Notice
Recent Posts
Recent Comments
Link
일 | 월 | 화 | 수 | 목 | 금 | 토 |
---|---|---|---|---|---|---|
1 | ||||||
2 | 3 | 4 | 5 | 6 | 7 | 8 |
9 | 10 | 11 | 12 | 13 | 14 | 15 |
16 | 17 | 18 | 19 | 20 | 21 | 22 |
23 | 24 | 25 | 26 | 27 | 28 |
Tags
- Quant
- 볼린저밴드
- 파트5
- 파이썬
- PolynomialFeatures
- GridSearchCV
- 비트코인
- 프로그래머스
- Python
- 데이터분석전문가
- lstm
- randomforest
- 실기
- ADP
- Crawling
- 데이터분석
- SQL
- TimeSeries
- 변동성돌파전략
- 빅데이터분석기사
- sarima
- 코딩테스트
- 파이썬 주식
- 주식
- docker
- Programmers
- hackerrank
- 백테스트
- 토익스피킹
- backtest
Archives
- Today
- Total
데이터 공부를 기록하는 공간
[hackerrank][SQL][Advanced] 본문
https://www.hackerrank.com/domains/sql?filters%5Bskills%5D%5B%5D=SQL%20%28Advanced%29
Solve SQL Code Challenges
A special-purpose language designed for managing data held in a relational database.
www.hackerrank.com
☆ Draw The Triangle 1
SET @num = 21;
SELECT REPEAT("* ", @num:=@num-1)
/* for without table */
FROM INFORMATION_SCHEMA.TABLES
WHERE @num > 0
Draw The Triangle 2
SET @num = 0;
SELECT REPEAT("* ", @num:=@num+1)
FROM INFORMATION_SCHEMA.TABLES
WHERE @num<20
☆Print Prime Numbers
-
☆☆☆15 Days of Learning SQL
▶ STEP 1
SELECT S.submission_date, COUNT(DISTINCT S.hacker_id) cnt
FROM Submissions S,
/* condition : submission_date - 2016.03.01(min of submission_date) */
WHERE (
SELECT COUNT(DISTINCT S2.submission_date)
FROM Submissions S2
WHERE S2.submission_date < S.submission_date
AND S2.hacker_id = S.hacker_id
)
= (S.submission_date - (SELECT MIN(S3.submission_date) FROM Submissions S3))
GROUP BY S.submission_date
ORDER BY S.submission_date
▶ STEP 2
SELECT
/* submission_date */
S.submission_date,
/* count_of_hacker_id's submit each day */
COUNT(DISTINCT S.hacker_id) cnt,
/* hacker_id_max_day */
(
SELECT S3.hacker_id
FROM Submissions S3
WHERE S.submission_date = S3.submission_date
GROUP BY S3.hacker_id
ORDER BY count(S3.submission_id) DESC, hacker_id LIMIT 1
) AS ID,
/* hacker_name_max_day */
(
SELECT H.Name FROM Hackers H WHERE H.hacker_id = ID
)
FROM Submissions S
/* condition : submission_date - 2016.03.01(min of submission_date) */
WHERE (
SELECT COUNT(DISTINCT S2.submission_date)
FROM Submissions S2
WHERE S2.submission_date < S.submission_date
AND S2.hacker_id = S.hacker_id
)
= (S.submission_date - (SELECT MIN(S3.submission_date) FROM Submissions S3))
GROUP BY S.submission_date
ORDER BY S.submission_date
'STUDY > SQL_HACKERRANK' 카테고리의 다른 글
[MySQL] 기본문법 코딩테스트 (0) | 2021.10.20 |
---|---|
[HackerRank][SQL][Advanced Join] Interviews ☆☆ (0) | 2021.10.19 |
[hackerrank][SQL][Intermediate]#3 (0) | 2021.10.18 |
[hackerrank][SQL][Intermediate]#2 (0) | 2021.10.18 |
[hackerrank][SQL][Intermediate] (0) | 2021.10.18 |
Comments