알고리즘
[SQL]프로그래머스 Lv.3 즐겨찾기가 가장 많은 식당 정보 출력하기
✅ 코드 SELECT T2.FOOD_TYPE AS FOOD_TYPE , T2.REST_ID AS REST_ID , T2.REST_NAME AS REST_NAME , T2.FAVORITES AS FAVORITES FROM (SELECT FOOD_TYPE, MAX(FAVORITES) AS FAVORITES FROM REST_INFO GROUP BY FOOD_TYPE) AS T1 INNER JOIN REST_INFO AS T2 ON T1.FOOD_TYPE = T2.FOOD_TYPE AND T1.FAVORITES = T2.FAVORITES ORDER BY FOOD_TYPE DESC; 서브쿼리를 사용한 풀이입니다. 🔎 참고 SELECT FOOD_TYPE , REST_ID , REST_NAME , MAX(FAVOR..
