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Problem solution · SQL

Find Top Performing Driver

Find Top Performing Driver: 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 Find Top Performing Driver, 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 3308.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 Top Performing Driver · SQLSQL
Use this to learn the idea, then write your own version.
WITH  DriverPerformance AS (    SELECT      Vehicles.fuel_type,      Vehicles.driver_id,      Drivers.accidents,      ROUND(AVG(Trips.rating), 2) AS rating,      SUM(Trips.distance) AS distance    FROM Vehicles    INNER JOIN Trips      USING (vehicle_id)    INNER JOIN Drivers      USING (driver_id)    GROUP BY 1, 2  ),  RankedDrivers AS (    SELECT      *,      RANK() OVER(        PARTITION BY fuel_type        ORDER BY rating DESC, distance DESC, accidents      ) AS `rank`    FROM DriverPerformance  )SELECT fuel_type, driver_id, rating, distanceFROM RankedDriversWHERE `rank` = 1ORDER BY 1;

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