Decode String: LeetCode 394 Guide

Leetcode 394 is an important coding problem that involves decoding a string of characters. It is an essential skill for data analysts and computer programmers who work with string manipulation. The problem involves decoding a string that has been encoded in a specific way by using Integers in square brackets to indicate the number of times a string must be repeated. In this article, we will explore the Leetcode 394 problem, and how to solve it using efficient methods.

Understanding Leetcode 394

What is Leetcode 394?

Leetcode 394 is a coding problem that requires the decoding of an encoded string. The encoding rule used for the string is k[encoded_string], where the original string is encoded by replacing each substring that has exactly k repeating characters with the substring followed by the integer k inside square brackets. The task is to decode the given encoded string and return the original string.

Breaking down the problem

The first step in solving the problem is to parse the given encoded string and separate it into its constituent parts. We need to identify the repeating substrings and the corresponding repetition count. This can be done using stack data structure by iterating over the given string character by character. Whenever we encounter a closing bracket, we pop elements from the stack until we get to the corresponding opening bracket. This gives us the repeating substring and its repetition count.

The next step is to use this information to construct the decoded string. We start with an empty string and iterate over the given encoded string character by character. Whenever we encounter a letter, we add it to the decoded string. Whenever we encounter an opening bracket, we push the current decoded string and the corresponding repetition count onto the stack. Whenever we encounter a closing bracket, we pop the previous decoded string and repetition count from the stack and repeat the current decoded string that many times.

Sample inputs and outputs

For example, if the given encoded string is “3[a]2[bc]”, the decoded string would be “aaabcbc”. Here, we have 3 repetitions of the substring “a” and 2 repetitions of the substring “bc”.

Solution Approaches

Recursive approach

One of the ways to solve the Leetcode 394 problem is through a recursive approach. This method involves first searching for the opening bracket in the encoded string, then finding the corresponding closing bracket, and finally, recursively processing the string enclosed in the brackets using the following formula:

Here, repeat_times represents the integer preceding the opening bracket, substring_in_brackets is the string inside the brackets, and result is the final decoded string. This process is continued until the entire string is processed, returning the final decoded string as the result.

Stack-based approach

Another approach to solve the Leetcode 394 problem is the stack-based approach, which involves using two stacks: one to store the integers and the other to store the strings. Starting from the beginning of the string, we iterate through each character, performing the following operations:

1. If the character is a digit, we update the repeat_times variable with this number.
2. If the character is an opening bracket, we push the current repeat_times and substring to the respective stacks and reset the repeat_times and substring variables.
3. If the character is a closing bracket, we retrieve the last repeat_times and substring from the stacks, and process them according to the following formula:
substring = string_stack.pop() + substring * repeat_times
4. If the character is a letter, we add it to the current substring.

The final result is the contents of the string stack, which represents the decoded string.

Comparing the approaches

Leetcode 394 coding problems can be solved using either a recursive or a stack-based approach. Both methods have their advantages and disadvantages.

In general, simple recursion problems are better solved using the stack-based approach as it avoids the overhead of function calls and call stack management. However, for problems involving multiple recursion, it is generally better to use the recursive approach for better code clarity and understanding.

Code performance analysis

When it comes to performance, both recursion and iteration have their own advantages and disadvantages. Iteration is generally preferred in imperative programming, especially for simple recursion since it avoids the additional overhead of function calls and call stack management. On the other hand, recursion is more suitable for multiple recursion. Given the problem under consideration and the programming language used, one of the approaches may be more efficient than the other.

For the Leetcode 394 problem, both recursive and stack-based iterative solutions have O(n) time complexity, where n is the length of the encoded string. However, the space complexity varies between the two approaches. The recursive solution has O(n) space complexity due to the call stack usage, while the iterative solution has O(kn) space complexity where k is the maximum number found in the string. This is because the stack stores the current temporary result multiple times in the iterative approach, for every inner bracket it enounters.

Therefore, if the length of the encoded string is significantly large, the recursive solution may lead to a stack overflow error, while the iterative solution could potentially consume a lot of memory. However, for shorter strings, the differences in performance may not be significant.

References

LeetCode 394 Problem Statement

Problem 394 in LeetCode involves decoding an encoded string based on a specific encoding rule. The rule is k[encoded_string], where the k represents the number of times the encoded_string should be repeated. The task is to return the decoded string.

Stack (abstract data type)

The solution to this problem involves the use of a stack – an abstract data type that follows the Last-In-First-Out (LIFO) principle. The stack is used to keep track of the counts and strings as we iterate through the input string. We push and pop elements from the stack depending on the value of the current character in the input string.

Recursion

The solution to this problem can also be approached using recursion. Recursion involves the process of calling a function within itself. In this case, we can recursively call a function to decode the encoded substring within a pair of square brackets.