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Problem solution · SQL

Market Analysis III

Market Analysis III: a SQL solution using relational aggregation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Relational aggregation
Source
walkccc LeetCode Solutions
Length
24 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Relational aggregation

For Market Analysis III, the query transforms and combines relational rows, then filters or aggregates them into the requested result.

  1. Identify the source rows and join keys.
  2. Apply filters before aggregation when possible.
  3. Group, rank, or project the final columns required by the result.

Code notes

  • 24 lines of SQL from the credited upstream file 2922.sql.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected.

Complexity

Review join cardinality, grouping keys, and available indexes when estimating query cost.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMarket Analysis III · SQLSQL
Use this to learn the idea, then write your own version.
WITH  Sellers AS(    SELECT      Users.seller_id,      COUNT(DISTINCT Orders.item_id) AS num_items    FROM Users    INNER JOIN Orders      USING (seller_id)    INNER JOIN Items      USING (item_id)    WHERE Items.item_brand != Users.favorite_brand    GROUP BY 1  ),  RankedSellers AS (    SELECT      seller_id,      num_items,      RANK() OVER(ORDER BY num_items DESC) AS `rank`    FROM Sellers  )SELECT seller_id, num_itemsFROM RankedSellersWHERE `rank` = 1;

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