Remove Comments

Remove Comments - Problem

Array String Medium

Given a C++ program, remove comments from it. The program source is an array of strings source where source[i] is the i-th line of the source code.

In C++, there are two types of comments:

Line comments: The string // denotes a line comment, which represents that it and the rest of the characters to the right of it in the same line should be ignored.
Block comments: The string /* denotes a block comment, which represents that all characters until the next occurrence of */ should be ignored.

Important rules:

The first effective comment takes precedence over others
If // occurs in a block comment, it is ignored
If /* occurs in a line or block comment, it is also ignored
If a line becomes empty after removing comments, do not output that line
Block comments can span multiple lines and delete implicit newline characters

Return the source code with comments removed in the same format.

Input & Output

Example 1 — Mixed Comments

$ Input: source = ["/*Test program */", "int main()", "{ ", " // variable declaration ", "int a, b, c;", "/* This is a long ", " comment to test ", " if the comment is", " removed properly.*/", "a = b + c;", "}"]

› Output: ["int main()","{ ","int a, b, c;","a = b + c;","}"]

💡 Note: Block comment /*Test program */ removed from first line making it empty. Line comment // variable declaration removes rest of line. Multi-line block comment removed completely. Only actual code lines remain.

Example 2 — Block Comment Across Lines

$ Input: source = ["a/*comment", "line", "more_comment*/b"]

› Output: ["ab"]

💡 Note: Block comment spans multiple lines from 'a/*comment' through 'more_comment*/b'. After removing the comment content, 'a' and 'b' are concatenated since the block comment removal eliminates the newline characters in between.

Example 3 — Line Comment Only

$ Input: source = ["class Test {", " public static void main() {", " // This line is comment", " System.out.println(\"Hello\");", " }", "}"]

› Output: ["class Test {"," public static void main() {"," System.out.println(\"Hello\");"," ","}"]

💡 Note: Line comment '// This line is comment' removes the entire line. All other lines containing actual code are preserved exactly as they are.

Constraints

1 ≤ source.length ≤ 100
0 ≤ source[i].length ≤ 80
source[i] consists of printable ASCII characters
Every open block comment will eventually be closed
There are no string literals with quotes

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8 f Facebook 6

The key insight is to use a state machine to track whether we're inside a block comment while processing each character. The optimal approach processes all characters in a single pass, maintaining block comment state across lines and filtering out comment content. Time: O(n×m), Space: O(n×m) where n is number of lines and m is average line length.

Common Approaches

Approach	Time	Space	Notes
✓ Single Pass with String Building	O(n × m)	O(n × m)	Efficiently process all lines in one pass with optimized string operations
Character-by-Character State Machine	O(n × m)	O(n × m)	Process each character while tracking comment state

Single Pass with String Building — Algorithm Steps

Step 1: Initialize block comment state and result builder
Step 2: Process each character, checking for comment delimiters
Step 3: Handle state transitions between normal code and comments
Step 4: Collect non-empty lines into final result

Visualization

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Step-by-Step Walkthrough

Scan Line

Process each character while tracking block comment state

Filter Content

Skip comment regions, keep code content

Build Result

Add non-empty processed lines to result

Code -

solution.c — C

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

char** solution(char** source, int sourceSize, int* returnSize) {
    char** result = (char**)malloc(sourceSize * sizeof(char*));
    *returnSize = 0;
    bool inBlockComment = false;
    
    for (int lineIdx = 0; lineIdx < sourceSize; lineIdx++) {
        char* line = source[lineIdx];
        char newLine[1000] = "";
        int newLineLen = 0;
        int i = 0;
        
        while (i < strlen(line)) {
            if (inBlockComment) {
                if (i <= strlen(line) - 2 && strncmp(&line[i], "*/", 2) == 0) {
                    inBlockComment = false;
                    i += 2;
                } else {
                    i++;
                }
            } else {
                if (i <= strlen(line) - 2 && strncmp(&line[i], "/*", 2) == 0) {
                    inBlockComment = true;
                    i += 2;
                } else if (i <= strlen(line) - 2 && strncmp(&line[i], "//", 2) == 0) {
                    break;
                } else {
                    newLine[newLineLen++] = line[i];
                    i++;
                }
            }
        }
        
        if (!inBlockComment) {
            newLine[newLineLen] = '\0';
            bool hasContent = false;
            for (int j = 0; j < newLineLen; j++) {
                if (newLine[j] != ' ' && newLine[j] != '\t') {
                    hasContent = true;
                    break;
                }
            }
            
            if (hasContent) {
                result[*returnSize] = (char*)malloc((newLineLen + 1) * sizeof(char));
                strcpy(result[*returnSize], newLine);
                (*returnSize)++;
            }
        }
    }
    
    return result;
}

int main() {
    char input[10000];
    fgets(input, sizeof(input), stdin);
    input[strcspn(input, "\n")] = 0;
    
    char* source[100];
    int sourceSize = 0;
    
    char* ptr = input + 1;
    while (*ptr && *ptr != ']') {
        if (*ptr == '"') {
            ptr++;
            char* start = ptr;
            while (*ptr && *ptr != '"') ptr++;
            
            int len = ptr - start;
            source[sourceSize] = (char*)malloc((len + 1) * sizeof(char));
            strncpy(source[sourceSize], start, len);
            source[sourceSize][len] = '\0';
            sourceSize++;
            
            if (*ptr == '"') ptr++;
        }
        if (*ptr == ',') ptr++;
        while (*ptr == ' ') ptr++;
    }
    
    int returnSize;
    char** result = solution(source, sourceSize, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("\"%s\"", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m)

Still need to process every character in every line

✓ Linear Growth

Space Complexity

O(n × m)

Result storage and string building operations

⚡ Linearithmic Space

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Input & Output

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Related Problems

Common Approaches

Single Pass with String Building — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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