[Algorithm /프로그래머스] 헤비 유저가 소유한 장소

[문제 설명]
PLACES 테이블은 공간 임대 서비스에 등록된 공간의 정보를 담은 테이블입니다. PLACES 테이블의 구조는 다음과 같으며 ID, NAME, HOST_ID는 각각 공간의 아이디, 이름, 공간을 소유한 유저의 아이디를 나타냅니다. ID는 기본키입니다.

NAME TYPE
ID INT
NAME VARCHAR
HOST_ID INT

[문제]
이 서비스에서는 공간을 둘 이상 등록한 사람을 "헤비 유저"라고 부릅니다. 헤비 유저가 등록한 공간의 정보를 아이디 순으로 조회하는 SQL문을 작성해주세요.

[예시]
예를 들어, PLACES 테이블이 다음과 같다면

ID NAME HOST_ID
4431977 BOUTIQUE STAYS - Somerset Terrace, Pet Friendly 760849
5194998 BOUTIQUE STAYS - Elwood Beaches 3, Pet Friendly 760849
16045624 Urban Jungle in the Heart of Melbourne 30900122
17810814 Stylish Bayside Retreat with a Luscious Garden 760849
22740286 FREE PARKING - The Velvet Lux in Melbourne CBD 30900122
22868779 ★ Fresh Fitzroy Pad with City Views! ★ 21058208
760849번 유저는 공간을 3개 등록했으므로 이 유저는 헤비유저입니다.
30900122번 유저는 공간을 2개 등록했으므로 이 유저는 헤비유저입니다.
21058208번 유저는 공간을 1개 등록했으므로 이 유저는 헤비유저가 아닙니다.
따라서 SQL 문을 실행하면 다음과 같이 나와야 합니다.


[문제 해결 -  JOIN]

SELECT
     P.ID AS ID
     , P.NAME AS NAME
     , T1.HOST_ID AS HOST_ID
FROM(
    SELECT
      HOST_ID
    FROM 
        PLACES 
    GROUP BY HOST_ID
      HAVING COUNT(ID) >=2
)T1
    INNER JOIN PLACES P
ON T1.HOST_ID = P.HOST_ID
    ORDER BY ID

 

 

[문제 해결 - 서브쿼리]

SELECT
    ID
    , NAME
    , HOST_ID
FROM
    PLACES 
WHERE (HOST_ID) IN (
    SELECT
        HOST_ID
    FROM PLACES
        GROUP BY HOST_ID
    HAVING COUNT(ID) > =2
)
ORDER BY ID