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Using a CTE, compute each album's track count, then return only albums whose track count exceeds the average track count across all albums.
Return AlbumId, Title, and TrackCount, ordered by TrackCount descending (ties broken by AlbumId ascending).
| Column | Type |
|---|---|
| AlbumId | INTEGER (Primary Key) |
| Title | TEXT |
| Column | Type |
|---|---|
| TrackId | INTEGER (Primary Key) |
| AlbumId | INTEGER (Foreign Key -> Album.AlbumId) |
Your result should show the above-average albums: | AlbumId | Title | TrackCount | |---------|------------------|------------| | 141 | Greatest Hits | 57 | | 23 | Minha Historia | 34 | | 73 | Unplugged | 30 | | ... | ... | ... |
Double-counting metrics by using COUNT(*) after joining parent and child tables. When joining an invoice table with an invoice lines table, a single invoice multiplies across all its line items, causing COUNT(invoice_id) to return line counts instead of unique invoice counts.
Interviewers check whether you notice 1-to-many cardinality multiplication and use COUNT(DISTINCT col) or pre-aggregate child records before joining.
Construct the solution logically from first principles to avoid typical edge case pitfalls.
Identify the base table and apply preliminary WHERE filters.
FROM TableName WHERE is_active = true
Group by primary business keys and compute aggregate expressions.
SELECT category, COUNT(DISTINCT item_id) AS total_items, SUM(amount) AS revenue GROUP BY category
Order by specified metrics descending and apply limit clauses.
ORDER BY revenue DESC LIMIT 10;
WITH album_track_counts AS ( SELECT AlbumId, COUNT(*) AS TrackCount FROM Track GROUP BY AlbumId ) SELECT al.AlbumId, al.Title, atc.TrackCount FROM album_track_counts atc JOIN Album al ON al.AlbumId = atc.AlbumId WHERE atc.TrackCount > (SELECT AVG(TrackCount) FROM album_track_counts) ORDER BY atc.TrackCount DESC, al.AlbumId;
Real code patterns candidates submit that fail the grading suite.
SELECT a.Name, COUNT(t.TrackId) FROM Artist a JOIN Album al ON a.ArtistId = al.ArtistId JOIN Track t ON al.AlbumId = t.AlbumId GROUP BY a.Name;
Three recurring syntax and semantic traps relevant to this problem domain.
Joining a fact table with child lines multiplies fact table rows, distorting sums and counts.
SELECT c.id, SUM(i.total) FROM customer c JOIN invoice i ON c.id = i.customer_id JOIN invoice_line il ON i.id = il.invoice_id -- ❌ Inflated SUM
SELECT c.id, SUM(i.total) FROM customer c JOIN invoice i ON c.id = i.customer_id GROUP BY c.id; -- ✅ Avoids line multiplication
Using COUNT(*) when duplicate rows exist due to joins counts duplicate records.
SELECT artist_id, COUNT(album_id) ... -- ❌ Counts duplicate occurrences
SELECT artist_id, COUNT(DISTINCT album_id) ... -- ✅ Distinct unique entities
In SQL, dividing integers like 5 / 10 results in 0. Cast at least one operand to FLOAT or NUMERIC.
SELECT solved_count / total_count AS rate ... -- ❌ Returns 0
SELECT CAST(solved_count AS FLOAT) / total_count AS rate ... -- ✅ Returns 0.5
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