{leetcode}/problems/minimum-edge-reversals-so-every-node-is-reachable/[LeetCode - 2858. Minimum Edge Reversals So Every Node Is Reachable ^]
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] = [u<sub>i</sub>, v<sub>i</sub>] represents a directed edge going from node u<sub>i</sub> to node v<sub>i</sub>.
An edge reversal changes the direction of an edge, i.e., a directed edge going from node u<sub>i</sub> to node v<sub>i</sub> becomes a directed edge going from node v<sub>i</sub> to node u<sub>i</sub>.
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:
<img height="246" src="https://assets.leetcode.com/uploads/2023/08/26/image-20230826221104-3.png" width="312" />
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:
<img height="217" src="https://assets.leetcode.com/uploads/2023/08/26/image-20230826225541-2.png" width="322" />
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 ⇐ u<sub>i</sub> == edges[i][0] < n -
0 ⇐ v<sub>i</sub> == edges[i][1] < n -
u<sub>i</sub> != v<sub>i</sub> -
The input is generated such that if the edges were bi-directional, the graph would be a tree.