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Meta·Software Engineer·Technical Phone Screen·Senior

SeniorPrefer not to say
Jun 2026

Summary

Meta data engineer SQL question, just one problem but it had enough layers to keep me busy for a while. The referrer logic tripped me up more than I expected.

Questions Asked (1)

Q1

Given a copies table (with copy_id and reserved_by_member_id) and a members table (with member_id and referred_by_member_id), find the member with the largest member_id. Return their member_id, referred_by_member_id, and a calculated column representing the difference between the number of copies they reserved and the number of copies their referrer reserved. If the referrer is NULL or has made no reservations, treat their reservation count as zero.

Data ModelingAlgorithms & Data Structures
Author's notes

I got the basic structure pretty quick, aggregate copies by member, join to members, grab the max member_id.

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AI HintsAI Generated

Suggested Approach

Break the problem into three parts: identify the target member (max member_id), compute reservation counts per member, and compute the referrer's reservation count. Use a subquery or CTE to get the max member_id, then join to members and left join to aggregated reservation counts for both the member and their referrer. Finally, calculate the difference, treating NULL as zero.

Pro tip: Mention that you would verify the result with edge cases, such as when the referrer is NULL or has no reservations, and ensure the query handles ties or missing data gracefully. Also, note that using a CTE improves readability and maintainability.

1. Identify the target member

Find the member with the largest member_id using a subquery or ORDER BY with LIMIT. This ensures you're focusing on the correct member.

2. Aggregate reservation counts

Create a derived table or CTE that counts the number of copies reserved by each member from the copies table, grouping by reserved_by_member_id.

3. Join member and referrer data

Join the members table to the aggregated counts for the member and also for the referrer (using referred_by_member_id). Use LEFT JOINs to handle cases where the referrer is NULL or has no reservations.

4. Calculate the difference

Compute the difference between the member's reservation count and the referrer's reservation count, using COALESCE or IFNULL to treat NULL as zero.

5. Select and format output

Return member_id, referred_by_member_id, and the calculated difference column. Ensure the query is efficient and readable.

Key Points to Mention

  • Use of LEFT JOIN to handle NULL referrers and missing reservations
  • Aggregation with GROUP BY and COUNT to get reservation counts
  • Subquery or CTE to find the max member_id
  • COALESCE or IFNULL to treat NULL as zero
  • Consideration of performance: indexing on member_id and reserved_by_member_id
  • Edge cases: referrer with no reservations, member with no reservations, multiple members with same max member_id (if applicable)

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