Problem solution · SQL

Analyze Subscription Conversion

Analyze Subscription Conversion : 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
30 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 Analyze Subscription Conversion , 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

  • 30 lines of SQL from the credited upstream file 3497.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.

Full codeAnalyze Subscription Conversion · SQLSQL
Use this to learn the idea, then write your own version.
WITH  FreeTrial AS (    SELECT user_id, AVG(activity_duration) AS avg_free_trial_duration    FROM UserActivity    WHERE activity_type = 'free_trial'    GROUP BY 1  ),  Paid AS (    SELECT user_id, AVG(activity_duration) AS avg_paid_duration    FROM UserActivity    WHERE activity_type = 'paid'    GROUP BY 1  ),  ConvertedUsers AS (    SELECT DISTINCT FreeTrial.user_id    FROM FreeTrial    INNER JOIN Paid      USING (user_id)  )SELECT  ConvertedUsers.user_id,  ROUND(FreeTrial.avg_free_trial_duration, 2) AS trial_avg_duration,  ROUND(Paid.avg_paid_duration, 2) AS paid_avg_durationFROM ConvertedUsersINNER JOIN FreeTrial  USING (user_id)INNER JOIN Paid  USING (user_id)ORDER BY 1; 

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