LeetCode 2227: Encrypt and Decrypt Strings
Problem Description
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.
Solutions
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
}
}Loading editor...