# Find Eventual Safe States Problem

## Description

LeetCode Problem 802.

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.

Return an array containing all the safe nodes of the graph. The answer should be sorted in ascending order.

Example 1:

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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:

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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 <= 10^4
- 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 * 10^4].

## Sample C++ Code

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enum Color {WHITE, GRAY, BLACK};
class Solution {
public:
bool dfs(int i, vector<int> & color, vector<vector<int>> & graph) {
if (color[i])
return color[i] == BLACK;
color[i] = GRAY;
for (int & v : graph[i]) {
if (color[v] == GRAY)
return false;
if (color[v] == WHITE && !dfs(v, color, graph))
return false;
}
color[i] = BLACK;
return true;
}
vector<int> eventualSafeNodes(vector<vector<int>>& graph) {
int n = int(graph.size());
vector<int> color(n, WHITE);
vector<int> result;
for (int i = 0; i < n; ++i)
if (dfs(i, color, graph))
result.push_back(i);
return result;
}
};