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Problem solution · SQL

Active Users

Active Users: 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
29 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 Active Users, 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

  • 29 lines of SQL from the credited upstream file 1454.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 codeActive Users · SQLSQL
Use this to learn the idea, then write your own version.
WITH  DistinctLogins AS (    SELECT DISTINCT * FROM Logins  ),  RankedLogins AS (    SELECT      *,      DENSE_RANK() OVER(        PARTITION BY id        ORDER BY login_date      ) AS `rank`    FROM DistinctLogins  ),  RankedLoginsWithGroupId AS (    SELECT      *,      DATE_ADD(login_date, INTERVAL -`rank` DAY) AS group_id    FROM RankedLogins  )SELECT DISTINCT  id,  Accounts.nameFROM RankedLoginsWithGroupIdINNER JOIN Accounts  USING (id)GROUP BY Accounts.id, RankedLoginsWithGroupId.group_idHAVING COUNT(*) >= 5ORDER BY 1;

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