2325. Decode the Message

Problem

You are given the strings key and message, which represent a cipher key and a secret message, respectively. The steps to decode message are as follows:

• Use the first appearance of all 26 lowercase English letters in key as the order of the substitution table.

• Align the substitution table with the regular English alphabet.

• Each letter in message is then substituted using the table.

• Spaces ' ' are transformed to themselves.

• For example, given key = "**hap**p**y** **bo**y" (actual key would have at least one instance of each letter in the alphabet), we have the partial substitution table of ('h' -> 'a', 'a' -> 'b', 'p' -> 'c', 'y' -> 'd', 'b' -> 'e', 'o' -> 'f').

Return the decoded message.

  Example 1:

Input: key = "the quick brown fox jumps over the lazy dog", message = "vkbs bs t suepuv"
Output: "this is a secret"
Explanation: The diagram above shows the substitution table.
It is obtained by taking the first appearance of each letter in "the quick brown fox jumps over the lazy dog".

Example 2:

Input: key = "eljuxhpwnyrdgtqkviszcfmabo", message = "zwx hnfx lqantp mnoeius ycgk vcnjrdb"
Output: "the five boxing wizards jump quickly"
Explanation: The diagram above shows the substitution table.
It is obtained by taking the first appearance of each letter in "eljuxhpwnyrdgtqkviszcfmabo".

  Constraints:

• 26 <= key.length <= 2000

• key consists of lowercase English letters and ' '.

• key contains every letter in the English alphabet ('a' to 'z') at least once.

• 1 <= message.length <= 2000

• message consists of lowercase English letters and ' '.

Solution (Java)

class Solution {
    public String decodeMessage(String key, String message) {
        StringBuilder sb = new StringBuilder();
        Map<Character, Character> temp = new HashMap<>();
        char[] alphabet = new char[26];
        int itr = 0;
        for (char c = 'a'; c <= 'z'; ++c) {
            alphabet[c - 'a'] = c;
        }
        for (int i = 0; i < key.length(); i++) {
            if (!temp.containsKey(key.charAt(i)) && key.charAt(i) != ' ') {
                temp.put(key.charAt(i), alphabet[itr++]);
            }
        }
        for (int j = 0; j < message.length(); j++) {
            if (message.charAt(j) == ' ') {
                sb.append(' ');
            } else {
                char result = temp.get(message.charAt(j));
                sb.append(result);
            }
        }
        return sb.toString();
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).