Smallest cycle node
A build system has n targets (0..n-1) and `edges` `[a, b]` meaning target a depends on target b. A circular dependency makes the build impossible. Return the smallest-numbered node that participates in any dependency cycle, or -1 if the dependency graph is acyclic. A self-loop `[a, a]` counts as a cycle on node a.
Implement
find_cycle_node(n: int, edges: list[list[int]]) → intExamples
in
[4,[[0,1],[1,2],[2,0],[2,3]]]out0What 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 25 min
solution.py
InputExpectedGot
[4,[[0,1],[1,2],[2,0],[2,3]]]0not run yetsample