Databricks·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Databricks coding round, one question the whole time. It was a networking/bit-manipulation problem that looks deceptively clean on paper but has enough edge cases to make you sweat if you haven't seen it before.

Questions Asked (1)

Given a starting IP address and a positive integer n, return the smallest list of CIDR blocks that exactly covers n consecutive IP addresses beginning at that IP. Each block must be aligned to its size, which must be a power of 2.

Algorithms & Data Structures

Author's notes

The core loop isn't too bad once you see it: figure out the largest aligned power-of-2 block you can emit without overshooting n, write out the CIDR, subtract from n, move the pointer forward, repeat.

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AI HintsAI Generated

Suggested Approach

Use a greedy algorithm that repeatedly finds the largest CIDR block aligned to the current IP address that does not exceed the remaining number of addresses. Convert the IP to an integer, then at each step compute the maximum block size based on the lowest set bit of the current address and the remaining count, add that block, and advance the address and remaining count.

Pro tip: Clarify that the greedy choice is optimal because any valid cover must include a block starting at the current address, and choosing the largest possible aligned block minimizes the total number of blocks. Also, handle edge cases like n=0 and IPv6 if applicable.

1. Understand the problem and constraints

Restate the problem: given a start IP and n, produce a minimal list of aligned CIDR blocks covering exactly n consecutive IPs. Clarify that blocks must be power-of-2 sized and aligned to their size.

2. Convert IP to integer and plan greedy approach

Convert the starting IP to a 32-bit integer. Recognize that a greedy strategy of taking the largest possible aligned block at each step yields an optimal solution.

3. Compute the largest valid block at each step

For the current integer address and remaining count, determine the maximum block size: it must be a power of 2, not exceed the remaining count, and the address must be divisible by the block size (alignment).

4. Add block, update state, and repeat

Add the CIDR block (address/size) to the result, increment the address by the block size, and decrement the remaining count. Repeat until remaining count is zero.

5. Verify and handle edge cases

Check that the total addresses covered equals n and blocks are non-overlapping. Handle n=0 (return empty list) and consider IPv6 if required.

Key Points to Mention

Greedy algorithm is optimal: at each step, the largest aligned block that fits within the remaining range must be chosen to minimize the number of blocks.
Alignment condition: a block of size 2^k is aligned if the starting IP integer is divisible by 2^k.
Block size determination: the maximum size is the largest power of 2 that is ≤ remaining count and divides the current IP integer.
Efficiency: O(log n) time and O(log n) space in the worst case, since each step reduces the remaining count by at least half.
Edge cases: n=0 returns empty list; n=1 returns a /32 block; handle IP address overflow if n exceeds the address space.
Implementation details: use bitwise operations to compute the lowest set bit and the largest power of 2, and convert integers back to dotted-decimal notation.

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