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Problem solution · SQL

User Purchase Platform

User Purchase Platform: 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
33 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 User Purchase Platform, 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

  • 33 lines of SQL from the credited upstream file 1127.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 codeUser Purchase Platform · SQLSQL
Use this to learn the idea, then write your own version.
WITH  UserToAmount AS (    SELECT      user_id,      spend_date,      CASE        WHEN COUNT(DISTINCT platform) = 2 THEN 'both'        ELSE platform      END AS platform,      SUM(amount) AS amount    FROM Spending    GROUP BY 1, 2  ),  DateAndPlatforms AS (    SELECT DISTINCT(spend_date), 'desktop' AS platform    FROM Spending    UNION ALL    SELECT DISTINCT(spend_date), 'mobile' AS platform    FROM Spending    UNION ALL    SELECT DISTINCT(spend_date), 'both' AS platform    FROM Spending  )SELECT  DateAndPlatforms.spend_date,  DateAndPlatforms.platform,  IFNULL(SUM(UserToAmount.amount), 0) AS total_amount,  COUNT(DISTINCT UserToAmount.user_id) AS total_usersFROM DateAndPlatformsLEFT JOIN UserToAmount  USING (spend_date, platform)GROUP BY 1, 2;

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