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Loyalty wants every customer bucketed by total lifetime spend so they can target perks.
| Column | Type |
|---|---|
| CustomerId | INTEGER (PK) |
| FirstName | TEXT NOT NULL |
| LastName | TEXT NOT NULL |
| Company | TEXT |
| Country | TEXT |
| TEXT NOT NULL | |
| SupportRepId | INTEGER (FK → Employee) |
| Column | Type |
|---|---|
| InvoiceId | INTEGER (PK) |
| CustomerId | INTEGER (FK) |
| InvoiceDate | TIMESTAMP NOT NULL |
| BillingCountry | TEXT |
| Total | NUMERIC(10,2) NOT NULL |
FullName, LifetimeSpend (rounded to 2 decimals), and TierTier: 'A_Elite' if spend ≥ 45, 'B_Premium' if ≥ 40, 'C_Standard' if ≥ 35, else 'D_Basic'Your query should return 15 rows with 3 columns: | fullname | lifetimespend | tier | |--------------------|---------------|---------| | Helena Holý | 49.62 | A_Elite | | Richard Cunningham | 47.62 | A_Elite | | Luis Rojas | 46.62 | A_Elite | | Ladislav Kovács | 45.62 | A_Elite | | Hugh O'Reilly | 45.62 | A_Elite | | ... | ... | ... |
Putting aggregated filter conditions in the WHERE clause instead of HAVING, or including un-aggregated columns in SELECT without listing them in GROUP BY. Postgres strictly enforces that every non-aggregated projection column must appear in the GROUP BY expression.
Interviewers verify whether you understand the distinction between row-level filtering (WHERE) versus post-aggregation partition filtering (HAVING), as well as SQL standard group syntax.
Construct the solution logically from first principles to avoid typical edge case pitfalls.
Determine the attributes that define unique summary rows (e.g. Artist, Country, or Category).
GROUP BY entity_id, entity_name
Apply SUM, AVG, COUNT, or conditional aggregations over each bucket.
SELECT entity_name, COUNT(*) AS total_items, SUM(amount) AS total_revenue
Filter only the groups that satisfy minimum aggregate thresholds.
HAVING COUNT(*) >= 10 ORDER BY total_revenue DESC;
SELECT c.FirstName || ' ' || c.LastName AS FullName,
ROUND(SUM(i.Total), 2) AS LifetimeSpend,
CASE
WHEN SUM(i.Total) >= 45 THEN 'A_Elite'
WHEN SUM(i.Total) >= 40 THEN 'B_Premium'
WHEN SUM(i.Total) >= 35 THEN 'C_Standard'
ELSE 'D_Basic'
END AS Tier
FROM Customer c
JOIN Invoice i ON c.CustomerId = i.CustomerId
GROUP BY c.CustomerId, c.FirstName, c.LastName
ORDER BY LifetimeSpend DESC
LIMIT 15;Real code patterns candidates submit that fail the grading suite.
SELECT country, SUM(total) FROM Invoice WHERE COUNT(InvoiceId) > 10 GROUP BY country;
Three recurring syntax and semantic traps relevant to this problem domain.
WHERE operates on individual rows before grouping occurs. Aggregate functions like COUNT(), SUM(), AVG() can only be filtered in HAVING.
SELECT genre_id, COUNT(*) FROM tracks WHERE COUNT(*) > 50 GROUP BY genre_id; -- ❌ Syntax Error
SELECT genre_id, COUNT(*) FROM tracks GROUP BY genre_id HAVING COUNT(*) > 50; -- ✅ Correct
Every non-aggregated column in the SELECT list must appear in the GROUP BY clause.
SELECT artist_id, artist_name, COUNT(album_id) FROM albums GROUP BY artist_id; -- ❌ artist_name missing
SELECT artist_id, artist_name, COUNT(album_id) FROM albums GROUP BY artist_id, artist_name; -- ✅ Correct
COUNT(*) counts every row in the group including NULLs. COUNT(column) counts only non-null instances.
SELECT department, COUNT(commission_pct) FROM employees GROUP BY department; -- ❌ Ignores 0-commission staff
SELECT department, COUNT(*) FROM employees GROUP BY department; -- ✅ Accurate total count
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