Algorithm

Problem Name: 802. Find Eventual Safe States

Problem Link: https://leetcode.com/problems/find-eventual-safe-states/

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].

 

Code Examples

#1 Code Example with Javascript Programming

Code - Javascript Programming


class Solution {
public:
    vector<int> eventualSafeNodes(vector < vector<int>>& graph) {
        vector<int>res;
        int n = graph.size();
        vector<int>loop(n), safe(n), visited(n);
        for(int i = 0; i  <  n; i++)
            if(isSafe(graph, visited, loop, safe, i)) res.push_back(i);
        return res;
    }
    
    bool isSafe(vector<vector < int>>& graph, vector<int>& visited, vector<int>& loop, vector<int>& safe, int node){
        if(safe[node]) return true;
        if(loop[node] || visited[node]) return false;
        visited[node] = 1;
        bool b = true;
        for(int neigh: graph[node]) b &= isSafe(graph, visited, loop, safe, neigh>;
        visited[node] = 0;
        b ? safe[node] = 1 : loop[node] = 1;
        return b;
    }
};
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Input

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graph = [[1,2],[2,3],[5],[0],[5],[],[]]

#2 Code Example with Python Programming

Code - Python Programming


class Solution:
    def eventualSafeNodes(self, graph):
        def explore(i):
            visited[i] = 0
            for v in graph[i]:
                if visited[v] == 0 or (visited[v] == -1 and explore(v)): return True
            visited[i] = 1
            res.append(i)
            return False
        visited, res = [-1] * len(graph), []
        for i in range(len(graph)):
            if visited[i] == -1: explore(i)
        return sorted(res)
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Input

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graph = [[1,2],[2,3],[5],[0],[5],[],[]]

Output

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[2,4,5,6][2,4,5,6]
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