Databricks·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Databricks data engineering interview with a coding problem focused on revenue aggregation and heap-based selection. Pretty standard algorithmic stuff but the edge case discussion was more involved than I expected.

Questions Asked (1)

Given a list of (customer_id, revenue) records, aggregate total revenue per customer and return the bottom K customers by total revenue, breaking ties by customer_id ascending.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I went with heapq.nsmallest after building the aggregation dict in one pass, which felt right.

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AI HintsAI Generated

Suggested Approach

Clarify the input format and constraints, then propose an efficient two-phase solution: aggregate revenue per customer using a hash map, then select the bottom K using a min-heap of size K (or sort if K is large). Discuss trade-offs between sorting and heap-based selection, and handle ties by customer_id ascending.

Pro tip: Mention that if K is small relative to the number of customers, a max-heap of size K is optimal for finding the bottom K, but if K is large, sorting all customers may be simpler and faster due to lower constant factors. Also, note that tie-breaking by customer_id ascending requires a total order, so ensure your comparator includes customer_id.

1. Clarify requirements and constraints

Ask about input size, memory limits, whether the list fits in memory, and if K is known to be small or large. Confirm that ties are broken by customer_id ascending.

2. Aggregate revenue per customer

Use a hash map to sum revenues for each customer_id. This gives O(N) time and O(U) space, where U is the number of unique customers.

3. Select bottom K customers

Choose between sorting all customers (O(U log U)) or using a max-heap of size K (O(U log K)). Discuss trade-offs based on U and K.

4. Handle tie-breaking

When revenues are equal, order by customer_id ascending. Ensure the comparator used in sorting or heap respects this.

5. Analyze complexity and edge cases

State time and space complexity, and discuss edge cases: K=0, K > U, negative revenues, duplicate records, and empty input.

Key Points to Mention

Hash map for aggregation: O(N) time, O(U) space.
Heap vs. sorting trade-off: heap is O(U log K) and better when K << U; sorting is O(U log U) and simpler when K is close to U.
Tie-breaking by customer_id ascending requires a total order comparator.
Edge cases: K=0, K > U, negative revenues, empty input.
Memory considerations: if data doesn't fit in memory, discuss external sorting or streaming with a heap.
Potential for parallelization or distributed processing (e.g., MapReduce) if data is large.

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