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Problem solution · SQL

Consecutive Transactions with Increasing Amounts

Consecutive Transactions with Increasing Amounts: 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
37 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 Consecutive Transactions with Increasing Amounts, 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

  • 37 lines of SQL from the credited upstream file 2701.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 codeConsecutive Transactions with Increasing Amounts · SQLSQL
Use this to learn the idea, then write your own version.
WITH  IncreasingTransactions AS (    SELECT      Curr.customer_id,      Curr.transaction_date    FROM Transactions AS Curr    LEFT JOIN Transactions AS Next      USING (customer_id)    WHERE      Curr.amount < Next.amount      AND DATEDIFF(Next.transaction_date, Curr.transaction_date) = 1  ),  IncreasingTransactionsWithGroupId AS (    SELECT      *,      TO_DAYS(transaction_date) - ROW_NUMBER() OVER(        PARTITION BY customer_id        ORDER BY transaction_date      ) AS group_id    FROM IncreasingTransactions  ),  IncreasingTransactionsWithCountDays AS (    SELECT      customer_id,      MIN(transaction_date) AS consecutive_start,      COUNT(*) AS count_days    FROM IncreasingTransactionsWithGroupId    GROUP BY customer_id, group_id  )SELECT  customer_id,  consecutive_start,  DATE_ADD(consecutive_start, INTERVAL count_days DAY) AS consecutive_endFROM IncreasingTransactionsWithCountDaysWHERE count_days >= 2ORDER BY 1;

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