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Openai·Backend Engineer·Technical Phone Screen·Senior

Senior
May 2026

Summary

Got a coding question from OpenAI for a backend role that was more IP networking than I expected. One question, fairly self-contained, but the edge cases are where it gets you.

Questions Asked (1)

Q1

Write a function that takes a starting IPv4 address and a count, then returns the smallest list of CIDR blocks that covers exactly that many consecutive addresses starting from the given IP, no more and no less.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

The core idea clicked pretty fast: convert the IP to a 32-bit int, then greedily pick the largest valid aligned block that fits within the remaining count.

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AI HintsAI Generated

Suggested Approach

Start by converting the IPv4 address to a 32-bit integer and clarifying that the count represents the number of consecutive addresses. Then, greedily select the largest CIDR block that fits within the remaining range, aligning to block boundaries, and repeat until the count is exhausted. Finally, convert each block back to CIDR notation and verify the total coverage.

Pro tip: Mention that the greedy algorithm is optimal because CIDR blocks are power-of-two aligned, and discuss how to handle edge cases like count=0 or overflow beyond 255.255.255.255.

1. Clarify requirements and edge cases

Confirm that 'count' is the number of addresses and that the range must be exactly covered. Ask about handling count=0, invalid IPs, or ranges exceeding the IPv4 space.

2. Convert IP to integer and define range

Convert the starting IP to a 32-bit unsigned integer and compute the end address as start + count - 1. This simplifies arithmetic and boundary checks.

3. Greedily select largest aligned CIDR blocks

While the remaining count > 0, find the largest block size (power of two) that is aligned to the current address and does not exceed the remaining count. Add that block to the result and advance the address and count.

4. Convert blocks back to CIDR notation

For each selected block, compute the prefix length from the block size and format the network address as 'a.b.c.d/prefix'.

5. Verify and discuss complexity

Check that the total number of addresses covered equals the original count. Explain that the algorithm runs in O(log n) time and produces the minimal number of blocks.

Key Points to Mention

  • IPv4 addresses as 32-bit integers for easy arithmetic and alignment checks.
  • CIDR blocks are power-of-two sized and must be aligned to their size (e.g., /24 blocks start at multiples of 256).
  • Greedy algorithm: at each step, take the largest possible block that fits and is aligned, which yields the minimal number of blocks.
  • Alignment condition: the block size must divide the current address (i.e., address % block_size == 0).
  • Edge cases: count=0 (return empty list), count exceeding available addresses (handle overflow or error), and starting IP near the end of the address space.
  • Time complexity: O(log n) because each step reduces the remaining count by at least half, and space complexity O(log n) for the result.

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