Leetcode #722: Remove Comments
In this guide, we solve Leetcode #722 Remove Comments in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a C++ program, remove comments from it. The program source is an array of strings source where source[i] is the ith line of the source code.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, String
Intuition
We need to scan characters while tracking positions or counts.
A simple state machine keeps the logic precise.
Approach
Iterate through the string once and update the state for each character.
Use a map or array if you need fast lookups.
Steps:
- Iterate through characters.
- Maintain necessary state.
- Build or validate the output.
Example
Input: source = ["/*Test program */", "int main()", "{ ", " // variable declaration ", "int a, b, c;", "/* This is a test", " multiline ", " comment for ", " testing */", "a = b + c;", "}"]
Output: ["int main()","{ "," ","int a, b, c;","a = b + c;","}"]
Explanation: The line by line code is visualized as below:
/*Test program */
int main()
{
// variable declaration
int a, b, c;
/* This is a test
multiline
comment for
testing */
a = b + c;
}
The string /* denotes a block comment, including line 1 and lines 6-9. The string // denotes line 4 as comments.
The line by line output code is visualized as below:
int main()
{
int a, b, c;
a = b + c;
}
Python Solution
class Solution:
def removeComments(self, source: List[str]) -> List[str]:
ans = []
t = []
block_comment = False
for s in source:
i, m = 0, len(s)
while i < m:
if block_comment:
if i + 1 < m and s[i : i + 2] == "*/":
block_comment = False
i += 1
else:
if i + 1 < m and s[i : i + 2] == "/*":
block_comment = True
i += 1
elif i + 1 < m and s[i : i + 2] == "//":
break
else:
t.append(s[i])
i += 1
if not block_comment and t:
ans.append("".join(t))
t.clear()
return ans
Complexity
The time complexity is , and the space complexity is , where is the total length of the source code. The space complexity is , where is the total length of the source code.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.