Detect a Cycle in Directed Hate Relationships
Reported by candidates from Arcesium's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The edge case that breaks a naive solution on this Arcesium OA, reported in August 2023, is a graph that isn't connected. Run DFS from node 0 only and you'll miss a cycle hiding in another component. The task is plain: given n politicians and directed hate pairs, return true if the graph has a directed cycle. It's cycle detection in a directed graph, a classic graph problem. Example 2 looks scary because two paths converge on node 3, but that's a diamond, not a cycle. If you blank on the visited-state logic during the live assessment, StealthCoder is the safety net running invisibly on your screen.
The problem
There are n politicians labeled from 0 to n - 1. You are given a list of directed hate relationships hatePairs, where each pair [u, v] means that politician u hates politician v. Return true if the directed graph formed by these relationships contains a directed cycle. Otherwise, return false. A directed cycle exists when it is possible to start at a politician, follow one or more hate relationships in their stated direction, and return to the starting politician. Function hasDirectedHateCycle(n: int, hatePairs: int[][]) → boolean Examples Example 1 n = 4 hatePairs = [[0,1],[1,2],[2,0],[2,3]] return = true The relationships 0 -> 1, 1 -> 2, and 2 -> 0 form a directed cycle, so the result is true. Example 2 n = 4 hatePairs = [[0,1],[0,2],[1,3],[2,3]] return = false Every relationship moves toward politician 3, and no path returns to an earlier politician. The graph has no directed cycle, so the result is false.
Reported by candidates. Source: FastPrep
Pattern and pitfall
Two clean approaches work. First, DFS with three states per node: unvisited, in the current recursion path, and done. If you reach a node that's in the current path, that's a back edge, so return true. Second, Kahn's algorithm: compute indegrees, peel off zero-indegree nodes with a queue, and if you process fewer than n nodes, a cycle exists. The pitfall is using a single visited set. That flags the diamond in Example 2 as a cycle, because node 3 gets reached twice. Also loop over all n nodes as start points, since the graph can be disconnected, and watch for self-loops like [u,u], which count as a cycle. Both approaches run in O(n + E). If you freeze live, StealthCoder can hand you the three-state version so you can adapt it quickly.
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Detect a Cycle in Directed Hate Relationships FAQ
What's the trick to the Arcesium directed hate cycle problem?+
Track node state, not just visited. Use three states: unvisited, in progress, finished. A cycle exists only when DFS hits a node that's still in progress. Alternatively use Kahn's topological sort and check whether every node got processed.
Why does a plain visited set give the wrong answer?+
Because reaching an already-visited node doesn't mean a cycle in a directed graph. In Example 2, nodes 1 and 2 both lead to node 3. A single visited set would flag node 3 on the second arrival, returning true when the right answer is false.
Do I need to start DFS from every node?+
Yes. The graph can have several disconnected pieces, and a cycle might live in one you never reach from node 0. Loop through all n nodes and start a DFS from each unvisited one. Isolated nodes with no edges are fine and just return quickly.
How hard is this one really?+
It's a standard medium, equivalent to checking whether a course schedule is feasible. If you know DFS coloring or Kahn's algorithm, it's about 20 lines. The difficulty is in the edge cases: disconnected components, self-loops, and diamond shapes that look like cycles.
How do I prepare in 48 hours?+
Write the DFS three-state version and the Kahn's version from memory, once each. Test them on both examples plus a self-loop and a disconnected graph. Build the adjacency list first. That covers nearly everything this problem can throw at you.