Package g2801_2900.s2858_minimum_edge_reversals_so_every_node_is_reachable

Class Solution

java.lang.Object
g2801_2900.s2858_minimum_edge_reversals_so_every_node_is_reachable.Solution
public class Solution extends java.lang.Object
2858 - Minimum Edge Reversals So Every Node Is Reachable.

Hard

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.

  • Constructor Summary

    Constructors
    Constructor
    Description
    Solution()
     

  • Method Summary

    Modifier and Type
    Method
    Description
    int[]
    minEdgeReversals(int n, int[][] edges)
     

    Methods inherited from class java.lang.Object

    clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • minEdgeReversals

      public int[] minEdgeReversals(int n, int[][] edges)