Leetcode #2192: All Ancestors of a Node in a Directed Acyclic Graph
In this guide, we solve Leetcode #2192 All Ancestors of a Node in a Directed Acyclic Graph in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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).
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Depth-First Search, Breadth-First Search, Graph, Topological Sort
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
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.
Python Solution
class Solution:
def getAncestors(self, n: int, edges: List[List[int]]) -> List[List[int]]:
def bfs(s: int):
q = deque([s])
vis = {s}
while q:
i = q.popleft()
for j in g[i]:
if j not in vis:
vis.add(j)
q.append(j)
ans[j].append(s)
g = defaultdict(list)
for u, v in edges:
g[u].append(v)
ans = [[] for _ in range(n)]
for i in range(n):
bfs(i)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.