Count articulation points
Given an undirected graph with n nodes (0..n-1) and a list of edges [u, v], count the number of articulation points (cut vertices): vertices whose removal increases the number of connected components. Handle the DFS-root special case correctly (a root is an articulation point iff it has more than one DFS child). Assume 1 <= n <= 10000; the graph may be disconnected.
Implement
count_articulation_points(n: int, edges: list[list[int]]) → intExamples
in
[5,[[0,1],[1,2],[2,0],[1,3],[3,4]]]out2What a strong answer looks like
State your approach and its time/space complexity out loud before you optimize. Handle the edge cases (empty input, duplicates, overflow), and say why you chose this over the brute force. Green tests are the floor, not the grade.
0:00 of about 30 min
solution.py
InputExpectedGot
[5,[[0,1],[1,2],[2,0],[1,3],[3,4]]]2not run yetsample