Leetcode #1948: Delete Duplicate Folders in System
In this guide, we solve Leetcode #1948 Delete Duplicate Folders in System 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
Due to a bug, there are many duplicate folders in a file system. You are given a 2D array paths, where paths[i] is an array representing an absolute path to the ith folder in the file system.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Trie, Array, Hash Table, String, Hash Function
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: paths = [["a"],["c"],["d"],["a","b"],["c","b"],["d","a"]]
Output: [["d"],["d","a"]]
Explanation: The file structure is as shown.
Folders "/a" and "/c" (and their subfolders) are marked for deletion because they both contain an empty
folder named "b".
Python Solution
class Trie:
def __init__(self):
self.children: Dict[str, "Trie"] = defaultdict(Trie)
self.deleted: bool = False
class Solution:
def deleteDuplicateFolder(self, paths: List[List[str]]) -> List[List[str]]:
root = Trie()
for path in paths:
cur = root
for name in path:
if cur.children[name] is None:
cur.children[name] = Trie()
cur = cur.children[name]
g: Dict[str, Trie] = {}
def dfs(node: Trie) -> str:
if not node.children:
return ""
subs: List[str] = []
for name, child in node.children.items():
subs.append(f"{name}({dfs(child)})")
s = "".join(sorted(subs))
if s in g:
node.deleted = g[s].deleted = True
else:
g[s] = node
return s
def dfs2(node: Trie) -> None:
if node.deleted:
return
if path:
ans.append(path[:])
for name, child in node.children.items():
path.append(name)
dfs2(child)
path.pop()
dfs(root)
ans: List[List[str]] = []
path: List[str] = []
dfs2(root)
return ans
Complexity
The time complexity is O(n). The space complexity is O(n).
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.