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Leetcode daily trainning by using C/C++/RUST programming.
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leetcode2325: Decode the Message #382

Open carloscn opened 1 year ago

carloscn commented 1 year ago

Description

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 = "happy boy" (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:

image

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:

image

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 ' '.

carloscn commented 1 year ago

Analysis

static int32_t decode_message(const char *key, char *message)
{
    int32_t ret = 0;
    char *table = NULL;
    size_t key_len, mes_len;

    UTILS_CHECK_PTR(key);
    UTILS_CHECK_PTR(message);
    UTILS_CHECK_LEN(key_len = strlen(key));
    UTILS_CHECK_LEN(mes_len = strlen(message));

    table = (char *)calloc(key_len + 1, sizeof(char));
    UTILS_CHECK_PTR(table);

    for (size_t count = 0, i = 0; i < key_len; i ++) {
        if (key[i] == ' ') {
            table[i] = ' ';
            continue;
        }
        for (size_t j = 0; j < i; j ++) {
            if (key[i] == key[j]) {
                table[i] = '@';
                break;
            }
        }
        table[i] = table[i] == '@' ?
                   table[i] :
                   'a' + count ++;
    }

    for (size_t i = 0; i < mes_len; i ++) {
        char e = message[i];
        if (message[i] == ' ') {
            continue;
        }
        for (size_t j = 0; j < key_len; j ++) {
            if (key[j] == e) {
                message[i] = table[j];
                break;
            }
        }
    }

finish:
    UTILS_SAFE_FREE(table);
    return ret;
}
carloscn commented 1 year ago

Code

https://review.gerrithub.io/c/carloscn/structstudy/+/1170724 https://github.com/carloscn/structstudy/commit/86e399a3c1e20bb4bed256c6cae9c5d209e1b03b