#2858

Hard

auto_awesomeGraph traversal with depth-first search

LeetCode Problem Workspace

Minimum Edge Reversals So Every Node Is Reachable

This problem requires solving a graph traversal with edge reversals to ensure every node is reachable in a tree-like structure.

dynamic programming depth first search breadth first search graph

Problem Statement

You are given a directed graph with n nodes labeled from 0 to n - 1. The graph is such that if its edges were bidirectional, it would form a tree. Each edge is represented by a pair [ui, vi], meaning there is a directed edge from node ui to node vi.

Your task is to determine the minimum number of edge reversals required to make every node in the graph reachable from any other node. The output should be an array of size n, where each element represents the minimum edge reversals needed for that node.

Examples

Example 1

Input: n = 4, edges = [[2,0],[2,1],[1,3]]

Output: [1,1,0,2]

The image above shows the graph formed by the edges. For node 0: after reversing the edge [2,0], it is possible to reach any other node starting from node 0. So, answer[0] = 1. For node 1: after reversing the edge [2,1], it is possible to reach any other node starting from node 1. So, answer[1] = 1. For node 2: it is already possible to reach any other node starting from node 2. So, answer[2] = 0. For node 3: after reversing the edges [1,3] and [2,1], it is possible to reach any other node starting from node 3. So, answer[3] = 2.

Example 2

Input: n = 3, edges = [[1,2],[2,0]]

Output: [2,0,1]

The image above shows the graph formed by the edges. For node 0: after reversing the edges [2,0] and [1,2], it is possible to reach any other node starting from node 0. So, answer[0] = 2. For node 1: it is already possible to reach any other node starting from node 1. So, answer[1] = 0. For node 2: after reversing the edge [1, 2], it is possible to reach any other node starting from node 2. So, answer[2] = 1.

Constraints

2 <= n <= 105
edges.length == n - 1
edges[i].length == 2
0 <= ui == edges[i][0] < n
0 <= vi == edges[i][1] < n
ui != vi
The input is generated such that if the edges were bi-directional, the graph would be a tree.

Solution Approach

Graph Traversal with DFS

The problem can be approached by performing Depth-First Search (DFS) from each node, checking if a path exists from the node to all other nodes. For nodes that are not directly reachable, edge reversals are needed to establish the connection.

Dynamic Programming on Trees

By leveraging tree DP, we can efficiently calculate the minimum reversals required for each node. This involves dynamically updating the reversals based on traversals and keeping track of reversals needed to reach each node.

Breadth-First Search for Optimizing Traversals

Breadth-First Search (BFS) can be used to further optimize the solution by calculating the minimum reversals in a level-order manner, ensuring the solution is scalable for larger inputs.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the graph traversal and the approach used for dynamic programming. A DFS traversal has a time complexity of O(n), and applying DP could add another O(n) for each node. The space complexity is similarly tied to the storage of the graph and DP arrays, which would require O(n).

What Interviewers Usually Probe

Candidates should demonstrate an understanding of graph traversal techniques, especially DFS and BFS.
Look for candidates who can break down the problem into manageable subproblems, particularly using dynamic programming on trees.
Watch for efficient solutions that scale well, given the problem's constraints of up to 10^5 nodes.

Common Pitfalls or Variants

Common pitfalls

Failing to handle cases where reversing an edge is necessary for connectivity between nodes.
Overcomplicating the problem by using more complex algorithms than necessary for tree-based graphs.
Not considering the optimal way to track and minimize edge reversals during DFS traversal.

Follow-up variants

What if the graph were not a tree, but still had some cycles? How would this affect the approach?
How would the solution change if the edges could be reversed multiple times, rather than just once?
What happens if the graph is initially a fully connected directed graph? Can the solution be simplified?

FAQ

What is the primary algorithmic technique used in 'Minimum Edge Reversals So Every Node Is Reachable'?

The primary technique used is graph traversal with Depth-First Search (DFS) combined with dynamic programming to minimize edge reversals.

How does dynamic programming apply to this problem?

Dynamic programming is used to calculate the minimum reversals for each node, storing intermediate results to avoid redundant calculations.

What is the time complexity of solving this problem?

The time complexity depends on the graph traversal and DP implementation, typically O(n) for DFS and additional O(n) for DP updates.

How does GhostInterview assist in solving graph traversal problems?

GhostInterview helps by explaining step-by-step approaches to DFS and DP techniques, providing code templates and offering insights into edge reversal strategies.

What makes this problem difficult?

The difficulty lies in efficiently managing the edge reversals and ensuring that each node can reach all other nodes with the minimum number of reversals.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def minEdgeReversals(self, n: int, edges: List[List[int]]) -> List[int]:
        ans = [0] * n
        g = [[] for _ in range(n)]
        for x, y in edges:
            g[x].append((y, 1))
            g[y].append((x, -1))

        def dfs(i: int, fa: int):
            for j, k in g[i]:
                if j != fa:
                    ans[0] += int(k < 0)
                    dfs(j, i)

        dfs(0, -1)

        def dfs2(i: int, fa: int):
            for j, k in g[i]:
                if j != fa:
                    ans[j] = ans[i] + k
                    dfs2(j, i)

        dfs2(0, -1)
        return ans

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