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Problem solution · SQL

Friday Purchases II

Friday Purchases II: a SQL solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
20 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

Depth-first search

For Friday Purchases II, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 20 lines of SQL from the credited upstream file 2994.sql.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 codeFriday Purchases II · SQLSQL
Use this to learn the idea, then write your own version.
WITH  Recursive Fridays AS (     -- the first Friday of November 2023    SELECT '2023-11-03' AS friday    UNION ALL    SELECT DATE_ADD(friday, INTERVAL 7 DAY)    FROM Fridays    -- the last day of November 2023 - 7 days    WHERE friday < '2023-11-23'  )SELECT  WEEK(Fridays.friday, 1) - WEEK('2023-11-01', 1) + 1 AS week_of_month,  Fridays.friday AS purchase_date,  IFNULL(SUM(Purchases.amount_spend), 0) AS total_amountFROM FridaysLEFT JOIN Purchases  ON (Fridays.friday = Purchases.purchase_date)GROUP BY Fridays.fridayORDER BY week_of_month;

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