Databricks·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Databricks SWE interview with a data aggregation problem that looked simple on the surface but had enough edge cases to keep you busy. Nothing too wild, but worth knowing your heap/sort tradeoffs going in.

Questions Asked (1)

Given a list of purchase events (each with a customer ID and an amount), find the k customers with the lowest total revenue. Handle ties and the case where there are fewer than k customers.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The core grouping part was fine, just aggregate amounts by customer.

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AI HintsAI Generated

Suggested Approach

Start by clarifying requirements: confirm the definition of 'lowest total revenue' (e.g., ascending order), how to handle ties (e.g., by customer ID), and the expected output format. Then propose an efficient algorithm: aggregate revenue per customer using a hash map, then use a min-heap or quickselect to find the k smallest totals, ensuring O(n + m log k) time where m is the number of unique customers. Discuss trade-offs between sorting all customers (O(m log m)) and using a heap for better scalability when k is small.

Pro tip: Mention that in distributed systems like Databricks, this problem can be solved with a groupBy and then a global sort or a heap-based approach, but be mindful of data skew and memory constraints. Also, explicitly state how you handle ties (e.g., by customer ID) and the edge case where there are fewer than k customers (return all).

1. Clarify requirements and edge cases

Ask about tie-breaking rules, output format, and whether k can be larger than the number of unique customers. Confirm that 'lowest total revenue' means ascending order.

2. Aggregate revenue per customer

Use a hash map to sum amounts for each customer ID. This takes O(n) time and O(m) space, where m is the number of unique customers.

3. Select k smallest totals

Use a min-heap of size k or quickselect to find the k smallest totals. If k is close to m, sorting may be simpler; otherwise, a heap is more efficient.

4. Handle ties and fewer than k customers

If ties occur at the k-th position, decide whether to include all tied customers or break ties by customer ID. If m < k, return all customers.

5. Analyze complexity and trade-offs

Discuss time and space complexity: O(n + m log k) with heap, O(n + m log m) with sorting. Mention scalability and potential optimizations for large datasets.

Key Points to Mention

Hash map for aggregation: O(n) time, O(m) space.
Heap vs. sorting: heap is O(m log k) for selection, sorting is O(m log m); choose based on k relative to m.
Tie-breaking strategy: e.g., by customer ID to ensure deterministic output.
Edge case: fewer than k unique customers, return all.
Scalability: consider distributed processing (e.g., Spark) for large datasets, but watch for data skew.
Output format: list of customer IDs with their total revenue, sorted ascending.

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