LeetCode-in-All

394. Decode String

Medium

Given an encoded string, return its decoded string.

The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.

You may assume that the input string is always valid; there are no extra white spaces, square brackets are well-formed, etc. Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there will not be input like 3a or 2[4].

The test cases are generated so that the length of the output will never exceed 105.

Example 1:

Input: s = “3[a]2[bc]”

Output: “aaabcbc”

Example 2:

Input: s = “3[a2[c]]”

Output: “accaccacc”

Example 3:

Input: s = “2[abc]3[cd]ef”

Output: “abcabccdcdcdef”

Constraints:

• 1 <= s.length <= 30
• s consists of lowercase English letters, digits, and square brackets '[]'.
• s is guaranteed to be a valid input.
• All the integers in s are in the range [1, 300].

Solution

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

// Helper function to repeat a string a given number of times
char* repeatString(const char* str, int count) {
    int len = strlen(str);
    char* result = (char*)malloc(len * count + 1);
    result[0] = '\0'; // Start with an empty string
    
    for (int i = 0; i < count; i++) {
        strcat(result, str);
    }
    return result;
}

// Helper function for recursive decoding
char* decodeHelper(char* s, int* pos) {
    char* result = (char*)malloc(10000); // Allocate sufficient space for result
    result[0] = '\0'; // Initialize as an empty string
    int count = 0;

    while (s[*pos] != '\0') {
        char c = s[*pos];
        (*pos)++;

        if (isalpha(c)) {
            // If it's a letter, append it to the result
            int len = strlen(result);
            result[len] = c;
            result[len + 1] = '\0';
        } else if (isdigit(c)) {
            // If it's a digit, accumulate it as part of count
            count = count * 10 + (c - '0');
        } else if (c == '[') {
            // Start of a nested encoded string
            char* decodedSubstring = decodeHelper(s, pos);
            char* repeatedSubstring = repeatString(decodedSubstring, count);
            strcat(result, repeatedSubstring);
            free(decodedSubstring);
            free(repeatedSubstring);
            count = 0; // Reset count after using it
        } else if (c == ']') {
            // End of the current encoded substring
            break;
        }
    }
    return result;
}

// Main function to decode the string
char* decodeString(char* s) {
    int pos = 0;
    return decodeHelper(s, &pos);
}

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