820. Find Eventual Safe States¶

820. 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 (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:

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¶

class Solution {
    public List<Integer> eventualSafeNodes(int[][] graph) {
        int n = graph.length;
        int m = graph[0].length;
        ArrayList<ArrayList<Integer>> adj = new ArrayList<>();
        for(int i = 0; i <= n + 1; i++) adj.add(new  ArrayList<>());
        for(int i = 0; i < graph.length; i++) {
            for(int j = 0; j < graph[i].length; j++) {
               adj.get(graph[i][j]).add(i);
            }
        }
        int indegree[] = new int[n + 1];
        for(ArrayList<Integer> current : adj) {
            for(int ele : current) indegree[ele]++;
        }
        List<Integer> ans = new ArrayList<>();
        ans = solve(n,adj,indegree);
        Collections.sort(ans);
        return ans;
    }
    private static List<Integer> solve(int n , ArrayList<ArrayList<Integer>> adj , int indegree[]) {
        Queue<Integer> q = new LinkedList<>();
        for(int i = 0; i < n; i++) {
            if(indegree[i] == 0) q.offer(i);
        }
        List<Integer> ans = new ArrayList<>();
        while(!q.isEmpty()) {
            int current = q.peek();
            q.poll();
            ans.add(current);
            for(int child : adj.get(current)) {
                indegree[child]--;
                if(indegree[child] == 0) q.offer(child);
            }
        }
        return ans;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Explanation¶

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