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
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