Find Duplicate File in System

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 leasttwofiles that have exactly the same content.

A single directory info string in theinputlist has the following format:

"root/d1/d2/.../dm f1.txt(f1_content) f2.txt(f2_content) ... fn.txt(fn_content)"

It means there arenfiles (f1.txt,f2.txt...fn.txtwith contentf1_content,f2_content...fn_content, respectively) in directoryroot/d1/d2/.../dm. Note that n >= 1 and m >= 0. If m = 0, it means the directory is just the root directory.

Theoutputis 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:

1. No order is required for the final output.
2. 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].
3. The number of files given is in the range of [1,20000].
4. You may assume no files or directories share the same name in the same directory.
5. You may assume each given directory info represents a unique directory. Directory path and file info are separated by a single blank space.

Brainstorm

We are really just seeing which files have the same content, which is a many-to-one (or one-to-many?) relationship. This presents an opportunity to use a hash map with key-value pairs.

We could choose to use the filename as the key, and its content as the value, but as we return the answer, it will be cumbersome to iterate through the map and compare the key-value pairs to each other to check which values are duplicates.

Therefore, it is more worthwhile to reverse the key-value pair: use the content as the key. Since the same content exists in multiple files, we'll need all those filenames stored in an array/vector as the value. After processing the entire input, we flush out the map and only add the vectors with a size greater than 1 to our return vector.

Approach #1: HashMap

vector<vector<string>> findDuplicate(vector<string>& paths) {
    unordered_map<string, vector<string>> myMap;

    for (string s : paths) {
        string dir = s.substr(0, s.find_first_of(' ')) + "/";
        for (int i = s.find_first_of(' ') + 1; i < s.length(); ) {
            string filename = "", content = "";
            int leftParenthese = s.find('(', i), rightParenthese = s.find(')', i);
            filename = dir + s.substr(i, leftParenthese - i);
            content = s.substr(leftParenthese + 1, rightParenthese - leftParenthese - 1);
            myMap[content].push_back(filename);
            i = rightParenthese + 2;
        }
    }

    vector<vector<string>> ret;
    for (auto it : myMap) {
        if (it.second.size() > 1)
            ret.push_back(it.second);
    }

    return ret;
}

Time complexity: $$O(n*x)$$

Assuming there are n strings with average length x, we would have to parse through all of them.

Space complexity: $$O(n*x)$$

The map may contain all n*x strings, as does the return vector.