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task 56
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‎README.md‎

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53. Нарисуй треугольник 1 - [Draw The Triangle 1](./tasks/53.md)
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54. Нарисуй треугольник 2 - [Draw The Triangle 2](./tasks/54.md)
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55. Печать простых чисел - [Print Prime Numbers](./tasks/55.md)
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56. 15 дней изучения SQL - [15 Days of Learning SQL](./tasks/56.md)
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## В процессе... , будет дополнено)

‎tasks/56.md‎

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<img src="./art/56.png" alt="solution" >
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#### eng:
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Julia conducted a days of learning SQL contest. The start date of the contest was March 01, 2016 and the end date was March 15, 2016.
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Write a query to print total number of unique hackers who made at least submission each day (starting on the first
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day of the contest), and find the hacker_id and name of the hacker who made maximum number of submissions each day.
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If more than one such hacker has a maximum number of submissions, print the lowest hacker_id. The query should print
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this information for each day of the contest, sorted by the date.
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#### рус:
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Юлия провела конкурс дней обучения SQL. Дата начала конкурса 01.03.2016, дата окончания 15.03.2016.
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Напишите запрос для вывода общего количества уникальных хакеров, которые делали как минимум сабмиты каждый день
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(начиная с первого дня конкурса), и найдите hacker_id и имя хакера, который сделал максимальное количество сабмитов
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каждый день. Если более одного такого хакера имеют максимальное количество представлений, выведите наименьший
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hacker_id. Запрос должен вывести эту информацию для каждого дня конкурса, отсортированного по дате.
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#### код с коментариями:
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```sql
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SELECT
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b.submission_date, COALESCE(a.contSub,0), b.id, c.name
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FROM (
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SELECT o.submission_date, MIN(o.hacker_id) AS id
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FROM (
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SELECT n.submission_date, n.hacker_id,COUNT(n.submission_id)
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FROM Submissions n
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GROUP BY submission_date, hacker_id
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HAVING COUNT(submission_id)>= ALL(
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SELECT COUNT(m.submission_id)
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FROM Submissions m
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GROUP BY m.submission_date, m.hacker_id
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HAVING m.submission_date=n.submission_date
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)
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) AS o
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GROUP BY submission_date
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) AS b
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LEFT JOIN (
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SELECT aa.submission_date, COUNT(DISTINCT aa.hacker_id) AS contSub
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FROM Submissions AS aa
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WHERE aa.submission_date='2016-03-01'
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OR submission_date<ALL(
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SELECT bb.submission_date
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FROM (
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SELECT tdy.hacker_id, tdy.submission_date, ytd.submission_date AS SubmittedYtd
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FROM Submissions tdy LEFT JOIN Submissions ytd
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ON DATE_ADD(tdy.submission_date, INTERVAL -1 DAY)= ytd.submission_date
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AND ytd.hacker_id=tdy.hacker_id
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) AS bb
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WHERE aa.hacker_id=bb.hacker_id AND bb.submission_date<> STR_TO_DATE('2016-03-01', '%Y-%m-%d') AND bb.SubmittedYtd IS NULL
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)
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GROUP BY aa.submission_date
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) AS a
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ON a.submission_date=b.submission_date
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JOIN Hackers c ON b.id=c.hacker_id
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ORDER BY a.submission_date ASC
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```
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#### код для hackerrank:
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```sql
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SELECT
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b.submission_date, COALESCE(a.contSub,0), b.id, c.name
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FROM (
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SELECT o.submission_date, MIN(o.hacker_id) AS id
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FROM (
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SELECT n.submission_date, n.hacker_id,COUNT(n.submission_id)
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FROM Submissions n
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GROUP BY submission_date, hacker_id
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HAVING COUNT(submission_id)>= ALL(
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SELECT COUNT(m.submission_id)
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FROM Submissions m
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GROUP BY m.submission_date, m.hacker_id
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HAVING m.submission_date=n.submission_date
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)
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) AS o
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GROUP BY submission_date
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) AS b
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LEFT JOIN (
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SELECT aa.submission_date, COUNT(DISTINCT aa.hacker_id) AS contSub
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FROM Submissions AS aa
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WHERE aa.submission_date='2016-03-01'
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OR submission_date<ALL(
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SELECT bb.submission_date
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FROM (
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SELECT tdy.hacker_id, tdy.submission_date, ytd.submission_date AS SubmittedYtd
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FROM Submissions tdy LEFT JOIN Submissions ytd
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ON DATE_ADD(tdy.submission_date, INTERVAL -1 DAY)= ytd.submission_date
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AND ytd.hacker_id=tdy.hacker_id
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) AS bb
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WHERE aa.hacker_id=bb.hacker_id AND bb.submission_date<> STR_TO_DATE('2016-03-01', '%Y-%m-%d') AND bb.SubmittedYtd IS NULL
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)
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GROUP BY aa.submission_date
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) AS a
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ON a.submission_date=b.submission_date
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JOIN Hackers c ON b.id=c.hacker_id
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ORDER BY a.submission_date ASC
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```
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‎tasks/56.sql‎

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SELECT
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b.submission_date, COALESCE(a.contSub,0), b.id, c.name
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FROM (
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SELECT o.submission_date, MIN(o.hacker_id) AS id
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FROM (
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SELECT n.submission_date, n.hacker_id,COUNT(n.submission_id)
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FROM Submissions n
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GROUP BY submission_date, hacker_id
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HAVING COUNT(submission_id)>= ALL(
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SELECT COUNT(m.submission_id)
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FROM Submissions m
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GROUP BY m.submission_date, m.hacker_id
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HAVING m.submission_date=n.submission_date
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)
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) AS o
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GROUP BY submission_date
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) AS b
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LEFT JOIN (
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SELECT aa.submission_date, COUNT(DISTINCT aa.hacker_id) AS contSub
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FROM Submissions AS aa
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WHERE aa.submission_date='2016-03-01'
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OR submission_date<ALL(
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SELECT bb.submission_date
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FROM (
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SELECT tdy.hacker_id, tdy.submission_date, ytd.submission_date AS SubmittedYtd
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FROM Submissions tdy LEFT JOIN Submissions ytd
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ON DATE_ADD(tdy.submission_date, INTERVAL -1 DAY)= ytd.submission_date
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AND ytd.hacker_id=tdy.hacker_id
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) AS bb
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WHERE aa.hacker_id=bb.hacker_id AND bb.submission_date<> STR_TO_DATE('2016-03-01', '%Y-%m-%d') AND bb.SubmittedYtd IS NULL
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)
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GROUP BY aa.submission_date
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) AS a
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ON a.submission_date=b.submission_date
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JOIN Hackers c ON b.id=c.hacker_id
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ORDER BY a.submission_date ASC

‎tasks/art/56.png‎

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