Introduction

In this blog post, we will explore how to solve the LeetCode problem "609" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given a list paths of directory info, including the directory path, and all the files with contents in this directory, return all the duplicate files in the file system in terms of their paths. You may return the answer in any order. A group of duplicate files consists of at least two files that have the same content. A single directory info string in the input list has the following format: "root/d1/d2/.../dm f1.txt(f1_content) f2.txt(f2_content) ... fn.txt(fn_content)" It means there are n files (f1.txt, f2.txt ... fn.txt) with content (f1_content, f2_content ... fn_content) respectively in the directory "root/d1/d2/.../dm". Note that n >= 1 and m >= 0. If m = 0, it means the directory is just the root directory. The output is a list of groups of duplicate file paths. For each group, it contains all the file paths of the files that have the same content. A file path is a string that has the following format: "directory_path/file_name.txt" Example 1: Input: paths = ["root/a 1.txt(abcd) 2.txt(efgh)","root/c 3.txt(abcd)","root/c/d 4.txt(efgh)","root 4.txt(efgh)"] Output: [["root/a/2.txt","root/c/d/4.txt","root/4.txt"],["root/a/1.txt","root/c/3.txt"]] Example 2: Input: paths = ["root/a 1.txt(abcd) 2.txt(efgh)","root/c 3.txt(abcd)","root/c/d 4.txt(efgh)"] Output: [["root/a/2.txt","root/c/d/4.txt"],["root/a/1.txt","root/c/3.txt"]] Constraints: 1 <= paths.length <= 2 104 1 <= paths[i].length <= 3000 1 <= sum(paths[i].length) <= 5 105 paths[i] consist of English letters, digits, '/', '.', '(', ')', and ' '. You may assume no files or directories share the same name in the same directory. You may assume each given directory info represents a unique directory. A single blank space separates the directory path and file info. Follow up: Imagine you are given a real file system, how will you search files? DFS or BFS? If the file content is very large (GB level), how will you modify your solution? If you can only read the file by 1kb each time, how will you modify your solution? What is the time complexity of your modified solution? What is the most time-consuming part and memory-consuming part of it? How to optimize? How to make sure the duplicated files you find are not false positive?

Explanation

• High-Level Approach: Parse each path string to extract the directory, filename, and content. Use a hash map to store content as keys and a list of file paths as values. Finally, iterate through the hash map and return the list of file paths for each content that appears in at least two files.
  • Complexity: Time: O(N L), where N is the number of paths and L is the average length of each path. Space: O(N L) in the worst case where all files have unique content.
  • Source Code:

Code

    def findDuplicate(paths):
    """
    Finds duplicate files in a file system.

    Args:
        paths: A list of directory info strings.

    Returns:
        A list of lists of duplicate file paths.
    """

    content_map = {}
    for path in paths:
        parts = path.split(" ")
        root = parts[0]
        for i in range(1, len(parts)):
            file_info = parts[i]
            filename, content = file_info.split("(")
            content = content[:-1]  # Remove the closing parenthesis
            file_path = root + "/" + filename
            if content not in content_map:
                content_map[content] = []
            content_map[content].append(file_path)

    result = []
    for content, file_paths in content_map.items():
        if len(file_paths) > 1:
            result.append(file_paths)

    return result