LeetCode //C - 394. Decode String

394. Decode String

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 $10^5$.
?

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.
• ll the integers in s are in the range [1, 300].

From: LeetCode
Link: 394. Decode String

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Solution:

Ideas：

1. Stack Implementation: Since the problem involves nested structures and the need to process the most recent open bracket first, a stack data structure is suitable. We’ll use two stacks: one for characters and one for integers.

2. Parsing the Input String: We iterate through the input string. When we encounter a digit, we calculate the whole number (as numbers can have more than one digit) and push it onto the integer stack. When we encounter a ‘[’, we push a marker onto the character stack. For letters, we simply push them onto the character stack.

3. Decoding: When we encounter a ‘]’, it means we need to decode the string within the last ‘[’ and ‘]’ encountered. We pop characters from the character stack until we reach the marker, build the string, then multiply it according to the top value of the integer stack, and push the result back onto the character stack.

4. Building the Result: After processing the entire input string, the character stack will contain the decoded string. We then pop all characters to build the final result.

5. Memory Management: Since we need to return a dynamically allocated string, we need to ensure proper memory allocation and deallocation to avoid memory leaks.

Code:

char* decodeString(char* s) {
    int numStack[30], numTop = -1;     // Stack for numbers
    char charStack[30000], *p = s;     // Stack for characters
    int charTop = -1, num = 0;

    while (*p) {
        if (isdigit(*p)) {             // If it's a digit, calculate the number
            num = num * 10 + (*p - '0');
        } else if (*p == '[') {
            numStack[++numTop] = num;  // Push the number onto the numStack
            charStack[++charTop] = '['; // Push a marker onto the charStack
            num = 0;                   // Reset the number for next use
        } else if (*p == ']') {
            int repeat = numStack[numTop--]; // Pop the top number for repetition
            char temp[30000], *tempPtr = temp;
            while (charStack[charTop] != '[') // Build the string to repeat
                *tempPtr++ = charStack[charTop--];
            charTop--;                       // Pop the '[' marker
            while (repeat-- > 0)             // Repeat the string
                for (char* k = tempPtr - 1; k >= temp; k--)
                    charStack[++charTop] = *k;
        } else {
            charStack[++charTop] = *p;      // Push the character onto the stack
        }
        p++;
    }
    
    charStack[++charTop] = '\0';            // Null-terminate the string
    char* result = (char*)malloc(charTop + 1);
    strcpy(result, charStack);              // Copy the stack content to result
    return result;
}