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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:

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].

Solutions

The point with zero out-degree is safe, and if a point can only reach the safe point, then it is also safe, so the problem can be converted to topological sorting.

  • class Solution {
    private int[] color;
    private int[][] g;

    public List<Integer> eventualSafeNodes(int[][] graph) {
        int n = graph.length;
        color = new int[n];
        g = graph;
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            if (dfs(i)) {
                ans.add(i);
            }
        }
        return ans;
    }

    private boolean dfs(int i) {
        if (color[i] > 0) {
            return color[i] == 2;
        }
        color[i] = 1;
        for (int j : g[i]) {
            if (!dfs(j)) {
                return false;
            }
        }
        color[i] = 2;
        return true;
    }
}

  • class Solution {
public:
    vector<int> color;

    vector<int> eventualSafeNodes(vector<vector<int>>& graph) {
        int n = graph.size();
        color.assign(n, 0);
        vector<int> ans;
        for (int i = 0; i < n; ++i)
            if (dfs(i, graph)) ans.push_back(i);
        return ans;
    }

    bool dfs(int i, vector<vector<int>>& g) {
        if (color[i]) return color[i] == 2;
        color[i] = 1;
        for (int j : g[i])
            if (!dfs(j, g)) return false;
        color[i] = 2;
        return true;
    }
};

  • class Solution:
    def eventualSafeNodes(self, graph: List[List[int]]) -> List[int]:
        def dfs(i):
            if color[i]:
                return color[i] == 2
            color[i] = 1
            for j in graph[i]:
                if not dfs(j):
                    return False
            color[i] = 2
            return True

        n = len(graph)
        color = [0] * n
        return [i for i in range(n) if dfs(i)]


  • func eventualSafeNodes(graph [][]int) []int {
	n := len(graph)
	color := make([]int, n)
	var dfs func(int) bool
	dfs = func(i int) bool {
		if color[i] > 0 {
			return color[i] == 2
		}
		color[i] = 1
		for _, j := range graph[i] {
			if !dfs(j) {
				return false
			}
		}
		color[i] = 2
		return true
	}
	ans := []int{}
	for i := range graph {
		if dfs(i) {
			ans = append(ans, i)
		}
	}
	return ans
}

  • /**
 * @param {number[][]} graph
 * @return {number[]}
 */
var eventualSafeNodes = function (graph) {
    const n = graph.length;
    const color = new Array(n).fill(0);
    function dfs(i) {
        if (color[i]) {
            return color[i] == 2;
        }
        color[i] = 1;
        for (const j of graph[i]) {
            if (!dfs(j)) {
                return false;
            }
        }
        color[i] = 2;
        return true;
    }
    let ans = [];
    for (let i = 0; i < n; ++i) {
        if (dfs(i)) {
            ans.push(i);
        }
    }
    return ans;
};

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