LeetCode 2227: Encrypt and Decrypt Strings Solution

Master LeetCode problem 2227 (Encrypt and Decrypt Strings), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2227. Encrypt and Decrypt Strings

Hard

Array Hash Table String Design Trie

Problem Explanation

Explanation

To solve this problem, we can use two maps: one for encryption (character to string mapping) and another for decryption (string to character mapping). When encrypting a string, we iterate through each character and replace it with its corresponding value from the encryption map. When decrypting a string, we iterate through each pair of characters and replace it with any key from the decryption map that matches the pair.

Time Complexity

Initialization: O(n), where n is the number of keys/values in the constructor.
Encryption: O(m), where m is the length of the input word1.
Decryption: O(m), where m is the length of the input word2.

Space Complexity

O(n) for storing the encryption and decryption maps.

Solution Code

import java.util.HashMap;

class Encrypter {
    HashMap<Character, String> encryptMap;
    HashMap<String, Character> decryptMap;

    public Encrypter(char[] keys, String[] values, String[] dictionary) {
        encryptMap = new HashMap<>();
        decryptMap = new HashMap<>();

        for (int i = 0; i < keys.length; i++) {
            encryptMap.put(keys[i], values[i]);
            decryptMap.put(values[i], keys[i]);
        }
    }

    public String encrypt(String word1) {
        StringBuilder encrypted = new StringBuilder();
        for (char c : word1.toCharArray()) {
            if (encryptMap.containsKey(c)) {
                encrypted.append(encryptMap.get(c));
            } else {
                return "";
            }
        }
        return encrypted.toString();
    }

    public int decrypt(String word2) {
        int count = 1;
        for (int i = 0; i < word2.length(); i += 2) {
            String pair = word2.substring(i, i + 2);
            if (decryptMap.containsKey(pair)) {
                char decryptedChar = decryptMap.get(pair);
                count *= 2; // Double the count for every valid decryption option
            }
        }
        return count / 2; // Divide by 2 to account for doubling in the loop
    }
}

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