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Java-based LeetCode algorithm problem solutions, regularly updated
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package g2301_2400.s2325_decode_the_message;

// #Easy #String #Hash_Table #2022_07_03_Time_7_ms_(42.86%)_Space_42.6_MB_(57.14%)

import java.util.HashMap;
import java.util.Map;

/**
 * 2325 - Decode the Message\.
 *
 * Easy
 *
 * 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:
 *
 * 1.  Use the **first** appearance of all 26 lowercase English letters in `key` as the **order** of the substitution table.
 * 2.  Align the substitution table with the regular English alphabet.
 * 3.  Each letter in `message` is then **substituted** using the table.
 * 4.  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:**
 *
 * ![](https://assets.leetcode.com/uploads/2022/05/08/ex1new4.jpg)
 *
 * **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**   **f** o **x**   **j** u **mps**  o **v** er the  **lazy**   **d** o **g** ".
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2022/05/08/ex2new.jpg)
 *
 * **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 `' '`.
**/
public class Solution {
    public String decodeMessage(String key, String message) {
        StringBuilder sb = new StringBuilder();
        Map 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();
    }
}