1233. Remove Sub-Folders from the Filesystem - Detailed Explanation

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Problem Statement

You are given a list of folder paths from a virtual filesystem. Your task is to remove all sub-folders and return only the root folders. A folder /a/b is considered a sub-folder if there exists a folder /a in the list.

Constraints

(1 \leq folders.length \leq 10^4)
(2 \leq folders[i].length \leq 100)
All folder paths are unique.

Examples

Example 1

Input

folders = ["/a","/a/b","/c/d","/c/d/e","/c/f"]

Output

["/a", "/c/d", "/c/f"]

Explanation

/a/b is a sub-folder of /a, so it is removed.
/c/d/e is a sub-folder of /c/d, so it is removed.
/c/f is not a sub-folder of any other folder, so it remains.

Example 2

Input

folders = ["/a","/a/b/c","/a/b/d"]

Output

["/a"]

Explanation

/a/b/c and /a/b/d are sub-folders of /a, so they are removed.

Example 3

Input

folders = ["/a/b/c","/a/b/ca","/a/b/d"]

Output

["/a/b/c", "/a/b/ca", "/a/b/d"]

Explanation

No folder is a sub-folder of another.

Hints

Sorting the folder list lexicographically might help in grouping sub-folders under parent folders.
Using a Trie or Set can help in efficiently filtering sub-folders.

Approach 1: Brute Force (Checking Each Folder Against All Others)

Idea

For each folder, check if it is a sub-folder of any other folder in the list.
A folder /a/b is a sub-folder of /a if /a/ is a prefix of /a/b.

Algorithm

Iterate through each folder and check if it starts with any other folder in the list.
If it is a sub-folder, discard it; otherwise, keep it in the result list.

Time Complexity

O(N²) in the worst case, where N is the number of folders.

Python Implementation

Python3

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Java Implementation

Java

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Approach 2: Sorting + Greedy Approach (Optimized)

Idea

Sorting the list lexicographically ensures that a parent folder always appears before its sub-folders.
Using a single pass, we can check if a folder is a sub-folder of the previous folder.

Algorithm

Sort the folder list lexicographically.
Use a single pass to check if each folder starts with the last added folder (result[-1] + "/").
If yes, it's a sub-folder, so we skip it; otherwise, we add it to the result list.

Time Complexity

O(N log N) due to sorting.
O(N) for scanning through the folders.
Overall Complexity: O(N log N).

Python Implementation

Python3

. . . .

Java Implementation

Java

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Approach 3: Using Trie (Efficient Approach)

Idea

Use a Trie (prefix tree) to store folder paths.
Only insert a folder into the result list if it’s not inside an existing parent folder in the Trie.

Algorithm

Sort the folder paths.
Insert each folder into the Trie.
While inserting, stop if a parent folder is already present in the Trie.
Keep only top-level folders in the result list.

Time Complexity

O(N log N) for sorting.
O(M) for inserting and checking folders, where M is the total number of characters in all folder paths.
Overall Complexity: O(N log N + M).

Python Implementation

Python3

. . . .

Java Implementation

Java

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Complexity Analysis for Each Approach

Approach	Sorting Complexity	Checking Complexity	Total Complexity
Brute Force	-	(O(N^2)) (Nested loop checking)	(O(N^2))
Sorting + Greedy	(O(N \log N))	(O(N)) (Single pass filtering)	(O(N \log N))
Trie-Based Approach	(O(N \log N))	(O(M)) (where M is total characters in all paths)	(O(N \log N + M))

Brute Force is inefficient for large inputs.
Sorting + Greedy is optimal for practical use cases.
Trie-Based Approach is useful when we frequently insert/update folders dynamically.

Step-by-Step Walkthroughs with Visual Examples

Example Input

folders = ["/a","/a/b","/c/d","/c/d/e","/c/f"]

Step 1: Lexicographic Sorting

Index	Folder
1	`/a`
2	`/a/b`
3	`/c/d`
4	`/c/d/e`
5	`/c/f`

Step 2: Iteration (Sorting + Greedy Approach)

Index	Folder	Last Added Folder	Is Sub-folder?	Result
1	`/a`	-	No	[`/a`]
2	`/a/b`	`/a`	Yes	[`/a`]
3	`/c/d`	`/a`	No	[`/a`, `/c/d`]
4	`/c/d/e`	`/c/d`	Yes	[`/a`, `/c/d`]
5	`/c/f`	`/c/d`	No	[`/a`, `/c/d`, `/c/f`]

Trie Insertion Step-by-Step

Insert /a → Root → a (end=True).
Insert /a/b → Stops at a, as end=True.
Insert /c/d → Root → c → d (end=True).
Insert /c/d/e → Stops at c/d, as end=True.
Insert /c/f → Root → c → f (end=True).

Final Trie Structure:

Root
├── a (end=True)
│   └── b
└── c
    ├── d (end=True)
    │   └── e
    └── f (end=True)

Final Output: ["/a", "/c/d", "/c/f"]

Common Mistakes

Not Sorting the Input
- Sorting ensures parents appear before sub-folders.
- Example: ["/c/d", "/c/d/e", "/a", "/a/b"] fails without sorting.
Using StartsWith() Incorrectly
- folder.startswith(parent) might incorrectly detect /a/b1 as a sub-folder of /a/b.
Ignoring Edge Cases
- Duplicate paths: Ensure unique folder paths.
- Deeply nested folders: Handle up to length 100.

Edge Cases

Scenario	Example Input	Expected Output
All folders are root	`["/a", "/b", "/c"]`	`["/a", "/b", "/c"]`
Deep nesting	`["/a", "/a/b", "/a/b/c/d/e"]`	`["/a"]`
Long paths	`["/a" + "/b" * 50]`	`["/a" + "/b" * 50]`
No sub-folders	`["/a/b/c", "/a/b/d", "/x/y/z"]`	`["/a/b/c", "/a/b/d", "/x/y/z"]`

Alternative Variations

Finding Common Parent Paths
- Input: ["/a/b/c", "/a/b", "/a/b/d"]
- Output: ["/a/b"]
- Variation: Find the shortest common path.
Handling Files Alongside Folders
- Input: ["/a/b/file.txt", "/a/b/c", "/a/b/d"]
- Output: ["/a/b/file.txt", "/a/b"]
Dynamic Insertion & Querying
- Instead of a static list, allow dynamic insertion of folders.
- Useful in file-system-like implementations.

Related Problems for Further Practice

Problem	Description
[Leetcode 208: Implement Trie (Prefix Tree)]	Learn Trie implementation for prefix searching.
[Leetcode 648: Replace Words]	Use a Trie to replace words efficiently.
[Leetcode 1268: Search Suggestions System]	Auto-complete using Trie and sorting.
[Leetcode 211: Design Add and Search Words Data Structure]	Extend Trie to handle regex-like queries.

Final Thoughts

Sorting + Greedy is the best approach for practical use.
Trie-based approach is great when folders are inserted dynamically.
Edge cases like deeply nested paths should be handled carefully.