How do you solve Minimum Edge Reversals So Every Node Is Reachable on LeetCode?

Minimum Edge Reversals So Every Node Is Reachable (LeetCode #2858) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the direction of edges is crucial for calculating reversals.

What is the time complexity of Minimum Edge Reversals So Every Node Is Reachable?

The optimal solution for Minimum Edge Reversals So Every Node Is Reachable runs in O(n) time and uses O(n) space. This complexity is due to the single DFS traversal through the graph, where each node and edge is processed once.

What is the brute force solution for Minimum Edge Reversals So Every Node Is Reachable?

The brute force approach for Minimum Edge Reversals So Every Node Is Reachable is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible starting node and exploring all paths to see how many edge reversals are needed to reach every other node. This method is straightforward but inefficient for larger graphs. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Minimum Edge Reversals So Every Node Is Reachable?

Always clarify the problem statement and constraints before diving into coding. Think about the data structures you will use to represent the graph effectively. Practice explaining your thought process as you code, as this helps in interviews.

What are common mistakes when solving Minimum Edge Reversals So Every Node Is Reachable?

Failing to account for already visited nodes can lead to incorrect counts of reversals. Not considering edge cases where nodes may already have paths to others without reversals.

#2858

Minimum Edge Reversals So Every Node Is Reachable

Hard

Dynamic ProgrammingDepth-First SearchBreadth-First SearchGraph TheoryGraph TraversalDynamic Programming

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 approach utilizes a tree dynamic programming strategy, treating the graph as a tree and calculating reversals in a bottom-up manner. This allows us to efficiently compute the minimum reversals needed for each node to reach all others.

⚙️

Algorithm

3 steps

1Step 1: Build a directed graph from the edges, noting the direction of each edge.
2Step 2: Use a DFS to calculate the minimum reversals needed for each node by considering its children.
3Step 3: Store the results in an array and return it.

solution.py20 lines

1# Full working Python code
2from collections import defaultdict
3
4def minEdgeReversals(n, edges):
5    graph = defaultdict(list)
6    for u, v in edges:
7        graph[u].append((v, 0))  # original direction
8        graph[v].append((u, 1))  # reversed direction
9
10    def dfs(node, parent):
11        total_reversals = 0
12        for neighbor, cost in graph[node]:
13            if neighbor != parent:
14                total_reversals += cost + dfs(neighbor, node)
15        return total_reversals
16
17    result = [0] * n
18    for i in range(n):
19        result[i] = dfs(i, -1)
20    return result

ℹ

Complexity note: This complexity is due to the single DFS traversal through the graph, where each node and edge is processed once.

1Understanding the direction of edges is crucial for calculating reversals.
2Using DFS allows us to explore the graph efficiently and compute reversals in a structured manner.

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