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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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