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Problem solution · SQL

Find Category Recommendation Pairs

Find Category Recommendation Pairs: 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
42 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 Find Category Recommendation Pairs, 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

  • 42 lines of SQL from the credited upstream file 3554.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 codeFind Category Recommendation Pairs · SQLSQL
Use this to learn the idea, then write your own version.
WITH  UserCategories AS (    SELECT DISTINCT      ProductPurchases.user_id,      ProductInfo.category    FROM ProductPurchases    INNER JOIN ProductInfo      USING (product_id)  ),  CategoryPairs AS (    SELECT      UserCategories.user_id,      LEAST(        UserCategories.category,        UserCategories2.category      ) AS category1,      GREATEST(        UserCategories.category,        UserCategories2.category      ) AS category2    FROM UserCategories    INNER JOIN UserCategories AS UserCategories2      ON (        UserCategories.user_id = UserCategories2.user_id        AND UserCategories.category < UserCategories2.category)  ),  PairCustomerCounts AS (    SELECT      category1,      category2,      COUNT(DISTINCT user_id) AS customer_count    FROM CategoryPairs    GROUP BY 1, 2  )SELECT  category1,  category2,  customer_countFROM PairCustomerCountsWHERE customer_count >= 3ORDER BY customer_count DESC, category1, category2;

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