What is the time complexity of Find Duplicate File in System?

The optimal solution has a time complexity of O(N*L).

What is the space complexity of Find Duplicate File in System?

The space complexity is O(N*L).

What data structures are used to solve Find Duplicate File in System?

This problem uses Array, Hash Table, String. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Find Duplicate File in System?

Insight 1: The core task is to group files by their content. Therefore, a hash table (dictionary or map) is the most suitable data structure to store content as keys and a list of file paths as values. Insight 2: The input string needs to be parsed effectively to extract the directory path, filename, and content. String manipulation techniques like splitting and finding substrings are essential.

Which companies ask LeetCode 609. Find Duplicate File in System in interviews?

This problem is frequently asked at Applied Intuition, Dropbox, Turing.

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Find Duplicate File in System - LeetCode 609 Solution

Find Duplicate File in System - Complete Solution Guide

Find Duplicate File in System is LeetCode problem 609, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 f

Detailed Explanation

The problem requires identifying duplicate files within a file system, given directory information as a list of strings. Each string represents a directory path followed by space-separated file entries in the format 'filename(content)'. The goal is to return a list of lists, where each inner list contains the full paths of files sharing the same content. A group of duplicate files must have at least two files.

Solution Approach

The provided solutions employ a hash table (dictionary/map) to group files based on their content. The algorithm iterates through the input `paths` list, parses each string to extract the directory path, filename, and content. The content is used as the key in the hash table, and the corresponding value is a list of full file paths that share that content. Finally, the algorithm iterates through the hash table and returns a list of lists containing file paths for any content that has more than one file associated with it (duplicates).

Step-by-Step Algorithm

Step 1: Initialize a hash table (dictionary/map) `content_map` to store file content as keys and a list of corresponding file paths as values.
Step 2: Iterate through each string `path_info` in the input list `paths`.
Step 3: Split each `path_info` string by spaces to separate the directory path from the file entries.
Step 4: Extract the directory path from the first element of the split string.
Step 5: Iterate through the file entries (remaining elements after the directory path).
Step 6: For each file entry, extract the filename and content by splitting the string at the '(' character or using `rpartition`.
Step 7: Construct the full file path by concatenating the directory path, '/', and the filename.
Step 8: Add the full file path to the list associated with the content in the `content_map`. If the content doesn't exist as a key, create a new entry with the content as the key and a new list containing the file path as the value.
Step 9: After processing all input strings, iterate through the `content_map`.
Step 10: For each entry in the `content_map`, if the list of file paths has more than one element, it means there are duplicate files with the same content. Add this list to the result list.
Step 11: Return the result list, which contains lists of duplicate file paths.
Step 12 (C Specific): Handle memory allocation and deallocation for each string created to avoid memory leaks.

Key Insights

Insight 1: The core task is to group files by their content. Therefore, a hash table (dictionary or map) is the most suitable data structure to store content as keys and a list of file paths as values.
Insight 2: The input string needs to be parsed effectively to extract the directory path, filename, and content. String manipulation techniques like splitting and finding substrings are essential.
Insight 3: The problem's constraints specify the formats and ranges of input data, guiding the choice of appropriate data structures and algorithms. Handling memory allocation and deallocation properly is critical in languages like C to avoid memory leaks.

Complexity Analysis

Time Complexity: O(N*L)

Space Complexity: O(N*L)

Topics

This problem involves: Array, Hash Table, String.

Companies

Asked at: Applied Intuition, Dropbox, Turing.

Find Duplicate File in System | LeetCode 609

Medium

Array Hash Table String

Time: O(N*L)Space: O(N*L)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public List<List<String>> findDuplicate(String[] paths) {
        Map<String, List<String>> contentMap = new HashMap<>();

        for (String pathInfo : paths) {
            String[] parts = pathInfo.split(" ");
            String directoryPath = parts[0];
            for (int i = 1; i < parts.length; i++) {
                String fileAndContent = parts[i];
                int lastIndexOfOpenParen = fileAndContent.lastIndexOf('(');
                String fileName = fileAndContent.substring(0, lastIndexOfOpenParen);
                String contentWithParen = fileAndContent.substring(lastIndexOfOpenParen);
                String content = contentWithParen.substring(1, contentWithParen.length() - 1);

                String fullPath = directoryPath + "/" + fileName;
                contentMap.computeIfAbsent(content, k -> new ArrayList<>()).add(fullPath);
            }
        }

        List<List<String>> result = new ArrayList<>();
        for (List<String> pathsList : contentMap.values()) {
            if (pathsList.size() > 1) {
                result.add(pathsList);
            }
        }

        return result;
    }
}

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Problem Info

Description

Problem 609: Find Duplicate File in System - Detailed explanation and solution approaches.

Topics

Array Hash Table String

Companies

Applied IntuitionDropboxTuring

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:609

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

12 steps

1
Step 1: Initialize a hash table (dictionary/map) `content_map` to store file content as keys and a list of corresponding file paths as values.
2
Step 2: Iterate through each string `path_info` in the input list `paths`.
3
Step 3: Split each `path_info` string by spaces to separate the directory path from the file entries.
4
Step 4: Extract the directory path from the first element of the split string.
5
Step 5: Iterate through the file entries (remaining elements after the directory path).
6
Step 6: For each file entry, extract the filename and content by splitting the string at the '(' character or using `rpartition`.
7
Step 7: Construct the full file path by concatenating the directory path, '/', and the filename.
8
Step 8: Add the full file path to the list associated with the content in the `content_map`. If the content doesn't exist as a key, create a new entry with the content as the key and a new list containing the file path as the value.
9
Step 9: After processing all input strings, iterate through the `content_map`.
10
Step 10: For each entry in the `content_map`, if the list of file paths has more than one element, it means there are duplicate files with the same content. Add this list to the result list.
11
Step 11: Return the result list, which contains lists of duplicate file paths.
12
Step 12 (C Specific): Handle memory allocation and deallocation for each string created to avoid memory leaks.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(N*L)

💾

Space Complexity

O(N*L)

📝

Why?

The time complexity arises from iterating through each path and then processing each file entry within the path, which involves string splitting and manipulation. The space complexity comes from storing the content and the full file paths in the hash table. The worst-case scenario happens when all files have distinct content, causing the hash table to store every file's path.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using File Hashing for Large Files

Instead of storing the entire file content, compute a hash (e.g., MD5, SHA-256) of the file's content and use the hash as the key in the hash table. This is especially useful for very large files, where storing the entire content in memory might be prohibitive.

Time: O(N*L + M*H), where N is the number of paths, L is the average path length, M is the number of files, and H is the average time to compute the hash of a file. Assuming H is significantly less than L, and the average length is short relative to number of files, then it could be simplified to O(M), number of total files.Space: O(M*K), where M is the number of files, and K is the size of the hash (e.g., 16 bytes for MD5).

Trade-offs: This approach is beneficial when dealing with very large files as it reduces memory usage. The downside is the added overhead of computing the hash for each file and the risk of hash collisions (although the probability is low with good hash functions).

Comparison using chunking for extremely large files

If the files are too large to even hash at once, they can be compared using small chunks. Only if the first chunk matches, the next one will be checked. If at any point, there is a mismatch, the program will skip that file. This would allow checking huge files without loading the entire content in RAM.

Time: O(N * C), N is the number of files, and C is the cost of chunk comparison. C will depend on chunk size and the point where the files differ.Space: O(chunk_size), The space complexity will depend on the chunk size, that is being stored in the RAM.

Trade-offs: Benefit is to check very large files, but time complexity depends on chunk size, frequency and location of mismatches.

✨

Key Insights

1Insight 1: The core task is to group files by their content. Therefore, a hash table (dictionary or map) is the most suitable data structure to store content as keys and a list of file paths as values.
2Insight 2: The input string needs to be parsed effectively to extract the directory path, filename, and content. String manipulation techniques like splitting and finding substrings are essential.
3Insight 3: The problem's constraints specify the formats and ranges of input data, guiding the choice of appropriate data structures and algorithms. Handling memory allocation and deallocation properly is critical in languages like C to avoid memory leaks.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty input list `paths`. The solution should return an empty list of lists, which the given code correctly handles as the loop will not execute.
Edge case 2: Paths with no duplicate files. The solution correctly identifies that the list of file paths for each unique content is of size 1, and these are not included in the final result.
Edge case 3: A directory path contains spaces. The splitting logic accounts for spaces within directory names.
Edge case 4: A file contains no content (empty content). The algorithm correctly extracts and handles this by creating an entry in the hashmap with empty content as the key.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly parsing the input string and failing to extract the filename and content accurately. Using the wrong delimiters or string manipulation methods can lead to errors.

✗Common mistake 2: Failing to handle the case where there are no duplicate files. The solution should still function correctly and return an empty list.

✗Common mistake 3: Not considering memory management in C. Failing to free allocated memory can result in memory leaks, especially with string duplication and dynamic array resizing.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Identifying duplicate files on a computer system or network drive to reclaim storage space. Utility programs often use this algorithm, sometimes incorporating file hashing, to locate redundant data.

🏦Real-world scenario 2: Detecting plagiarism in a collection of documents. By comparing the content of documents, plagiarism detection software can identify instances where text has been copied from other sources.

🎮Real-world scenario 3: Data deduplication in cloud storage services. Cloud providers use data deduplication techniques to avoid storing multiple copies of the same data, which can save significant storage costs.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 9, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Find Duplicate File in System"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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