solve: 17/17 Aggregation problems added.

hackerrank-solutions/03 - Aggregation/01-revising-aggregations-the-count-function.sql

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+-- Query a count of the number of cities in CITY having a Population larger than .
+
+SELECT COUNT(id)
+FROM   City
+WHERE  population > 100000

hackerrank-solutions/03 - Aggregation/02-revising-aggregations-the-sum-function.sql

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+-- Query the total population of all cities in CITY where District is California.
+
+SELECT SUM(population)
+FROM   City
+WHERE  district='California'

hackerrank-solutions/03 - Aggregation/03-revising-aggregations-averages.sql

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+-- Query the average population of all cities in CITY where District is California.
+
+SELECT AVG(population)
+FROM   City
+WHERE  district='California'

hackerrank-solutions/03 - Aggregation/04-average-population.sql

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+-- Query the average population for all cities in CITY, rounded down to the nearest
+
+SELECT ROUND(AVG(population))
+FROM   City

hackerrank-solutions/03 - Aggregation/05-japan-population.sql

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+-- Query the sum of the populations for all Japanese cities in CITY. The COUNTRYCODE for Japan is JPN.
+
+SELECT SUM(population)
+FROM   City
+WHERE  COUNTRYCODE='JPN'

hackerrank-solutions/03 - Aggregation/06-population-density-difference.sql

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+-- Query the difference between the maximum and minimum populations in CITY.
+
+SELECT MAX(population) - MIN(population)
+FROM City

hackerrank-solutions/03 - Aggregation/07-the-blunder.sql

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+-- Write a query calculating the amount of error (i.e.:  average monthly salaries), and round it up to the next integer.
+
+SELECT CEIL(AVG(salary) - AVG(REPLACE(salary, '0', '')))
+FROM Employees
+
+
+-- REMARKS : 
+-- did not read the question properly and was using ROUND instead of CEIL. 

hackerrank-solutions/03 - Aggregation/08-top-earners.sql

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+-- Write a query to find the maximum total earnings for all employees as well as the total number of employees who have maximum total earnings. Then print these values as  space-separated integers.
+
+SELECT   (salary * months) as total_salary,
+         Count(*)
+FROM     employee
+GROUP BY 1
+ORDER BY 1 DESC
+LIMIT    1; 

hackerrank-solutions/03 - Aggregation/09-weather-observation-station-2.sql

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+-- Query the following two values from the STATION table:
+
+-- The sum of all values in LAT_N rounded to a scale of  decimal places.
+-- The sum of all values in LONG_W rounded to a scale of  decimal places.
+
+
+SELECT ROUND(SUM(LAT_N), 2), ROUND(SUM(LONG_W), 2)
+FROM   Station

hackerrank-solutions/03 - Aggregation/10-weather-observation-station-13.sql

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+-- Query the sum of Northern Latitudes (LAT_N) from STATION having values greater than 38.7880 and less than 137.2345 . Truncate your answer to  decimal places.
+
+SELECT ROUND(SUM(LAT_N), 4)
+FROM   Station
+WHERE  LAT_N BETWEEN 38.7880 AND 137.2345

hackerrank-solutions/03 - Aggregation/11-weather-observation-station-14.sql

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+-- Query the greatest value of the Northern Latitudes (LAT_N) from STATION that is less than . Truncate your answer to  decimal places.
+
+SELECT ROUND(MAX(LAT_N), 4)
+FROM   Station
+WHERE  LAT_N < 137.2345

hackerrank-solutions/03 - Aggregation/12-weather-observation-station-15.sql

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+-- Query the greatest value of the Northern Latitudes (LAT_N) from STATION that is less than 137.2345 . Truncate your answer to  decimal places.
+
+SELECT   ROUND(LONG_W, 4)
+FROM     Station
+WHERE    LAT_N < 137.2345
+ORDER BY LAT_N DESC
+LIMIT    1

hackerrank-solutions/03 - Aggregation/13-weather-observation-station-16.sql

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+-- Query the smallest Northern Latitude (LAT_N) from STATION that is greater than 38.7780. Round your answer to  decimal places.
+
+
+SELECT   ROUND(LAT_N, 4)
+FROM     Station
+WHERE    LAT_N > 38.7780
+ORDER BY LAT_N 
+LIMIT    1

hackerrank-solutions/03 - Aggregation/14-weather-observation-station-17.sql

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+-- Query the Western Longitude (LONG_W)where the smallest Northern Latitude (LAT_N) in STATION is greater than 38.7780. Round your answer to  decimal places.
+
+
+SELECT   ROUND(LONG_W, 4)
+FROM     Station
+WHERE    LAT_N > 38.7780
+ORDER BY LAT_N 
+LIMIT    1

hackerrank-solutions/03 - Aggregation/15-weather-observation-station-18.sql

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+-- Query the Manhattan Distance between points P1 and P2 and round it to a scale of 4 decimal places.
+
+SELECT Round(Abs(Min(lat_n)-Max(lat_n))
+             + Abs(Min(long_w)-Max(long_w)), 4)
+FROM   station; 

hackerrank-solutions/03 - Aggregation/16-weather-observation-station-19.sql

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+-- Query the Euclidean Distance between points P1 and P2 and round it to a scale of 4 decimal places.
+
+SELECT Round(Sqrt(Power(Max(lat_n) - Min(lat_n), 2)
+                  + Power(Max(long_w) - Min(long_w), 2)), 4)
+FROM   station; 

hackerrank-solutions/03 - Aggregation/17-weather-observation-station-20.sql

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+-- A median is defined as a number separating the higher half of a data set from the lower half. Query the median of the Northern Latitudes (LAT_N) from STATION and round your answer to  decimal places.
+
+
+SELECT ROUND(MEDIAN(LAT_N), 4)
+FROM   Station;