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2192. All Ancestors of a Node in a Directed Acyclic Graph

Difficulty Topics

Description

You are given a positive integer n representing the number of nodes of a Directed Acyclic Graph (DAG). The nodes are numbered from 0 to n - 1 (inclusive).

You are also given a 2D integer array edges, where edges[i] = [fromi, toi] denotes that there is a unidirectional edge from fromi to toi in the graph.

Return a list answer, where answer[i] is the list of ancestors of the ith node, sorted in ascending order.

A node u is an ancestor of another node v if u can reach v via a set of edges.

 

Example 1:

Input: n = 8, edgeList = [[0,3],[0,4],[1,3],[2,4],[2,7],[3,5],[3,6],[3,7],[4,6]]
Output: [[],[],[],[0,1],[0,2],[0,1,3],[0,1,2,3,4],[0,1,2,3]]
Explanation:
The above diagram represents the input graph.
- Nodes 0, 1, and 2 do not have any ancestors.
- Node 3 has two ancestors 0 and 1.
- Node 4 has two ancestors 0 and 2.
- Node 5 has three ancestors 0, 1, and 3.
- Node 6 has five ancestors 0, 1, 2, 3, and 4.
- Node 7 has four ancestors 0, 1, 2, and 3.

Example 2:

Input: n = 5, edgeList = [[0,1],[0,2],[0,3],[0,4],[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]
Output: [[],[0],[0,1],[0,1,2],[0,1,2,3]]
Explanation:
The above diagram represents the input graph.
- Node 0 does not have any ancestor.
- Node 1 has one ancestor 0.
- Node 2 has two ancestors 0 and 1.
- Node 3 has three ancestors 0, 1, and 2.
- Node 4 has four ancestors 0, 1, 2, and 3.

 

Constraints:

  • 1 <= n <= 1000
  • 0 <= edges.length <= min(2000, n * (n - 1) / 2)
  • edges[i].length == 2
  • 0 <= fromi, toi <= n - 1
  • fromi != toi
  • There are no duplicate edges.
  • The graph is directed and acyclic.

Solution

all-ancestors-of-a-node-in-a-directed-acyclic-graph.py
class Solution:
    def getAncestors(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        graph = collections.defaultdict(list)
        visited = [False] * n
        p = collections.defaultdict(set)

        for x, y in edges:
            graph[y].append(x)

        def go(node):
            if visited[node]:
                return p[node]

            visited[node] = True
            parents = set()

            for parent in graph[node]:
                parents |= go(parent)
                parents.add(parent)

            p[node] = parents
            return parents


        for node in range(n):
            if not visited[node]:
                go(node)

        res = []

        for i in range(n):
            res.append(sorted(list(p[i])))

        return res