802. Find Eventual Safe States

Description

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].

 

Solutions

Solution: Depth-First Search

• Time complexity: O(n+E)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[][]} graph
 * @return {number[]}
 */
const eventualSafeNodes = function (graph) {
  const n = graph.length;
  const SAFE = 0;
  const CYCLE = 1;
  const visited = Array.from({ length: n }, () => false);
  const states = Array.from({ length: n }, () => -1);
  const result = [];

  const isSafeNode = node => {
    if (visited[node]) return false;
    if (states[node] !== -1) return states[node] === SAFE;

    visited[node] = true;

    for (const neighbor of graph[node]) {
      states[neighbor] = isSafeNode(neighbor) ? SAFE : CYCLE;

      if (states[neighbor] === CYCLE) return false;
    }
    visited[node] = false;
    states[node] = SAFE;

    return true;
  };

  for (let node = 0; node < n; node++) {
    if (isSafeNode(node)) result.push(node);
  }
  return result;
};