802

802. Find Eventual Safe States (Medium)

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].

Companies: Amazon, Bloomberg, Mindtickle

Related Topics:
Depth-First Search, Breadth-First Search, Graph, Topological Sort

Similar Questions:

• Build a Matrix With Conditions (Hard)

Solution 1. DFS

// OJ: https://leetcode.com/problems/find-eventual-safe-states
// Author: github.com/lzl124631x
// Time: O(V + E)
// Space: O(V)
class Solution {
public:
    vector<int> eventualSafeNodes(vector<vector<int>>& G) {
        int N = G.size();
        vector<int> state(N, -2), ans; // -2 unvisited, -1 visiting, 0 unsafe, 1 safe
        function<bool(int)> dfs = [&](int u) {
            if (state[u] == -1) return state[u] = 0;
            if (state[u] >= 0) return state[u];
            state[u] = -1;
            for (int v : G[u]) {
                if (!dfs(v)) return state[u] = 0;
            }
            return state[u] = 1;
        };
        for (int i = 0; i < N; ++i) {
            if (dfs(i)) ans.push_back(i);
        }
        return ans;
    }
};

Solution 2. Topological Sort (BFS) + Reverse Edges

// OJ: https://leetcode.com/problems/find-eventual-safe-states/
// Author: github.com/lzl124631x
// Time: O(V + E)
// Space: O(V + E)
class Solution {
public:
    vector<int> eventualSafeNodes(vector<vector<int>>& G) {
        int N = G.size();
        vector<vector<int>> R(N);
        vector<int> outdegree(N), safe(N), ans;
        queue<int> q;
        for (int i = 0; i < N; ++i) {
            for (int v : G[i]) {
                R[v].push_back(i);
            }
            outdegree[i] = G[i].size();
            if (outdegree[i] == 0) q.push(i);
        }
        while (q.size()) {
            int u = q.front();
            q.pop();
            safe[u] = 1;
            for (int v : R[u]) {
                if (--outdegree[v] == 0) q.push(v);
            }
        }
        for (int i = 0; i < N; ++i) {
            if (safe[i]) ans.push_back(i);
        }
        return ans;
    }
};