[Algorithm /프로그래머스] 즐겨찾기가 가장 많은 식당 정보 출력하기

[문제 설명]
다음은 식당의 정보를 담은 REST_INFO 테이블입니다. REST_INFO 테이블은 다음과 같으며 REST_ID, REST_NAME, FOOD_TYPE, VIEWS, FAVORITES, PARKING_LOT, ADDRESS, TEL은 식당 ID, 식당 이름, 음식 종류, 조회수, 즐겨찾기수, 주차장 유무, 주소, 전화번호를 의미합니다.

Column name Type Nullable
REST_ID VARCHAR(5) FALSE
REST_NAME VARCHAR(50) FALSE
FOOD_TYPE VARCHAR(20) TRUE
VIEWS NUMBER TRUE
FAVORITES NUMBER TRUE
PARKING_LOT VARCHAR(1) TRUE
ADDRESS VARCHAR(100) TRUE
TEL VARCHAR(100) TRUE


[문제]
REST_INFO 테이블에서 음식종류별로 즐겨찾기수가 가장 많은 식당의 음식 종류, ID, 식당 이름, 즐겨찾기수를 조회하는 SQL문을 작성해주세요. 이때 결과는 음식 종류를 기준으로 내림차순 정렬해주세요.


[문제 해결 - 조인]

SELECT 
    T1.FOOD_TYPE AS FOOD_TYPE
    , RIF.REST_ID AS REST_ID
    , RIF. REST_NAME AS REST_NAME
      , T1.FAVORITES AS FAVORITES
FROM 
( 
    SELECT
       MAX(FAVORITES) AS FAVORITES,
       FOOD_TYPE
    FROM REST_INFO 
    GROUP BY FOOD_TYPE
)T1
INNER JOIN REST_INFO RIF
ON T1.FOOD_TYPE = RIF.FOOD_TYPE
AND T1.FAVORITES = RIF.FAVORITES
ORDER BY T1.FOOD_TYPE DESC
;

 

[문제 해결 - 서브쿼리]

SELECT
    FOOD_TYPE
    , REST_ID
    , REST_NAME
    , FAVORITES
FROM
    REST_INFO
WHERE (FOOD_TYPE, FAVORITES) IN (
    SELECT
        FOOD_TYPE
        , MAX(FAVORITES)
    FROM 
        REST_INFO
    GROUP BY FOOD_TYPE
)
ORDER BY FOOD_TYPE DESC