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목록전체 글 (88)
민희의 코딩일지
https://school.programmers.co.kr/learn/courses/30/lessons/389479 프로그래머스SW개발자를 위한 평가, 교육의 Total Solution을 제공하는 개발자 성장을 위한 베이스캠프programmers.co.kr 풀이 24시간 동안만 확인하면 되기 때문에 시간 복잡도는 고려하지 않았다.s 라는 배열을 만들어 각각 시간에 사용할 수 있는 서버의 수를 입력해두고, 사람이 많아진다면 추가하는 방식으로 풀었다. 이용자의 수가 0보다 큰 경우, 증설해야 하는 서버의 수를 수식으로 구해서 s 배열의 인덱스 i 부터 i+k 전까지 더해주었다. 테스트 케이스 1 에서 m=3 이기 때문에 0~2 구간: 증설 x 3~5 구간: 1대 증설6~8 구간: 2대 증설 로 증설해야 하는..
https://solvesql.com/problems/volleyball-players-in-two-consecutive-olympics/select distinct t1.athlete_id as id, t1.name as namefrom ( select distinct a.id as athlete_id, a.name, g.year as game_year from records as r join events as e on r.event_id = e.id join teams as t on r.team_id = t.id join games as g on r.game_id = g.id join athletes as a on r.athlete_id = a.id where e.event = 'Vol..
https://solvesql.com/problems/mentor-mentee-list/select e.employee_id as mentee_id, e.name as mentee_name, o.employee_id as mentor_id, o.name as mentor_namefrom ( select employee_id, name, department from employees where join_date > date_sub('2021-12-31', interval 3 month) and join_date o.departmentorder by mentee_id, mentor_id;
https://solvesql.com/problems/multiple-medalist/select a.name from ( select a.id, count(distinct team_id) from records as r join games as g on g.id = r.game_id join athletes as a on r.athlete_id = a.id where r.medal is not null and g.year >= 2000 group by a.id having count(distinct team_id) > 1 ) as t join athletes as a on t.id = a.idorder by a.name;
https://solvesql.com/problems/artists-without-artworks/select m.artist_id, m.name from artists as m left join artworks_artists as a on m.artist_id = a.artist_idwhere m.death_year is not null and a.artist_id is null;
https://solvesql.com/problems/characteristics-of-orders/select region as Region, count(distinct(case when category = 'Furniture' then order_id end)) as Furniture, count(distinct(case when category = 'Office Supplies' then order_id end)) as "Office Supplies", count(distinct(case when category = 'Technology' then order_id end)) as Technologyfrom recordsgroup by regionorder by region;
https://solvesql.com/problems/installment-month/select payment_installments, count(distinct order_id) as order_count, min(payment_value) as min_value, max(payment_value) as max_value, sum(payment_value) / count(*) as avg_valuefrom olist_order_payments_datasetwhere payment_type = 'credit_card'group by payment_installments;
https://solvesql.com/problems/redefine-session/WITH t1 AS ( SELECT user_pseudo_id, event_timestamp_kst, LAG(event_timestamp_kst) OVER ( PARTITION BY user_pseudo_id ORDER BY event_timestamp_kst ) AS prev_time FROM ga WHERE user_pseudo_id = 'S3WDQCqLpK'),t2 AS ( SELECT user_pseudo_id, event_timestamp_kst, CASE ..