LeetCode-in-Java

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] = [u_i, v_i] represents a directed edge going from node u_i to node v_i.

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

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

Solution

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

@SuppressWarnings("unchecked")
public class Solution {
    public int[] minEdgeReversals(int n, int[][] edges) {
        List<int[]>[] nexts = new List[n];
        for (int i = 0; i < n; i++) {
            nexts[i] = new ArrayList<>();
        }
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            nexts[u].add(new int[] {1, v});
            nexts[v].add(new int[] {-1, u});
        }
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = -1;
        }
        res[0] = dfs(nexts, 0, -1);
        Queue<Integer> queue = new LinkedList<>();
        queue.add(0);
        while (!queue.isEmpty()) {
            Integer index = queue.remove();
            int val = res[index];
            List<int[]> next = nexts[index];
            for (int[] node : next) {
                if (res[node[1]] == -1) {
                    if (node[0] == 1) {
                        res[node[1]] = val + 1;
                    } else {
                        res[node[1]] = val - 1;
                    }
                    queue.add(node[1]);
                }
            }
        }
        return res;
    }

    private int dfs(List<int[]>[] nexts, int index, int pre) {
        int res = 0;
        List<int[]> next = nexts[index];
        for (int[] node : next) {
            if (node[1] != pre) {
                if (node[0] == -1) {
                    res++;
                }
                res += dfs(nexts, node[1], index);
            }
        }
        return res;
    }
}

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