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Problem solution · SQL

Find Overlapping Shifts II

Find Overlapping Shifts 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
41 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 Overlapping Shifts 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

  • 41 lines of SQL from the credited upstream file 3268.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 Overlapping Shifts II · SQLSQL
Use this to learn the idea, then write your own version.
WITH  EmployeeTimes AS (    SELECT DISTINCT employee_id, start_time AS `time`    FROM EmployeeShifts    UNION DISTINCT    SELECT DISTINCT employee_id, end_time AS `time`    FROM EmployeeShifts  ),  Segments AS (    SELECT      employee_id,      `time` AS start_time,      LEAD(`time`) OVER(PARTITION BY employee_id ORDER BY `time`) AS end_time    FROM EmployeeTimes  ),  SegmentsCount AS (    SELECT      Segments.*,      COUNT(*) AS concurrent_count    FROM Segments    INNER JOIN EmployeeShifts      USING (employee_id)    WHERE      Segments.start_time >= EmployeeShifts.start_time      AND Segments.end_time <= EmployeeShifts.end_time    GROUP BY 1, 2, 3  )SELECT  employee_id,  MAX(concurrent_count) AS max_overlapping_shifts,  SUM(    concurrent_count * (concurrent_count - 1) / 2 * TIMESTAMPDIFF(      MINUTE,      start_time,      end_time    )  ) AS total_overlap_duration -- C(concurrent_count, 2)FROM SegmentsCountGROUP BY 1ORDER BY 1;

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