Algorithm
Problem Name: 609. Find Duplicate File in System
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 * 1041 <= paths[i].length <= 30001 <= sum(paths[i].length) <= 5 * 105paths[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.
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public List();
for (String path : paths) {
String[] splits = path.split("\\s+");
String directoryName = splits[0];
for (int i = 1; i < splits.length; i++) {
int contentStartIdx = splits[i].indexOf('(');
String fileName = splits[i].substring(0, contentStartIdx);
String content = splits[i].substring(contentStartIdx + 1, splits[i].length() - 1);
contentToDirectoryMapping.computeIfAbsent(content, k -> new HashSet < >())
.add(directoryName + "/" + fileName);
}
}
return contentToDirectoryMapping.values().stream()
.filter(e -> e.size() > 1)
.map(ArrayList::new)
.collect(Collectors.toList());
}
}
Input
Output
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const findDuplicate = function (paths) {
const map = {}
for (let text of paths) {
for (let i = 1, files = text.split(' '); i < files.length; i++) {
const paren = files[i].indexOf('(')
const content = files[i].substring(paren + 1, files[i].length - 1)
map[content] = map[content] || []
map[content].push(files[0] + '/' + files[i].substr(0, paren))
}
}
return Object.values(map).filter((dups) => dups.length > 1)
}
Input
Output
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def findDuplicate(self, paths):
dic = collections.defaultdict(list)
for path in paths:
root, *f = path.split(" ")
for file in f:
txt, content = file.split("(")
dic[content] += root + "/" + txt,
return [dic[key] for key in dic if len(dic[key]) > 1]
Input
Output
#4 Code Example with C# Programming
Code -
C# Programming
using System.Collections.Generic;
namespace LeetCode
{
public class _0609_FindDuplicateFileInSystem
{
public IList < IList();
map[content].Add($"{dir}/{name}");
}
}
var result = new List < IList 1)
result.Add(files);
return result;
}
}
}
Input
Output