2858. Minimum Edge Reversals So Every Node Is Reachable

Description

There is a simple directed graph with n nodes labeled from 0 to n - 1. The graph would form a tree if its edges were bi-directional.

You are given an integer n and a 2D integer array edges, where edges[i] = [ui, vi] represents a directed edge going from node ui to node vi.

An edge reversal changes the direction of an edge, i.e., a directed edge going from node ui to node vi becomes a directed edge going from node vi to node ui.

For every node i in the range [0, n - 1], your task is to independently calculate the minimum number of edge reversals required so it is possible to reach any other node starting from node i through a sequence of directed edges.

Return an integer array answer, where answer[i] is the minimum number of edge reversals required so it is possible to reach any other node starting from node i through a sequence of directed edges.

 

Example 1:

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

 

Solutions

Solution: Dynamic Programming + Depth-First Search

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number} n
 * @param {number[][]} edges
 * @return {number[]}
 */
const minEdgeReversals = function (n, edges) {
  const graph = Array.from({ length: n }, () => []);
  const result = Array.from({ length: n }, () => 0);

  for (const [u, v] of edges) {
    graph[u].push({ node: v, isForward: true });
    graph[v].push({ node: u, isForward: false });
  }

  const getMinReversals = (node, prev) => {
    let result = 0;

    for (const { node: neighbor, isForward } of graph[node]) {
      if (neighbor === prev) continue;

      result += getMinReversals(neighbor, node) + (isForward ? 0 : 1);
    }

    return result;
  };

  const dfs = (node, prev) => {
    for (const { node: neighbor, isForward } of graph[node]) {
      if (neighbor === prev) continue;

      result[neighbor] = result[node] + (isForward ? 1 : -1);
      dfs(neighbor, node);
    }
  };

  result[0] = getMinReversals(0, -1);
  dfs(0, -1);

  return result;
};