? freq[s] : 0;

The problem didn’t say clearly whether a permitted original string might include unsupported keys. If it might include, then we need to consider the following testcase:



All Problems

All Solutions