Leetcode #1233: Remove Sub-Folders from the Filesystem
In this guide, we solve Leetcode #1233 Remove Sub-Folders from the Filesystem in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a list of folders folder, return the folders after removing all sub-folders in those folders. You may return the answer in any order.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Depth-First Search, Trie, Array, String
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: folder = ["/a","/a/b","/c/d","/c/d/e","/c/f"]
Output: ["/a","/c/d","/c/f"]
Explanation: Folders "/a/b" is a subfolder of "/a" and "/c/d/e" is inside of folder "/c/d" in our filesystem.
Python Solution
class Solution:
def removeSubfolders(self, folder: List[str]) -> List[str]:
folder.sort()
ans = [folder[0]]
for f in folder[1:]:
m, n = len(ans[-1]), len(f)
if m >= n or not (ans[-1] == f[:m] and f[m] == '/'):
ans.append(f)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.