802. Find Eventual Safe States

Problem

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

/**
 * @param {number[][]} graph
 * @return {number[]}
 */
var eventualSafeNodes = function(graph) {
    var map = Array(graph.length);
    var path = Array(graph.length);
    var res = [];
    for (var i = 0; i < graph.length; i++) {
        if (isSafeNode(i, graph, map, path)) {
            res.push(i);
        }
    }
    return res;
};

var isSafeNode = function(i, graph, map, path) {
    if (graph[i].length === 0 || map[i] === 1) return true;
    if (map[i] === 2 || path[i] === 1) return false;
    path[i] = 1;
    for (var j = 0; j < graph[i].length; j++) {
        var index = graph[i][j];
        if (!isSafeNode(index, graph, map, path)) {
            path[i] = 0;
            map[i] = 2;
            return false;
        }
    }
    path[i] = 0;
    map[i] = 1;
    return true;
};

Explain:

DFS (Depth First Search).

Complexity:

n nodes, m edges.

• Time complexity : O(n + m).
• Space complexity : O(n + m).