609. Find Duplicate File in System

Description

Given a list of directory info including directory path, and all the files with contents in this directory, you need to find out all the groups of duplicate files in the file system in terms of their paths.

A group of duplicate files consists of at least two files that have exactly 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 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 group 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:
[“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”]]
Note:
No order is required for the final output.
You may assume the directory name, file name and file content only has letters and digits, and the length of file content is in the range of [1,50].
The number of files given is in the range of [1,20000].
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. Directory path and file info are separated by a single blank space.
Follow-up beyond contest:
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?

Problem URL

---

Solution

给一个String数组，每个string为在这个文件夹下的所有文件的名字和内容，内容有括号括起来，找到所有的内容相同的文件的绝对路ing。

Using a hash map to perform a map from content to set of filenames. First, for each Stirng in paths, split them by white space “//s+”, then get the content start at strs[1] by find the index of ‘(’. Substring it, we could keep “)”, because it does not matter. Then get the absolute file name by Strs[0] + “/” + strs[s].substring(0, index).

After that, add it into a hash set and put it into the map. Second loop iterate the map’s key set, get the set with size > 1 and add them into the res list.

Code

class Solution {
    public List<List<String>> findDuplicate(String[] paths) {
        List<List<String>> res = new ArrayList<>();
        if (paths == null || paths.length == 0){
            return res;
        }
        Map<String, Set<String>> map = new HashMap<>();
        for (String path : paths){
            String[] strs = path.split("\\s+");
            for (int i = 1; i < strs.length; i++){
                int index = strs[i].indexOf("(");
                String content = strs[i].substring(index);
                String fileName = strs[0] + "/" + strs[i].substring(0, index);
                Set<String> fileNames = map.getOrDefault(content, new HashSet<String>());
                fileNames.add(fileName);
                map.put(content, fileNames);
            }
        }
        
        for (String key : map.keySet()){
            if (map.get(key).size() > 1){
                res.add(new ArrayList<String>(map.get(key)));
            }
        }
        return res;
    }
}

Time Complexity: O()
Space Complexity: O()

---

Review