LeetCode //C - 1309. Decrypt String from Alphabet to Integer Mapping

1309. Decrypt String from Alphabet to Integer Mapping

You are given a string s formed by digits and '#'. We want to map s to English lowercase characters as follows:

• Characters ('a' to 'i') are represented by ('1' to '9') respectively.
• Characters ('j' to 'z') are represented by ('10#' to '26#') respectively.

Return the string formed after mapping.

The test cases are generated so that a unique mapping will always exist.

Example 1:

Input: s = "10#11#12"

Output: "jkab"

Explanation: "j" -> "10#" , "k" -> "11#" , "a" -> "1" , "b" -> "2".

Example 2:

Input: s = "1326#"

Output: "acz"

Constraints:

• 1 <= s.length <= 1000
• s consists of digits and the '#' letter.
• s will be a valid string such that mapping is always possible.

From: LeetCode

Link: 1309. Decrypt String from Alphabet to Integer Mapping

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Solution:

Ideas:

Scan from left to right; when the character two positions ahead is '#', decode two digits, otherwise decode one digit.

Code:

#include <stdlib.h>
#include <string.h>

char* freqAlphabets(char* s) {
    int n = strlen(s);
    char* result = (char*)malloc((n + 1) * sizeof(char));
    int i = 0;
    int index = 0;

    while (i < n) {
        int value;

        if (i + 2 < n && s[i + 2] == '#') {
            value = (s[i] - '0') * 10 + (s[i + 1] - '0');
            i += 3;
        } else {
            value = s[i] - '0';
            i++;
        }

        result[index++] = 'a' + value - 1;
    }

    result[index] = '\0';
    return result;
}