CPMagical CP Library

Problem solution · SQL

Merge Overlapping Events in the Same Hall

Merge Overlapping Events in the Same Hall: 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
34 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 Merge Overlapping Events in the Same Hall, 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

  • 34 lines of SQL from the credited upstream file 2494.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 codeMerge Overlapping Events in the Same Hall · SQLSQL
Use this to learn the idea, then write your own version.
WITH  HallEventsWithIsNewEvent AS (    SELECT      hall_id,      start_day,      end_day,      IFNULL(        start_day > MAX(end_day) OVER(          PARTITION BY hall_id          ORDER BY start_day,            end_day DESC ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING        ),        1      ) AS is_new_event    FROM HallEvents  ),  HallEventsWithGroupId AS (    SELECT      hall_id,      start_day,      end_day,      SUM(is_new_event) OVER(        PARTITION BY hall_id        ORDER BY start_day, end_day DESC      ) AS group_id    FROM HallEventsWithIsNewEvent  )SELECT  hall_id,  MIN(start_day) AS start_day,  MAX(end_day) AS end_dayFROM HallEventsWithGroupIdGROUP BY hall_id, group_id;

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 ↗