All Ancestors of a Node in a Directed Acyclic Graph

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:

class Solution:
    def getAncestors(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        graph = {}
        for a, b in edges:
            graph[b] = graph.get(b, []) + [a]
        op = [[] for i in range(n)]
        for a in graph:
            visited = set()
            paths = [a]
            while len(paths) > 0:
                curr = paths.pop()
                for b in graph.get(curr, []):
                    if b not in visited:
                        visited.add(b)
                        paths.append(b)
            op[a] = sorted(visited)
        return op