Two metrics in one row sounds easy until you realize you need a conditional percentage with a safe divide.
Use conditional aggregation to compute both metrics in a single query. Filter for active checkouts with copy_condition = 'good', then count all such rows and calculate the percentage of those with renew_count > 2, handling division by zero with NULLIF or CASE. Round the percentage to 2 decimal places and ensure it returns 0.00 when the denominator is zero.
Pro tip: Always consider edge cases like zero denominators and use NULLIF to avoid division errors, showing attention to detail. Also, clarify the definition of 'active' checkouts if ambiguous, demonstrating thoroughness.
Identify the two metrics: count of active checkouts with copy_condition 'good', and percentage of those with renew_count > 2. Note the rounding and zero-denominator handling.
Apply WHERE clause to select only active checkouts with copy_condition = 'good'.
Use COUNT(*) to get the total number of filtered rows.
Use conditional aggregation: SUM(CASE WHEN renew_count > 2 THEN 1 ELSE 0 END) * 100.0 / COUNT(*), with NULLIF to handle zero denominator, then ROUND to 2 decimals.
Select both metrics in one row without GROUP BY, ensuring the percentage returns 0.00 when count is zero.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Filtered books in a subquery using HAVING COUNT(*) > 10 on the copies table, then joined to checkouts and computed return_date minus checkout_date, ignoring NULLs.
Break the problem into two parts: first, filter books with more than 10 copies, then compute the maximum completed checkout duration per book. Use a subquery or CTE to aggregate durations, then rank by max duration descending and book_id ascending, and finally limit to top 3.
Pro tip: Clarify what 'completed checkout' means (e.g., return_date is not null) and whether duration is in days as an integer or requires date arithmetic. Also, confirm if ties should be broken by book_id ascending only after sorting by max duration descending.
Identify tables (e.g., books, checkouts) and columns (book_id, copies, checkout_date, return_date). Clarify that 'completed checkout' means return_date IS NOT NULL and duration is DATEDIFF(return_date, checkout_date) in days.
Use a subquery or join to select only books where copies > 10. This reduces the dataset for subsequent aggregation.
For each book, calculate the maximum completed checkout duration using GROUP BY book_id and MAX(duration). Ensure only completed checkouts are included.
Order the aggregated results by max_duration DESC, then book_id ASC. Use LIMIT 3 to return the top 3 books.
Consider books with no completed checkouts (should they be excluded?), ties in max duration, and ensure the query handles NULLs appropriately.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Self-join on the members table plus two separate aggregations on reservations.
First, clarify the schema and relationships: identify the members table with referrer_id and the reservations table with member_id. Then, aggregate reservation counts per member, join members to their referrers to compute absolute differences, and finally order by difference descending and member_id ascending, limiting to one row.
Pro tip: Mention that you would validate edge cases like members with no reservations (count = 0) and self-referrals or cycles, and confirm whether ties should be broken by the referred member's ID or the referrer's ID.
Ask about table structures: members (member_id, referrer_id) and reservations (reservation_id, member_id). Confirm that 'reservation count' means total reservations per member, and that referred members are those with a non-null referrer_id.
Compute the number of reservations for each member using a GROUP BY on member_id, and ensure members with zero reservations are included (e.g., via LEFT JOIN or COALESCE).
Self-join the members table to link each referred member to their referrer, then join both to the reservation counts to get each member's and referrer's counts.
Calculate ABS(referred_count - referrer_count) for each referred member, then order by this difference descending and member_id ascending, and select the top row.
Check for members with no referrer, referrers with no reservations, and ties. Confirm that the output is a single row and that the tie-breaking rule is correctly applied.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Straightforward once you break it into steps: aggregate max per category, sort by (-max_score, category_name), take first three, sum them.
First, clarify the problem requirements and edge cases. Then, design an algorithm that computes the total sum and the top 3 category maxima efficiently, using a hash map to aggregate scores per category and a heap or sorting to find the top 3. Finally, implement the solution with clean code and analyze its time and space complexity.
Pro tip: Discuss how to handle ties in the top 3 category maxima: since ties are broken alphabetically by category name, you can sort the categories by score descending and then by name ascending, or use a custom comparator in a heap. This shows attention to detail and robustness.
Ask questions to confirm input format, possible empty list, negative scores, and tie-breaking rules. Clarify that 'top 3 category maxima' means the three highest maximum scores among all categories, with ties broken by category name alphabetically.
Plan to iterate through the list once to compute the total sum and to build a dictionary mapping each category to its maximum score. Then, extract the top 3 maxima from the dictionary, handling ties by sorting or using a heap with a custom comparator.
Write code that initializes total_sum and a dictionary. For each (category, score), add to total_sum and update the category's max if needed. Then, sort the dictionary items by score descending and category name ascending, take the first three, and sum their scores. Return the total sum and the top-3 sum.
Test with edge cases: empty list, fewer than 3 categories, ties in scores, negative scores. Analyze time complexity: O(n + m log m) where n is number of tuples and m is number of unique categories, or O(n + m) if using a heap of size 3. Space complexity: O(m).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a hash map to track the last action (checkout or return) for each book. Iterate through the log entries once, validating each entry against the stored state and updating the state. This ensures O(n) time and O(n) space.
Pro tip: Clarify upfront whether the log is guaranteed to be well-formed (e.g., no unknown book IDs) and whether the function should return a boolean or throw an exception. This shows attention to API design and edge cases.
Ask if the log can contain invalid book IDs, if the function should return a boolean or throw an error, and if the log is empty or has a single entry.
Use a hash map (dictionary) to store the last action for each book ID. The key is book_id, and the value is a boolean or enum indicating checkout (true) or return (false).
For each log entry, check if the book exists in the map. If not, it must be a checkout; otherwise, validate that the action is not the same as the last action (e.g., two checkouts or two returns in a row).
After validation, update the map with the current action. If any validation fails, return false or throw an exception immediately.
Confirm O(n) time and O(n) space. Walk through examples: valid sequence, invalid sequence with return before checkout, consecutive checkouts, consecutive returns.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.