# 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

``````
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]:

res = []
while queue:
node = queue.popleft()
res.append(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}