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:
- For each character
cin the string, we find the indexisatisfyingkeys[i] == cinkeys. - Replace
cwithvalues[i]in the string.
A string is decrypted with the following process:
- For each substring
sof length 2 occurring at an even index in the string, we find anisuch thatvalues[i] == s. If there are multiple validi, we choose any one of them. This means a string could have multiple possible strings it can decrypt to. - Replace
swithkeys[i]in the string.
Implement the Encrypter class:
Encrypter(char[] keys, String[] values, String[] dictionary)Initializes theEncrypterclass withkeys, values, anddictionary.String encrypt(String word1)Encryptsword1with the encryption process described above and returns the encrypted string.int decrypt(String word2)Returns the number of possible stringsword2could decrypt to that also appear indictionary.
Example 1:
Input ["Encrypter", "encrypt", "decrypt"] [[['a', 'b', 'c', 'd'], ["ei", "zf", "ei", "am"], ["abcd", "acbd", "adbc", "badc", "dacb", "cadb", "cbda", "abad"]], ["abcd"], ["eizfeiam"]] Output
[null, “eizfeiam”, 2]
Explanation 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.
Constraints:
1 <= keys.length == values.length <= 26values[i].length == 21 <= dictionary.length <= 1001 <= dictionary[i].length <= 100- All
keys[i]anddictionary[i]are unique. 1 <= word1.length <= 20001 <= word2.length <= 200- All
word1[i]appear inkeys. word2.lengthis even.keys,values[i],dictionary[i],word1, andword2only contain lowercase English letters.- At most
200calls will be made toencryptanddecryptin total.
Solution:
For encryption, follow the instruction. Time complexity: O(len(word)) = O(2000)
For decryption, try all words in the dictionary and encrypt them and compare the encrypted string with the word to decrypt. Time complexity: O(sum(len(word_in_dict))) = O(100*100)
Worst case: 200 calls to decryption, T = 200 * O(100 * 100) = O(2*106)
C++
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// Author: Huahua class Encrypter { public: Encrypter(vector<char>& keys, vector<string>& values, vector<string>& dictionary): vals(26), dict(dictionary) { for (size_t i = 0; i < keys.size(); ++i) vals[keys[i] - 'a'] = values[i]; } string encrypt(string word1) { string ans; for (char c : word1) ans += vals[c - 'a']; return ans; } int decrypt(string word2) { return count_if(begin(dict), end(dict), [&](const string& w){ return encrypt(w) == word2; }); } private: vector<string> vals; vector<string> dict; }; |
Optimization
Pre-compute answer for all the words in dictionary.
decrypt: Time complexity: O(1)
C++
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// Author: Huahua class Encrypter { public: Encrypter(vector<char>& keys, vector<string>& values, vector<string>& dictionary): vals(26) { for (size_t i = 0; i < keys.size(); ++i) vals[keys[i] - 'a'] = values[i]; for (const string& w : dictionary) ++counts[encrypt(w)]; } string encrypt(string word1) { string ans; for (char c : word1) ans += vals[c - 'a']; return ans; } int decrypt(string word2) { auto it = counts.find(word2); return it == counts.end() ? 0 : it->second; } private: vector<string> vals; unordered_map<string, int> counts; }; |
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