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Problem solution · SQL

The Airport With the Most Traffic

The Airport With the Most Traffic: 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
19 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 The Airport With the Most Traffic, 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

  • 19 lines of SQL from the credited upstream file 2112.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 codeThe Airport With the Most Traffic · SQLSQL
Use this to learn the idea, then write your own version.
WITH  AirportToCount AS (    SELECT departure_airport AS airport_id, flights_count    FROM Flights    UNION ALL    SELECT arrival_airport, flights_count    FROM Flights  ),  RankedAirports AS (    SELECT      airport_id,      RANK() OVER(ORDER BY SUM(flights_count) DESC) AS `rank`    FROM AirportToCount    GROUP BY 1  )SELECT airport_idFROM RankedAirportsWHERE `rank` = 1;

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