How do you solve Remove Comments on LeetCode?

Remove Comments (LeetCode #722) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the difference between line comments and block comments is crucial for parsing correctly.

What is the time complexity of Remove Comments?

The optimal solution for Remove Comments runs in O(n) time and uses O(n) space. The complexity is O(n) because we traverse each character in the input once, making it efficient for larger inputs.

What is the brute force solution for Remove Comments?

The brute force approach for Remove Comments is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves scanning each line of the source code and checking for comments. If a comment is found, we simply ignore the rest of the line or the block until the end of the comment. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Remove Comments?

Remove Comments uses the following concepts: Array, String. The recommended patterns to study are: State Management, String Parsing.

What are the interview tips for Remove Comments?

Always clarify the rules around comments before starting to code. Think about edge cases, such as comments at the beginning or end of lines. Discuss your thought process with the interviewer to show your understanding of the problem.

What are common mistakes when solving Remove Comments?

Not properly handling the case where block comments are nested or improperly terminated. Forgetting to check for empty lines after comment removal.

#722

Remove Comments

Medium

ArrayStringState ManagementString Parsing

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution processes the input in a single pass, maintaining a state to track whether we are inside a block comment. This avoids unnecessary string operations and improves efficiency.

⚙️

Algorithm

6 steps

1Step 1: Initialize an empty result list and a boolean to track if we are in a block comment.
2Step 2: Loop through each line in the source code.
3Step 3: For each character in the line, check for the start or end of comments.
4Step 4: If we encounter '/*', set the block comment state to true; if we encounter '*/', set it to false.
5Step 5: If we are not in a block comment, add valid characters to a temporary line.
6Step 6: After processing the line, if it is not empty, add it to the result list.

solution.py25 lines

1# Full working Python code
2
3def removeComments(source):
4    result = []
5    in_block = False
6    for line in source:
7        temp_line = ''
8        i = 0
9        while i < len(line):
10            if not in_block:
11                if line[i:i+2] == '/*':
12                    in_block = True
13                    i += 2
14                elif line[i:i+2] == '//':
15                    break
16                else:
17                    temp_line += line[i]
18            else:
19                if line[i:i+2] == '*/':
20                    in_block = False
21                    i += 2
22            i += 1
23        if temp_line:
24            result.append(temp_line)
25    return result

ℹ

Complexity note: The complexity is O(n) because we traverse each character in the input once, making it efficient for larger inputs.

1Understanding the difference between line comments and block comments is crucial for parsing correctly.
2Maintaining a state (in or out of a block comment) simplifies the logic for handling nested comments.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.