One Leader per Category

#978

Each category card must show exactly one most expensive product. If prices tie, the product with the smaller id wins. Fix the report so it returns category and name, one row per category, sorted by category.

Buggy query

SELECT category, name
FROM (
  SELECT category, name,
         RANK() OVER (PARTITION BY category ORDER BY price DESC) rn
  FROM products
) t
WHERE rn = 1
ORDER BY category, name;

Laukiamo rezultato pavyzdys

Taip atrodo teisingas atsakymas — jo eilučių skaičius yra savitas, jis neprivalo sutapti su schemos lentelėmis.

name	category
Lamp	home
Mug	kitchen
Notebook	office

Prisijunkite, kad pateiktumėte

Jūsų užklausos rezultatas bus rodomas čia