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Problem solution · SQL

The Most Recent Three Orders

The Most Recent Three Orders: 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
23 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 Most Recent Three Orders, 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

  • 23 lines of SQL from the credited upstream file 1532.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 Most Recent Three Orders · SQLSQL
Use this to learn the idea, then write your own version.
WITH  OrdersWithRowNumber AS (    SELECT      order_id,      order_date,      customer_id,      ROW_NUMBER() OVER(        PARTITION BY customer_id        ORDER BY order_date DESC      ) AS `row_number`    FROM Orders  )SELECT  Customers.name AS customer_name,  Customers.customer_id,  OrdersWithRowNumber.order_id,  OrdersWithRowNumber.order_dateFROM OrdersWithRowNumberINNER JOIN Customers  USING (customer_id)WHERE `row_number` <= 3ORDER BY customer_name, customer_id, order_date DESC;

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