지방에서 서울까지

-- 코드를 입력하세요 SELECT A.REST_ID, B.REST_NAME, B.FOOD_TYPE, B.FAVORITES, B.ADDRESS, ROUND(AVG(REVIEW_SCORE), 2) AS SCORE FROM REST_REVIEW A INNER JOIN (SELECT REST_ID, REST_NAME, FOOD_TYPE, FAVORITES, ADDRESS FROM REST_INFO WHERE SUBSTR(ADDRESS, 1, 2) = '서울') B ON A.REST_ID = B.REST_ID GROUP BY 1 ORDER BY SCORE DESC , B.FAVORITES DESC

def solution(A: str, B: str) -> int: result = 0 while result != len(A): if A == B: return result A = A[-1] + A[:-1] result += 1 return -1

SELECT DISTINCT A.CART_ID FROM (SELECT CART_ID, NAME FROM CART_PRODUCTS WHERE NAME = 'Yogurt') A LEFT JOIN (SELECT CART_ID, NAME FROM CART_PRODUCTS WHERE NAME = 'Milk') B ON A.CART_ID = B.CART_ID WHERE A.NAME IS NOT NULL AND B.NAME IS NOT NULL

SELECT ID, NAME, HOST_ID FROM PLACES WHERE PLACES.HOST_ID IN (SELECT HOST_ID FROM (SELECT HOST_ID,COUNT(HOST_ID) AS H_CNT FROM PLACES GROUP BY HOST_ID HAVING H_CNT >= 2)A)

def solution(s): answer = s num_s = {'zero':0, 'one':1, 'two':2, 'three':3, 'four':4, 'five':5, 'six':6, 'seven':7, 'eight':8, 'nine':9} for item in num_s.items(): answer = answer.replace(item[0], str(item[1])) return int(answer)

def solution(strings, n): strings = sorted(strings) answer = sorted(strings, key=lambda x:x[n]) return answer

def solution(array,commands): answer=[] for i in range(len(commands)): a = array[commands[i][0]-1:commands[i][1]] a.sort() answer.append(a[commands[i][2]-1]) return answer

def solution(s): answer = [] ch = {} for i in range(len(s)): if s[i] in ch: answer.append(i - ch[s[i]]) ch[s[i]] = i else: ch[s[i]] = i answer.append(-1) return answer

def solution(food): answer = '' for fd in enumerate(food[1:]): if fd[1] % 2 == 0: answer += str(fd[0]+1) * (fd[1] // 2) else: answer += str(fd[0]+1) * ((fd[1] - 1) // 2) answer += ('0' + answer[::-1]) return answer

def solution(numbers): answer = list() for i in range(len(numbers)): for j in range(i+1, len(numbers)): if numbers[i] + numbers[j] not in answer: answer.append(numbers[i] + numbers[j]) answer.sort() return answer