Molocoads·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Interviewed for a software engineer role at Molocoads and got a graph traversal problem about unlocking rooms. Pretty standard algorithmic question but it's the kind that trips you up if you overthink the setup.

Questions Asked (1)

You have n rooms labeled 0 to n-1, all locked except room 0. Each room contains keys to other rooms. Starting from room 0, can you visit every room? Given the list of keys per room, return true or false.

Algorithms & Data Structures

Author's notes

Classic graph reachability problem.

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AI HintsAI Generated

Suggested Approach

Model the rooms and keys as a directed graph where each room is a node and each key is a directed edge. Use DFS or BFS starting from room 0 to traverse all reachable rooms, and check if the count of visited rooms equals n. Alternatively, use Union-Find to connect rooms and verify all are in the same component as room 0.

Pro tip: Clarify that the graph is directed and may contain cycles; using an iterative DFS or BFS avoids recursion depth issues for large n. Also, mention that if the graph is disconnected, early termination can save time.

1. Clarify the problem

Confirm that keys are directed edges and that you start at room 0. Ask about constraints like n size and whether rooms can have duplicate keys.

2. Choose a traversal method

Select DFS, BFS, or Union-Find based on constraints and your comfort. Explain why you chose it (e.g., BFS for shortest path not needed, DFS for simplicity).

3. Implement traversal

Initialize a visited set or array. Start from room 0, mark visited, and explore all keys from each visited room, adding unvisited rooms to the queue/stack.

4. Check completeness

After traversal, compare the number of visited rooms to n. If equal, return true; otherwise, false.

5. Analyze complexity

State time complexity O(n + E) where E is total number of keys, and space complexity O(n) for visited set and queue/stack.

Key Points to Mention

Graph representation: adjacency list from rooms to keys.
Directed graph traversal: DFS or BFS with visited tracking.
Union-Find alternative: union each room with its keys, then check if all rooms share the same root as room 0.
Handling cycles: visited set prevents infinite loops.
Edge cases: n=1 (always true), empty key lists, disconnected components.
Time and space complexity analysis.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.