Databricks·Software Engineer·Technical Phone Screen·Senior

SeniorPrefer not to say

Jun 2026Remote

Summary

Databricks technical phone screen, probably for a backend or infrastructure-adjacent role. The whole thing was one meaty networking/algorithms problem that took most of the session to get through properly.

Questions Asked (1)

Given a list of CIDR blocks, merge any overlapping or adjacent ranges and return the minimal set of CIDR blocks that covers exactly the same IP address space.

Algorithms & Data StructuresSystem DesignTechnical Trade-offs

Author's notes

This one has layers and I did not appreciate that until I was already in trouble.

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AI HintsAI Generated

Suggested Approach

Convert each CIDR block to a numeric range [start, end], sort by start, then merge overlapping or adjacent ranges by comparing the current end + 1 with the next start. Finally, convert each merged range back to the minimal set of CIDR blocks using bitwise operations.

Pro tip: Mention that you handle both IPv4 and IPv6 by using 128-bit integers, and discuss how to avoid integer overflow when adding 1 to the end of a range. Also, note that the output should be the minimal set, which may require splitting a range into multiple CIDRs if it's not a single CIDR block.

1. Clarify requirements and edge cases

Ask about IP version (IPv4/IPv6), input size, whether CIDRs are well-formed, and if the output must be minimal. Confirm that adjacent ranges should be merged.

2. Convert CIDRs to numeric ranges

For each CIDR, compute the start and end IP addresses as integers. Use bitwise operations: start = ip & mask, end = start | ~mask.

3. Sort and merge ranges

Sort ranges by start. Iterate and merge if the current range's end + 1 >= next range's start. Keep track of the merged range's end.

4. Convert merged ranges back to CIDRs

For each merged range, repeatedly find the largest CIDR block that fits within the range, starting from the current start, and add it to the result. Update the start to the end of that block + 1.

5. Analyze complexity and trade-offs

Discuss time complexity O(n log n) due to sorting, and space O(n). Mention alternative approaches like using a trie, and trade-offs between simplicity and performance.

Key Points to Mention

Bitwise operations for CIDR to range conversion and back
Handling of adjacent ranges (end + 1 == next start)
Sorting and merging algorithm with O(n log n) time
Minimal CIDR representation using greedy largest block fitting
Edge cases: single IP, full range, non-overlapping, IPv6
Potential integer overflow when adding 1 to end (use 128-bit or handle max)

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