How do you solve Remove Sub-Folders from the Filesystem on LeetCode?

Remove Sub-Folders from the Filesystem (LeetCode #1233) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting the folders helps in efficiently identifying sub-folders.

What is the time complexity of Remove Sub-Folders from the Filesystem?

The optimal solution for Remove Sub-Folders from the Filesystem runs in O(n log n) time and uses O(n) space. The sorting step is O(n log n), and we traverse the sorted list once, leading to a linear pass afterward.

What is the brute force solution for Remove Sub-Folders from the Filesystem?

The brute force approach for Remove Sub-Folders from the Filesystem is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each folder against all others to see if it is a sub-folder. This is straightforward but inefficient, especially with a larger number of folders. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Remove Sub-Folders from the Filesystem?

Remove Sub-Folders from the Filesystem uses the following concepts: Array, String, Depth-First Search, Trie. The recommended patterns to study are: Sorting, Greedy.

What are the interview tips for Remove Sub-Folders from the Filesystem?

Always consider sorting when dealing with hierarchical data. Think about edge cases, such as folders that are exactly the same except for the last part. Practice explaining your thought process while coding.

What are common mistakes when solving Remove Sub-Folders from the Filesystem?

Not sorting the folders before checking for sub-folders. Confusing the conditions for identifying sub-folders.

#1233

Remove Sub-Folders from the Filesystem

Medium

ArrayStringDepth-First SearchTrieSortingGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

By sorting the folders first, we can efficiently check for sub-folders in a single pass. This way, we only need to compare each folder with the last added folder in the result.

⚙️

Algorithm

3 steps

1Step 1: Sort the folder list lexicographically.
2Step 2: Initialize an empty list to store the result.
3Step 3: Iterate through the sorted list and add a folder to the result if it does not start with the last added folder followed by a '/'.

solution.py7 lines

1def removeSubfolders(folder):
2    folder.sort()
3    result = []
4    for f in folder:
5        if not result or not f.startswith(result[-1] + '/'): 
6            result.append(f)
7    return result

ℹ

Complexity note: The sorting step is O(n log n), and we traverse the sorted list once, leading to a linear pass afterward.

1Sorting the folders helps in efficiently identifying sub-folders.
2Using a single pass after sorting reduces the time complexity significantly.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.