LeetCode-in-Java

802. Find Eventual Safe States

Medium

There is a directed graph of n nodes with each node labeled from 0 to n - 1. The graph is represented by a 0-indexed 2D integer array graph where graph[i] is an integer array of nodes adjacent to node i, meaning there is an edge from node i to each node in graph[i].

A node is a terminal node if there are no outgoing edges. A node is a safe node if every possible path starting from that node leads to a terminal node.

Return an array containing all the safe nodes of the graph. The answer should be sorted in ascending order.

Example 1:

Input: graph = [[1,2],[2,3],[5],[0],[5],[],[]]

Output: [2,4,5,6]

Explanation: The given graph is shown above.

Nodes 5 and 6 are terminal nodes as there are no outgoing edges from either of them.

Every path starting at nodes 2, 4, 5, and 6 all lead to either node 5 or 6.

Example 2:

Input: graph = [[1,2,3,4],[1,2],[3,4],[0,4],[]]

Output: [4]

Explanation: Only node 4 is a terminal node, and every path starting at node 4 leads to node 4.

Constraints:

• n == graph.length
• 1 <= n <= 104
• 0 <= graph[i].length <= n
• 0 <= graph[i][j] <= n - 1
• graph[i] is sorted in a strictly increasing order.
• The graph may contain self-loops.
• The number of edges in the graph will be in the range [1, 4 * 104].

Solution

import java.util.ArrayList;
import java.util.List;

public class Solution {
    public List<Integer> eventualSafeNodes(int[][] graph) {
        List<Integer> res = new ArrayList<>();
        int[] vis = new int[graph.length];
        for (int i = 0; i < graph.length; i++) {
            if (dfs(graph, i, vis)) {
                res.add(i);
            }
        }
        return res;
    }

    private boolean dfs(int[][] graph, int src, int[] vis) {
        if (vis[src] != 0) {
            return vis[src] == 2;
        }
        vis[src] = 1;
        for (int x : graph[src]) {
            if (!dfs(graph, x, vis)) {
                return false;
            }
        }
        vis[src] = 2;
        return true;
    }
}

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