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Minimum Edge Reversals So Every Node Is Reachable

Difficulty: Hard

Problem Description

Given a simple directed graph with n nodes and a list of directed edges, determine the minimum number of edge reversals required for every node i to reach all other nodes in the graph. Return an array where each element represents the minimum reversals needed for the corresponding node.

Key Insights

  • The graph represents a tree when edges are bidirectional.
  • Each edge can either be left as is or reversed, affecting reachability.
  • The problem can be approached using graph traversal techniques to explore paths and calculate reversals.
  • Since the graph is a directed tree, the number of edges is n - 1, which allows for efficient traversal.

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(n)

Solution

To solve the problem, we can perform a Depth-First Search (DFS) or Breadth-First Search (BFS) from each node to calculate the reversals needed. The approach involves:

  1. Constructing a graph representation using an adjacency list.
  2. For each node, traversing the graph to count the number of reversals required to reach all other nodes.
  3. Keeping track of visited nodes to avoid redundant calculations.

We'll utilize a modified DFS that counts reversals when navigating through the directed edges and considers the reversals necessary for edges not pointing in the desired direction.

Code Solutions

def min_edge_reversals(n, edges):
    from collections import defaultdict, deque
    
    graph = defaultdict(list)
    reverse_graph = defaultdict(list)
    
    for u, v in edges:
        graph[u].append(v)
        reverse_graph[v].append(u)
    
    def bfs(start):
        queue = deque([start])
        visited = [False] * n
        visited[start] = True
        reversals = 0
        while queue:
            node = queue.popleft()
            for neighbor in graph[node]:
                if not visited[neighbor]:
                    visited[neighbor] = True
                    queue.append(neighbor)
            for neighbor in reverse_graph[node]:
                if not visited[neighbor]:
                    reversals += 1  # counting the reversal
                    visited[neighbor] = True
                    queue.append(neighbor)
        return reversals
    
    result = []
    for i in range(n):
        result.append(bfs(i))
    
    return result
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