Find Duplicate File in System (Leetcode 609)

Difficulty: Medium Link: Day 18: May Leetcode Challenge

Problem Description

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"
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"]]
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?

Solution

After going through the description and examples, you can easily notice that problem is nothing but related to key-value mapping and string manipulation. The content is the criteria on which we have to decide if the files are duplicate. Which data structure would allow us to make mapping of unique keys with multiple values ? Dictionary. We will create a dictionary[String:[String]] for mapping list of files that point to the same content. Also, notice that notice that input is given in the form directory file1(content_1) file2(content) … but output is returned in the form [dir/file1 , dir/file2], so while inserting them in the dictionary, we will format the path string before hand. Another thing is that path could be just a filename or multiple files separated by space (” “). So we also need to iterate the full file and extract all the file names. Finally, we can iterate and filter the duplicate values or count > 1.

Swift

class Solution {
func findDuplicate(_ paths: [String]) -> [[String]] {
var map = [String:[String]]()
for path in paths {
var dir = ""
let components = path.split(separator : " ")
if components.count > 1 { dir = String(components[0])}
for i in 1..<components.count {
if let fileNamePair = getFileNameContentPair(String(components[i])){
let key = fileNamePair.0
let value = "\(dir)/\(fileNamePair.1)"
if let existing = map[key] {
map[key] = existing + [value]
}
else{
map[key] = [value]
}
}
}
}
return map.values.filter{ $0.count > 1}
}
// filename(content)
func getFileNameContentPair(_ path:String) -> (String,String)?{
if let firstIdx = path.indexInt(of:"("), let lastIdx = path.indexInt(of:")") {
let fileName = String(path.prefix(firstIdx))
let startIdx = path.index(path.startIndex, offsetBy: firstIdx 1)
let last = path.index(path.startIndex, offsetBy: lastIdx)
let content = String(path[startIdx..<last])
return (content,fileName)
}
return nil
}
}
public extension String {
func indexInt(of char: Character) -> Int? {
return firstIndex(of: char)?.utf16Offset(in: self)
}
}

Python

class Solution:
def findDuplicate(self, paths: List[str]) -> List[List[str]]:
contMap, ans = defaultdict(list), []
for pStr in paths:
sep = pStr.split(" ")
for i in range(1, len(sep)):
parts = sep[i].split('(')
cont = parts[1][:1]
contMap[cont].append(sep[0] + '/' + parts[0])
for v in contMap.values():
if len(v) > 1: ans.append(v)
return ans

Java

class Solution {
public List<List<String>> findDuplicate(String[] paths) {
Map<String, List<String>> contMap = new HashMap<>();
StringBuilder pathfile = new StringBuilder();
for (String pStr : paths) {
int i = 0;
pathfile.setLength(0);
while (pStr.charAt(i) != ' ') i++;
pathfile.append(pStr.substring(0,i)).append('/');
int pLen = ++i;
for (int j = i, k = 0; i < pStr.length(); i++)
if (pStr.charAt(i) == '(') {
pathfile.append(pStr.substring(j,i));
k = i + 1;
} else if (pStr.charAt(i) == ')') {
String cont = pStr.substring(k, i);
if (!contMap.containsKey(cont))
contMap.put(cont, new ArrayList<>());
contMap.get(cont).add(pathfile.toString());
j = i + 2;
pathfile.setLength(pLen);
}
}
List<List<String>> ans = new ArrayList<>();
for (List<String> v : contMap.values())
if (v.size() > 1) ans.add(v);
return ans;
}
}

Complexity Analysis

We are iterating over all the paths and then extracting all the file names from the path. If we take average length of every path to be L and there are N paths then total operations are O(N*L). Same is the number of values that would be in the dictionary.

Time = O(N * L) // N = number of paths , L = Average Length of each path

Space = O(L * N)

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