solve: Advanced-Select - #1,#4,#5 + Aggregation - #1-#8

interviewbit-solutions/02 - Advanced Select/1-conditional-work.sql

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+-- Given a table NUMBERS, determine for each row, if the sum of the numbers are ‘Positive’, ‘Negative’, or ‘Zero’.
+-- Note: Output should be printed with column name A.
+
+SELECT CASE WHEN A+B+C > 0 THEN 'Positive' WHEN A+B+C < 0 THEN 'Negative' ELSE 'Zero' END AS A
+FROM NUMBERS
+
+
+-- REMARKS: 
+-- didnt see input table properly.

interviewbit-solutions/02 - Advanced Select/4-role-counter.sql

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+-- Given a table GAMERS, query for the count of each Role of the Players in the given table. Sort the result based on ascending order of the count and print the result in the following format:
+
+-- [Role] Count is [Role_Count]
+
+
+SELECT   CONCAT(Role, ' Count is ', COUNT(Role)) as COUNT
+FROM     GAMERS
+GROUP BY Role
+ORDER BY COUNT(Role)
+

interviewbit-solutions/02 - Advanced Select/5-vowel-country.sql

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+-- Given a table PLACES, count the number of Countries which end with a vowel.
+
+SELECT COUNT(Country)
+FROM   PLACES 
+WHERE  Country REGEXP '[aeiou]$'
+
+
+-- similar to hackerrank --> Basic select --> last 7-8 problems, all based on REGEX. 

interviewbit-solutions/03 - Aggregation/01-firefighters-average.sql

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+-- Given a table FIREFIGHTERS, query the average of all the people saved by the Firefighters whose CountryCode is PM.
+
+
+SELECT AVG(PeopleSaved)
+FROM   FIREFIGHTERS
+WHERE  CountryCode='PM'

interviewbit-solutions/03 - Aggregation/02-firefighters-sum.sql

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+-- Given a table FIREFIGHTERS, query the sum of all the people saved by the Firefighters whose CountryCode is PG.
+
+SELECT SUM(PeopleSaved)
+FROM   FIREFIGHTERS
+WHERE  CountryCode='PG'

interviewbit-solutions/03 - Aggregation/03-largest-difference.sql

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+-- Given a table FIREFIGHTERS, find the largest difference between the number of people saved by 2 firefighters.
+
+
+SELECT MAX(a.PeopleSaved - b.PeopleSaved) as A
+FROM   FIREFIGHTERS a, FIREFIGHTERS b

interviewbit-solutions/03 - Aggregation/04-distance-maximizer.sql

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+-- Given a table HOUSES, find the manhattan distance of the house which is the farthest from John’s house which lies at coordinates (20, 4).
+
+-- Basically you have to find the manhattan distance of some house which is the maximum amongst all the other houses from John’s house.
+
+-- Manhattan Distance between 2 points P(x1, y1), Q(x2, y2) is given as |x1 - x2| + |y1 - y2|.
+
+
+SELECT MAX(ABS(XCoordinate - 20) + ABS(YCoordinate - 4)) as A
+FROM   HOUSES

interviewbit-solutions/03 - Aggregation/05-euclids-distance.sql

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+-- Given a table HOUSES, find the euclidean distance between points with the largest X and Y coordinate, and the points with the smallest X and Y coordinate.
+
+-- The Euclidean Distance between 2 points P(x1, y1) and Q(x2, y2) is defined as: sqrt((x1 - x2)2 + (y1 - y2)2).
+
+
+SELECT SQRT(POW(MAX(XCoordinate) - MIN(XCoordinate), 2) + POW(MAX(YCoordinate) - MIN(YCoordinate), 2)) as A
+FROM   HOUSES

interviewbit-solutions/03 - Aggregation/06-big-salary.sql

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+-- Given a table WORKERS, find how many workers have the maximum total earnings among all the WORKERS.
+
+-- The total earnings of a worker is calculated as Daily Wage * Number of Days Worked.
+
+
+-- Solution : 
+
+SELECT Count(*) AS A
+FROM   WORKERS
+WHERE  (DailyWage * DaysWorked) = (SELECT Max(DailyWage * DaysWorked)
+                                   FROM   WORKERS) 
+
+
+-- my approaches : 
+
+SELECT MAX(DailyWage * DaysWorked)
+FROM (
+    SELECT COUNT(*) as A
+    FROM WORKERS
+) x
+
+SELECT   COUNT(*) as A
+FROM     WORKERS
+ORDER BY (DailyWage*DaysWorked) DESC
+LIMIT    1

interviewbit-solutions/03 - Aggregation/07-5th-highest-mark.sql

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+-- Given the ‘STUDENTS’ table. Write an SQL query to find the 5’th highest marks in the students table.
+
+SELECT   MARKS
+FROM     STUDENTS
+ORDER BY 1 DESC 
+LIMIT    1
+OFFSET   4
+
+
+-- other approaches :
+
+SELECT MARKS
+FROM   (
+        SELECT * , RANK() OVER(ORDER BY MARKS DESC) AS rank
+        FROM STUDENTS
+        ) AS x
+WHERE  rank=5
+
+
+-- remarks : to remember --> asked many times in interview.
+
+

interviewbit-solutions/03 - Aggregation/08-famous-books.sql

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+-- You are given 2 tables, One containing the available Books and the other containing the Books that have been bought by a customer. You have to find the Id’s of all the ‘Famous’ Books. A book is called ‘Famous’ if it is bought by at least 3 customers.
+
+-- my approach : 
+
+SELECT   b.Id as Id
+FROM     Books b JOIN BoughtBooks bb ON b.Id=bb.BooksId
+GROUP BY bb.BooksId
+HAVING   COUNT(bb.Id) >= 3
+
+
+-- more easy solution: no need for the first table. 
+
+SELECT   BooksId as Id
+FROM     BoughtBooks
+GROUP BY 1
+HAVING   count(BooksId)>=3