802. 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:

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.

JAVA

• 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;
      }
  }

C++

• 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;
      }
  };

PYTHON

• class Solution:
      def eventualSafeNodes(self, graph):
          n = len(graph)
          color = [0] * n  # 0: unvisited, 1: visiting, 2: safe
          result = []
  
          def dfs(node):
              if color[node] > 0:
                  return color[node] == 2
              color[node] = 1  # Mark as visiting
              for neighbor in graph[node]:
                  if not dfs(neighbor):
                      return False
              color[node] = 2  # Mark as safe
              return True
  
          for i in range(n):
              if dfs(i):
                  result.append(i)
  
          return result