Problem Description

You are given a list of strings where each string represents directory information. Each string contains:

• A directory path
• One or more files with their contents

The format of each string is: "root/d1/d2/.../dm f1.txt(f1_content) f2.txt(f2_content) ... fn.txt(fn_content)"

This means:

• The directory path is root/d1/d2/.../dm
• There are n files (f1.txt, f2.txt, ..., fn.txt) in this directory
• Each file has its content shown in parentheses (f1_content, f2_content, ..., fn_content)

Your task is to find all duplicate files in the file system. Duplicate files are files that have the exact same content, regardless of their names or locations.

You need to return groups of duplicate file paths, where:

• Each group contains at least 2 files with identical content
• Each file path should be in the format: "directory_path/file_name.txt"

For example, if the input is: ["root/a 1.txt(abcd) 2.txt(efgh)", "root/c 3.txt(abcd)", "root 4.txt(efgh)"]

The output would be: [["root/a/1.txt","root/c/3.txt"], ["root/a/2.txt","root/4.txt"]]

This is because:

• Files 1.txt and 3.txt both contain "abcd"
• Files 2.txt and 4.txt both contain "efgh"

The solution uses a hash map to group files by their content. It parses each input string to extract the directory path, file names, and contents, then builds the complete file paths and groups them by content. Finally, it returns only the groups that contain more than one file (duplicates).

Intuition

The key insight is that we need to identify files with identical content, regardless of their location or name. This immediately suggests using a hash map (dictionary) where we can group files by their content.

Think of it like organizing documents in a filing system - instead of organizing by name or location, we want to organize by what's written inside them. Every document with the same content goes into the same folder.

The natural approach is:

1. Use the file content as the key in our hash map
2. Store all file paths that have this content as the value (a list of paths)

This way, when we encounter a file, we can quickly check if we've seen this content before and add the current file path to the appropriate group.

The parsing strategy becomes clear when we look at the input format. Each string has:

• First part: the directory path (everything before the first space)
• Remaining parts: files in format filename.txt(content)

For each file, we need to:

• Extract the filename (everything before the ()
• Extract the content (everything between ( and ))
• Combine the directory path with the filename to get the full path

After processing all files, we'll have a hash map where each key (content) maps to a list of all file paths with that content. Since we only want duplicates, we filter out any groups with just one file.

This approach is efficient because:

• We process each file exactly once: O(n) where n is the total number of files
• Hash map operations (insert/lookup) are O(1) on average
• We avoid comparing every file with every other file, which would be O(n²)

Solution Approach

The implementation uses a hash map (Python's defaultdict) to group files by their content:

1. Initialize the data structure: Create a defaultdict(list) where each key will be file content and each value will be a list of file paths with that content.

2. Parse each path string:

   • Split each path string by spaces using p.split()
   • The first element ps[0] is the directory path
   • All remaining elements ps[1:] are files with their contents

3. Extract file information: For each file string in format filename.txt(content):

   • Find the position of ( using f.find('(')
   • Extract the filename: f[:i] (everything before the ()
   • Extract the content: f[i + 1 : -1] (everything between ( and ), excluding the parentheses)

4. Build the complete file path:

   • Combine directory and filename: ps[0] + '/' + name
   • Add this complete path to the list for the corresponding content: d[content].append(path)

5. Filter for duplicates:

   • Return only groups with more than one file: [v for v in d.values() if len(v) > 1]
   • This list comprehension filters out any content that appears in only one file

Example walkthrough: If input is ["root/a 1.txt(abc) 2.txt(xyz)", "root/b 3.txt(abc)"]

• First string: splits into ["root/a", "1.txt(abc)", "2.txt(xyz)"]
  • 1.txt(abc) → filename: 1.txt, content: abc, path: root/a/1.txt
  • 2.txt(xyz) → filename: 2.txt, content: xyz, path: root/a/2.txt
• Second string: splits into ["root/b", "3.txt(abc)"]
  • 3.txt(abc) → filename: 3.txt, content: abc, path: root/b/3.txt

Hash map after processing:

• "abc": ["root/a/1.txt", "root/b/3.txt"]
• "xyz": ["root/a/2.txt"]

Final output: [["root/a/1.txt", "root/b/3.txt"]] (only groups with 2+ files)

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Input: ["root/a 1.txt(hello) 2.txt(world)", "root/b 3.txt(hello)", "root/c/d 4.txt(hello) 5.txt(foo)"]

Step 1: Initialize our hash map Create an empty hash map to group files by content:

content_map = {}

Step 2: Process first string "root/a 1.txt(hello) 2.txt(world)"

Split by spaces: ["root/a", "1.txt(hello)", "2.txt(world)"]

• Directory: root/a
• Files to process: ["1.txt(hello)", "2.txt(world)"]

For 1.txt(hello):

• Find ( at position 5
• Filename: 1.txt (characters before position 5)
• Content: hello (between parentheses)
• Full path: root/a/1.txt
• Update map: content_map["hello"] = ["root/a/1.txt"]

For 2.txt(world):

• Filename: 2.txt
• Content: world
• Full path: root/a/2.txt
• Update map: content_map["world"] = ["root/a/2.txt"]

Step 3: Process second string "root/b 3.txt(hello)"

Split: ["root/b", "3.txt(hello)"]

• Directory: root/b

For 3.txt(hello):

• Filename: 3.txt
• Content: hello
• Full path: root/b/3.txt
• Update map: content_map["hello"] = ["root/a/1.txt", "root/b/3.txt"]

Step 4: Process third string "root/c/d 4.txt(hello) 5.txt(foo)"

Split: ["root/c/d", "4.txt(hello)", "5.txt(foo)"]

• Directory: root/c/d

For 4.txt(hello):

• Full path: root/c/d/4.txt
• Update map: content_map["hello"] = ["root/a/1.txt", "root/b/3.txt", "root/c/d/4.txt"]

For 5.txt(foo):

• Full path: root/c/d/5.txt
• Update map: content_map["foo"] = ["root/c/d/5.txt"]

Step 5: Filter for duplicates

Final hash map state:

• "hello": ["root/a/1.txt", "root/b/3.txt", "root/c/d/4.txt"] (3 files - duplicates!)
• "world": ["root/a/2.txt"] (1 file - not duplicate)
• "foo": ["root/c/d/5.txt"] (1 file - not duplicate)

Return only groups with 2+ files:

Output: [["root/a/1.txt", "root/b/3.txt", "root/c/d/4.txt"]]

The algorithm successfully identified that three files contain "hello" and grouped them together, while filtering out the unique files.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n * m)

• n is the total number of paths in the input list
• m is the average length of each path string
• The outer loop iterates through all n paths
• For each path, we split the string which takes O(m) time
• For each file in the path (let's say k files on average), we:
  • Find the '(' character: O(length of filename)
  • Slice the string to extract name and content: O(length of filename)
  • Append to the dictionary list: O(1) amortized
• The final list comprehension iterates through all unique contents in the dictionary, which is at most O(total number of files)
• Overall: O(n * m) where the string operations dominate

Space Complexity: O(n * m)

• The dictionary d stores all file paths grouped by content
• In the worst case, every file has unique content, so we store all file paths
• Each stored path includes the directory path plus filename
• The total space used is proportional to the sum of all characters in all file paths
• The output list could contain all file paths if they all share the same content
• Overall: O(n * m) for storing all the path strings in the dictionary and output

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect String Parsing with Multiple Spaces

A common mistake is assuming the input string is always perfectly formatted with single spaces. If there are extra spaces or inconsistent spacing, using split() without parameters handles this correctly, but developers might try to use split(' ') which would create empty strings in the result.

Pitfall Example:

# Wrong approach
path_components = path_string.split(' ')  # Creates empty strings if multiple spaces exist
# If input is "root/a  1.txt(abc)" (double space), this gives ["root/a", "", "1.txt(abc)"]

Solution:

# Correct approach
path_components = path_string.split()  # Handles any whitespace correctly

2. Hardcoding Parenthesis Removal

Another pitfall is incorrectly extracting content by hardcoding indices without finding the parenthesis positions, which fails if filenames contain special characters or numbers.

Pitfall Example:

# Wrong approach - assumes parentheses are always at specific positions
content = file_info[-5:-1]  # Assumes content is always 4 characters

Solution:

# Correct approach - dynamically find parenthesis positions
paren_index = file_info.find('(')
filename = file_info[:paren_index]
content = file_info[paren_index + 1:-1]

3. Not Handling Edge Cases

Failing to consider edge cases like:

• Empty content in files: file.txt()
• Files with identical names in different directories
• Single file with unique content (should not be in output)

Pitfall Example:

# Wrong - returns all groups including single files
return list(content_to_paths.values())

Solution:

# Correct - only return groups with duplicates
return [file_paths for file_paths in content_to_paths.values() if len(file_paths) > 1]

4. Path Concatenation Issues

Using incorrect path concatenation that might create double slashes or missing slashes.

Pitfall Example:

# Wrong approaches
full_path = directory + filename  # Missing slash
full_path = directory + '//' + filename  # Double slash

Solution:

# Correct approach
full_path = f"{directory}/{filename}"
# Or using os.path.join for platform independence (though not needed for this problem)