CPMagical CP Library

Problem solution · SQL

Find Third Transaction

Find Third Transaction: 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
28 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 Third Transaction, 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

  • 28 lines of SQL from the credited upstream file 2986.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 Third Transaction · SQLSQL
Use this to learn the idea, then write your own version.
WITH  TransactionNeighbors AS (    SELECT      user_id,      spend,      transaction_date,      RANK() OVER(PARTITION BY user_id ORDER BY transaction_date) AS date_rank,      FIRST_VALUE(spend) OVER(        PARTITION BY user_id        ORDER BY transaction_date      ) AS first_spend,      LAG(spend) OVER(        PARTITION BY user_id        ORDER BY transaction_date      ) AS second_spend    FROM Transactions  )SELECT  user_id,  spend AS third_transaction_spend,  transaction_date AS third_transaction_dateFROM TransactionNeighborsWHERE  date_rank = 3  AND spend > first_spend  AND spend > second_spendORDER BY 1;

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗