CPMagical CP Library

Problem solution · SQL

The Number of Seniors and Juniors to Join the Company II

The Number of Seniors and Juniors to Join the Company II: 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
39 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 Number of Seniors and Juniors to Join the Company II, 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

  • 39 lines of SQL from the credited upstream file 2010.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 Number of Seniors and Juniors to Join the Company II · SQLSQL
Use this to learn the idea, then write your own version.
WITH  AccumualtedCandidates AS (    SELECT      employee_id,      experience,      ROW_NUMBER() OVER(        PARTITION BY experience        ORDER BY salary, employee_id      ) AS candidate_count,      SUM(salary) OVER(        PARTITION BY experience        ORDER BY salary, employee_id      ) AS accumulated_salary    FROM Candidates  ),  HiredSeniors AS (    SELECT      employee_id,      accumulated_salary    FROM AccumualtedCandidates    WHERE      experience = 'Senior'      AND accumulated_salary < 70000  )SELECT HiredSeniors.employee_idFROM HiredSeniorsUNION ALLSELECT Juniors.employee_idFROM AccumualtedCandidates AS JuniorsWHERE  experience = 'Junior'  AND Juniors.accumulated_salary < (    SELECT 70000 - IFNULL(MAX(accumulated_salary), 0)    FROM AccumualtedCandidates    WHERE      experience = 'Senior'      AND accumulated_salary < 70000  );

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 ↗