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Problem solution · SQL

Find Peak Calling Hours for Each City

Find Peak Calling Hours for Each City: 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
26 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 Peak Calling Hours for Each City, 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

  • 26 lines of SQL from the credited upstream file 2984.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 Peak Calling Hours for Each City · SQLSQL
Use this to learn the idea, then write your own version.
WITH  CityHourCount AS (    SELECT      city,      HOUR(call_time) AS call_hour,      COUNT(*) AS number_of_calls    FROM Calls    GROUP BY 1, 2  ),  RankedCityHourCount AS (    SELECT      *,      RANK() OVER(        PARTITION BY city        ORDER BY number_of_calls DESC      ) AS `rank`    FROM CityHourCount  )SELECT  city,  call_hour AS peak_calling_hour,  number_of_callsFROM RankedCityHourCountWHERE `rank` = 1ORDER BY 2 DESC, 1 DESC;

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