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https://www.hackerrank.com/challenges/harry-potter-and-wands/problem?isFullScreen=true Ollivander's Inventory | HackerRankHelp pick out Ron's new wand.www.hackerrank.com select a.id, b.age, a.coins_needed, a.powerfrom wands as ajoin wands_property as b on a.code=b.codewhere is_evil = 0and a.coins_needed = ( select min(w.coins_needed) from wands as w join wands_property as wp on w.code=w..

select h.hacker_id, h.namefrom submissions as s join hackers as h on s.hacker_id = h.hacker_id join challenges as c on s.challenge_id = c.challenge_id join difficulty as d on c.difficulty_level = d.difficulty_levelwhere d.score = s.scoregroup by h.hacker_id, h.namehaving count(s.hacker_id) > 1order by count(s.hacker_id) desc, s.hacker_id asc

select max(case when occupation = 'Doctor' then name end) as Doctor, max(case when occupation = 'Professor' then name end) as Professor, max(case when occupation = 'Singer' then name end) as Singer, max(case when occupation = 'Actor' then name end) as Actorfrom (select name, occupation, row_number() over(partition by occupation order by name as..

select case when grade >= 8 then name else "NULL" end as name, grade, marksfrom students as sjoin grades as g on s.marks between g.min_mark and g.max_markorder by grade desc, name asc, marks asc

select h.hacker_id, h.name, sum(b.score) as sum_scorefrom hackers as hjoin (select hacker_id, challenge_id, max(score) as scorefrom submissionsgroup by hacker_id, challenge_id ) as b on h.hacker_id = b.hacker_idgroup by 1,2having sum_score > 0order by sum_score desc, hacker_id asc

https://www.hackerrank.com/challenges/challenges/problem Challenges | HackerRankPrint the total number of challenges created by hackers.www.hackerrank.comwith t1 as(select hacker_id, count(distinct challenge_id) as c_cntfrom challengesgroup by hacker_id )select t1.hacker_id, hackers.name, c_cntfrom t1join hackers on t1.hacker_id = hackers.hacker_idwhere 1=1 and (t1.c_cnt = (select max(c_cn..

SELECT H.HISTORY_ID, FLOOR( (DATEDIFF(H.END_DATE, H.START_DATE) + 1) * C.DAILY_FEE * (1 - IFNULL(P.DISCOUNT_RATE, 0) / 100) ) AS FEEFROM CAR_RENTAL_COMPANY_CAR CJOIN CAR_RENTAL_COMPANY_RENTAL_HISTORY H ON C.CAR_ID = H.CAR_IDLEFT JOIN CAR_RENTAL_COMPANY_DISCOUNT_PLAN P ON C.CAR_TYPE = P.CAR_TYPE AND ( (DATEDIFF(H.END_DATE, H.START_DATE) + 1 >= 90..

-- 코드를 입력하세요SELECT MONTH(START_DATE) AS MONTH, CAR_ID,COUNT(*) AS RECORDSFROM CAR_RENTAL_COMPANY_RENTAL_HISTORYWHERE 1=1 AND START_DATE BETWEEN '2022-08-01' AND '2022-10-31' AND CAR_ID IN (SELECT CAR_ID FROM CAR_RENTAL_COMPANY_RENTAL_HISTORY WHERE 1=1 AND START_DATE BETWEEN '2022-08-01' AND '2022-10-31' GROUP BY CAR_ID..

order by, limit을 이용할 수 있지만 max 함수 이용해서 출력하기 SELECT *FROM FOOD_PRODUCTWHERE 1=1 AND PRICE = (SELECT MAX(PRICE) FROM FOOD_PRODUCT)

SELECT WAREHOUSE_ID, WAREHOUSE_NAME, ADDRESS, IFNULL(FREEZER_YN, 'N') AS FREEZER_YNFROM FOOD_WAREHOUSEWHERE 1=1 AND ADDRESS LIKE '%경기%'ORDER BY WAREHOUSE_ID ASC