p145.java

/*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 (or another safe node).

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

 

Example 1:

Illustration of graph
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].*/
class Solution {
    public List<Integer> eventualSafeNodes(int[][] graph) {
        List<Integer> safe=new ArrayList<>();
        int V=graph.length;
        boolean vis[]=new boolean[V];
        boolean rec[]=new boolean[V];
        boolean cyc[]=new boolean[V];
        for(int i=0;i<V;i++){
            if(vis[i]==false){
                cycle(graph,vis,i,rec,cyc);
            }
        }
        for(int i=0;i<V;i++){
            if(cyc[i]==false){
                safe.add(i);
            }
        }
        return safe;
    }
    public static boolean cycle(int[][] graph,boolean vis[],int curr,boolean rec[],boolean cyc[]){
        vis[curr]=true;
        rec[curr]=true;
        for(int i:graph[curr]){
            if(rec[i]==true){
                cyc[curr]=true;
                return true;
            }else if(vis[i]==false && cycle(graph,vis,i,rec,cyc)){
                cyc[curr]=true;
                return true;
            }
        }
        rec[curr]=false;
        return false;
    }
}