Netflix·Software Engineer·Onsite - Coding / Algorithms·Senior

SeniorPrefer not to say

Apr 2026

Summary

Netflix onsite coding round, one question but it had three layers to it and the interviewer kept pushing on runtimes until I could spit them out without re-deriving everything on the spot. Felt like a stress test more than a coding exercise.

Questions Asked (1)

Implement a topological sort on a DAG. Then extend it to handle tie-breaking when multiple nodes are available at the same time (e.g. by lexicographic order). Then extend it further to enumerate ALL valid topological orderings. For each variant, state the runtime explicitly.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The base implementation was fine, I went with Kahn's and got through it cleanly.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Start by clearly defining the problem and the three variants, then walk through each with a concrete algorithm, data structure choices, and runtime analysis. Emphasize the trade-offs between using Kahn's algorithm (BFS) versus DFS, and how tie-breaking and enumeration affect complexity. Conclude by discussing practical considerations like cycle detection and scalability.

Pro tip: Mention that Kahn's algorithm naturally supports tie-breaking with a priority queue, but for enumeration, backtracking with in-degree tracking is essential; also note that the number of topological orderings can be exponential, so output-sensitive complexity is key.

1. Clarify requirements and assumptions

Confirm that the graph is a DAG, discuss input representation (adjacency list), and define what 'tie-breaking' means (e.g., lexicographic order of node labels).

2. Implement basic topological sort

Use Kahn's algorithm (BFS with in-degree tracking) or DFS with post-order reversal. State runtime: O(V+E) for both.

3. Extend with tie-breaking

Replace the queue in Kahn's algorithm with a min-heap (or priority queue) to always pick the smallest available node. Runtime becomes O(V log V + E log V) or O(E + V log V) depending on implementation.

4. Enumerate all topological orderings

Use backtracking: at each step, choose any node with in-degree 0, mark it as visited, decrement in-degrees of its neighbors, recurse, then backtrack. Runtime is O(V * (V+E) * K) where K is the number of orderings, or more precisely O((V+E) * K) with careful implementation.

5. Analyze trade-offs and edge cases

Discuss cycle detection, memory usage, and when enumeration is feasible (small K). Mention that tie-breaking adds log factor but ensures deterministic output.

Key Points to Mention

Kahn's algorithm vs DFS for topological sort: both O(V+E), but Kahn's is easier to extend for tie-breaking.
Using a priority queue for tie-breaking changes runtime to O(V log V + E log V) or O(E + V log V) with a binary heap.
Enumeration via backtracking has output-sensitive complexity: O((V+E) * K) where K is the number of valid orderings.
Cycle detection: if the topological sort doesn't include all vertices, there's a cycle.
Space complexity: O(V+E) for basic sort, O(V) for recursion stack in DFS, and O(V) for in-degree array.
Practical considerations: enumeration is only feasible for small graphs or when K is small; otherwise, it's exponential.

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