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  • leetcode - 1126. Active Businesses
    SQL 문제풀이 2022. 7. 18. 16:24

    난이도 : Medium

     

    Table: Events

    +---------------+---------+
    | Column Name   | Type    |
    +---------------+---------+
    | business_id   | int     |
    | event_type    | varchar |
    | occurences    | int     | 
    +---------------+---------+
    (business_id, event_type) is the primary key of this table.
    Each row in the table logs the info that an event of some type occurred at some business for a number of times.
    

     

    The average activity for a particular event_type is the average occurences across all companies that have this event.

    An active business is a business that has more than one event_type such that their occurences is strictly greater than the average activity for that event.

    Write an SQL query to find all active businesses.

    Return the result table in any order.

    The query result format is in the following example.

     

    Example 1:

    Input: 
    Events table:
    +-------------+------------+------------+
    | business_id | event_type | occurences |
    +-------------+------------+------------+
    | 1           | reviews    | 7          |
    | 3           | reviews    | 3          |
    | 1           | ads        | 11         |
    | 2           | ads        | 7          |
    | 3           | ads        | 6          |
    | 1           | page views | 3          |
    | 2           | page views | 12         |
    +-------------+------------+------------+
    Output: 
    +-------------+
    | business_id |
    +-------------+
    | 1           |
    +-------------+
    Explanation:  
    The average activity for each event can be calculated as follows:
    - 'reviews': (7+3)/2 = 5
    - 'ads': (11+7+6)/3 = 8
    - 'page views': (3+12)/2 = 7.5
    The business with id=1 has 7 'reviews' events (more than 5) and 11 'ads' events (more than 8), so it is an active business.

     

    WITH t1 AS (
                SELECT business_id
                     , event_type
                     , SUM(occurences) occurences
                FROM Events
                GROUP BY business_id, event_type
                )
    SELECT business_id
    FROM t1
         JOIN (SELECT event_type
                    , AVG(occurences) avg
               FROM Events
               GROUP BY event_type
               ) t2 on t1.event_type = t2.event_type
                    AND t1.occurences > t2.avg
    GROUP BY business_Id
    HAVING COUNT(*) > 1

    Accepted (75.76%)

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