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# 00802-Find-Eventual-Safe-States

### Problem

<https://leetcode.com/problems/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].

### Solution

topological sort

```python

class Solution:
    def eventualSafeNodes(self, graph: List[List[int]]) -> List[int]:
        n = len(graph)
        adj, degree = defaultdict(list), [0] * n
        queue = deque()
        for i in range(n):
            degree[i] = len(graph[i])
            if degree[i] == 0:
                queue.append(i)
            for j in graph[i]:
                adj[j].append(i)
        
        res = []
        while queue:
            node = queue.popleft()
            res.append(node)
            for i in adj[node]:
                if degree[i] != 0:
                    degree[i] -= 1
                if degree[i] == 0:
                    queue.append(i)
        return sorted(res)

# degree = {0:2, 1:2, 2:1, 3:1, 4:1, 5:0, 6:0}
# adj = {
#       0: [4],
#       1: [0],
#       2: [0, 1],
#       3: [1],
#       5: [2, 4],
# }
         
```