leetcode - 802. 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:

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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:

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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:

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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].

Solution

Topological sort. If we start from the terminal node, and remove its edges, then the next terminal node would be safe node. So this is actually a topological sort problem.

Time complexity: o ( e d g e s + n o d e s ) o(edges + nodes) o(edges+nodes)

Space complexity: o ( e d g e s + n o d e s ) o(edges + nodes) o(edges+nodes)

Code

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class Solution:
    def eventualSafeNodes(self, graph: List[List[int]]) -> List[int]:
        def build_graph(edges: list) -> tuple:
            graph = {i: [] for i in range(len(edges))}
            outdegree = {i: 0 for i in range(len(edges))}
            for i in range(len(edges)):
                for next_node in edges[i]:
                    graph[next_node].append(i)
                    outdegree[i] += 1
            return graph, outdegree
        # new_graph: {node: [node that points to this node]}
        # outdegree: {node: out_degree}
        new_graph, outdegree = build_graph(graph)
        queue = collections.deque([])
        res = set()
        for each_node in new_graph:
            if outdegree[each_node] == 0:
                queue.append(each_node)
        while queue:
            node = queue.popleft()
            if node in res:
                continue
            res.add(node)
            for neighbor_node in new_graph[node]:
                outdegree[neighbor_node] -= 1
                if outdegree[neighbor_node] == 0:
                    queue.append(neighbor_node)
        return list(sorted(res))