Cats Who Code

-- USING OUTER JOIN SELECT o.ANIMAL_ID, o.NAME FROM ANIMAL_INS AS i RIGHT OUTER JOIN ANIMAL_OUTS AS oON i.ANIMAL_ID = o.ANIMAL_ID WHERE i.ANIMAL_ID is null ORDER BY o.ANIMAL_ID-- USING NOT EXIST SELECT o.ANIMAL_ID, o.NAME FROM ANIMAL_OUTS AS o WHERE NOT EXISTS (SELECT 1 FROM ANIMAL_INS AS i WHERE i.ANIMAL_ID = o.ANIMAL_ID) ORDER BY o.ANIMAL_ID prompt 이해하기 - OUTS 에는 있지만 INS에는 존재 ..

#DATE_FORMAT - NOT NEEDED FOR COMPARISON HERE. #IT CONVERTS TO A DATETIIME VALUE INTO A STRING: #A GOOD USE FOR STRING COMPARISON. SELECT i.ANIMAL_ID, i.NAME FROM ANIMAL_INS AS iJOIN ANIMAL_OUTS AS o ON i.ANIMAL_ID = o.ANIMAL_ID WHERE i.DATETIME > o.DATETIME ORDER BY i.DATETIME;SELECT i.ANIMAL_ID, i.NAME FROM ANIMAL_INS as iJOIN ANIMAL_OUTS as o ON i.ANIMAL_ID = o.ANIMAL_ID WHERE DATE_FORMAT(i.D..

DATE_FORMAT(OUT_DATE,'%Y-%m-%d') - needed formatted outputs for the requirement SELECT ORDER_ID, PRODUCT_ID, DATE_FORMAT(OUT_DATE,'%Y-%m-%d'), CASE WHEN OUT_DATE IS NULL THEN '출고미정' WHEN DATE_FORMAT(OUT_DATE,'%Y-%m-%d') # enter else statement OUT_DATE IS NULL implicitly >> explicitily # not necessarily handle null SELECT ORDER_ID, PRODUCT_ID, DATE_FORMAT(OUT_DATE,'%Y-%m-%d'..

SELECT i.ANIMAL_ID, i.NAME FROM ANIMAL_INS i JOIN ANIMAL_OUTS oON i.ANIMAL_ID = o.ANIMAL_ID #WHERE o.datetime IS NOT NULL - NOT NDDED - BOTH DATETIMES IS NOT NULLABLE ORDER BY DATEDIFF(o.DATETIME,i.DATETIME) DESC LIMIT 2

#from just what i had in mind to query SELECT DISTINCT h.CAR_ID FROM CAR_RENTAL_COMPANY_CAR AS cJOIN CAR_RENTAL_COMPANY_RENTAL_HISTORY AS hON c.CAR_ID = h.CAR_ID WHERE CAR_TYPE ='세단' AND START_DATE LIKE '2022-10-%'ORDER BY CAR_ID DESC#USING SUBQUERY - seems to be more adequate for what the question asks SELECT DISTINT CAR_ID FROM CAR_RENTAL_COMPANY_RENTAL_HISTORY WHERE MONTH(START_DATE) = 10 AND..

def solution(n, words): answer = [] wordSet = set() wordSet.add(words[0]) for k in range(1, len(words)): #row int division #col remainder i = (k % n) + 1 #1 based j = (k // n) + 1 if words[k] in wordSet: return [i, j] if words[k][0] != words[k-1][-1]: return [i, j] wordSet.a..

def convert(length): result = "" while length > 0: if length % 2 == 0: result = '0' + result else: result = '1' + result length //= 2 return resultdef solution(s): counter = 0 numOfOnes = 0 numOfZerosToRemove = 0 while len(s) > 1: counter += 1 numOfOnes = 0 for num in s: if num == '1': ..

def solution(s): answer = -1 stack = [] for char in s: if stack and char == stack[-1]: stack.pop() else: stack.append(char) return 1 if not stack else 0

def solution(A,B): answer = 0 n = len(A) A.sort() B.sort(reverse = True) for i in range(n): answer += A[i]*B[i] return answer

def solution(brown, yellow): ''' (w-2) * (h-2) = yellow w * h = brown + yellow // sum i * something = sum then ''' w,h = 0, 0 totalSum = brown + yellow divisors = [] for h in range(2, totalSum+1): if totalSum % h == 0: divisors.append(h) for h in divisors: #(h-2) * (w-2) == yellow # w = sum/h w = totalSum/h ..