Find Eventual Safe States

Question

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 <= 10^4
• 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 * 10^4].

Solution

Time Complexity: In the worst case, you may visit each node once, and for each node, you might traverse all its edges once. Hence, the overall time complexity is O(V + E)

Space Complexity:

1. State Array:
   • The state array takes O(V) space, where V is the number of vertices.
2. Call Stack:
   • The recursion stack in DFS can go as deep as the number of nodes in the worst case, so the space required for the stack is O(V).
3. Result List:
   • The result list will store the safe nodes, which can take up to O(V) space. The overall space complexity is O(V), which accounts for the state array and the call stack.

class Solution {
    public List<Integer> eventualSafeNodes(int[][] graph) {
        int n = graph.length;
        int[] state = new int[n]; //0 : unvisited, 1: processing, 2: safe
        List<Integer> result = new ArrayList<>();
 
        for(int i = 0; i < graph.length; i++){
            if(dfs(graph, i, state)){
                result.add(i);
            }
        }
        return result;
    }
 
    private boolean dfs(int[][] graph, int node, int[] state){
        if(state[node] == 1){ //cycle detected
            return false;
        }
        if(state[node] == 2){ //reached safe node
            return true;
        }
 
        state[node] = 1; // mark visited
 
        for(int nextNode: graph[node]){
            if(state[nextNode] == 2){
                continue; // If the next node is already safe, skip
            }
            if(state[nextNode] == 1 || !dfs(graph, nextNode, state)){
                return false; // Cycle detected or path to an unsafe node
            }
        }
 
        state[node] = 2; // Mark as safe
        return true;
    }
}