What is the time complexity of Reorder Routes to Make All Paths Lead to the City Zero?

The optimal solution for Reorder Routes to Make All Paths Lead to the City Zero runs in O(n) time and uses O(n) space. The complexity is linear because we visit each city and road exactly once during the DFS traversal.

What is the brute force solution for Reorder Routes to Make All Paths Lead to the City Zero?

The brute force approach for Reorder Routes to Make All Paths Lead to the City Zero is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each road and determining if it needs to be reversed to ensure all cities can reach city 0. This method is straightforward but inefficient as it examines all possible paths. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Reorder Routes to Make All Paths Lead to the City Zero?

Reorder Routes to Make All Paths Lead to the City Zero uses the following concepts: Depth-First Search, Breadth-First Search, Graph Theory. The recommended patterns to study are: Graph Traversal, Depth-First Search, Tree Structures.

What are the interview tips for Reorder Routes to Make All Paths Lead to the City Zero?

Clarify the problem requirements before coding. Think about edge cases, such as minimal inputs. Discuss your thought process with the interviewer to showcase your understanding.

What are common mistakes when solving Reorder Routes to Make All Paths Lead to the City Zero?

Not considering the direction of roads correctly. Failing to mark cities as visited, leading to infinite loops.

#1466

Reorder Routes to Make All Paths Lead to the City Zero

Medium

Depth-First SearchBreadth-First SearchGraph TheoryGraph TraversalDepth-First SearchTree Structures

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses Depth-First Search (DFS) to traverse the graph while counting the number of roads that need to be reversed. This method is efficient because it explores each road only once.

⚙️

Algorithm

3 steps

1Step 1: Build a graph representation of the cities and roads, marking each road with a flag indicating if it needs to be reversed.
2Step 2: Perform a DFS starting from city 0, counting the roads that need to be reversed as you traverse.
3Step 3: Return the total count of roads that need to be reversed.

solution.py22 lines

1# Full working Python code
2from collections import defaultdict
3
4def minReorder(n, connections):
5    graph = defaultdict(list)
6    for a, b in connections:
7        graph[a].append((b, 1))  # 1 indicates a road that needs to be reversed
8        graph[b].append((a, 0))  # 0 indicates a road that is fine
9
10    count = 0
11    visited = set()
12
13    def dfs(city):
14        nonlocal count
15        visited.add(city)
16        for neighbor, needs_reverse in graph[city]:
17            if neighbor not in visited:
18                count += needs_reverse
19                dfs(neighbor)
20
21    dfs(0)
22    return count

ℹ

Complexity note: The complexity is linear because we visit each city and road exactly once during the DFS traversal.

1Understanding the direction of roads is crucial.
2Graph traversal techniques like DFS are effective for tree structures.

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