Problem Statement A description of the problem can be found on Hackerrank. Solution 2 queries: One that selects cities that are ordered by city length ascending and printed first row containing city and its length. Second is almost the same as first but cities are ordered descending. In both queries I used second ordering by […]
Problem Statement A description of the problem can be found on Hackerrank. Solution Check if city starts with one of (‘a’, ‘i’, ‘e’, ‘o’, ‘u’) and orders them alphabetically. I created solution in: SQL All solutions are also available on my GitHub. SQL
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select distinct city from station where lower(substr(city, 1, 1)) in ('a', 'e', 'i', 'o', 'u') order by city asc; |
Problem Statement A description of the problem can be found on Hackerrank. Solution Almost the same principle as Weather Observation Station 6, only substring is different. I created solution in: SQL All solutions are also available on my GitHub. SQL
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select distinct city from station where substr(lower(city), length(city), 1) in ('a', 'e', 'i', 'o', 'u') order by city; |
Problem Statement A description of the problem can be found on Hackerrank. Solution Joins conditions of Weather Observation Station 6 and Weather Observation Station 7 I created solution in: SQL All solutions are also available on my GitHub. SQL
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select distinct city from station where substr(lower(city), 1, 1) in ('a', 'e', 'i', 'o', 'u') AND substr(lower(city), length(city), 1) in ('a', 'e', 'i', 'o', 'u') order by city; |
Problem Statement A description of the problem can be found on Hackerrank. Solution As Weather Observation Station 6, only the condition is opposite. I created solution in: SQL All solutions are also available on my GitHub. SQL
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select distinct city from station where lower(substr(city, 1, 1)) not in ('a', 'e', 'i', 'o', 'u') order by city asc; |
Problem Statement A description of the problem can be found on Hackerrank. Solution As Weather Observation Station 7, only the condition is opposite. I created solution in: SQL All solutions are also available on my GitHub. SQL
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select distinct city from station where substr(lower(city), length(city), 1) not in ('a', 'e', 'i', 'o', 'u') order by city; |
Problem Statement A description of the problem can be found on Hackerrank. Solution As Weather Observation Station 8, only the condition is opposite. I created solution in: SQL All solutions are also available on my GitHub. SQL
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select distinct city from station where substr(lower(city), 1, 1) not in ('a', 'e', 'i', 'o', 'u') or substr(lower(city), length(city), 1) not in ('a', 'e', 'i', 'o', 'u') order by city; |
Problem Statement A description of the problem can be found on Hackerrank. Solution Only difference between Weather Observation Station 11 is ‘AND’. I created solution in: SQL All solutions are also available on my GitHub. SQL
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select distinct city from station where substr(lower(city), 1, 1) not in ('a', 'e', 'i', 'o', 'u') and substr(lower(city), length(city), 1) not in ('a', 'e', 'i', 'o', 'u') order by city; |
Problem Statement A description of the problem can be found on Hackerrank. Solution I created solution in: SQL All solutions are also available on my GitHub. SQL
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select name from ( select name, substr(name, length(name)-2) as sub_name from Students where marks > 75 order by sub_name, id asc ); |
Problem Statement A description of the problem can be found on Hackerrank. Solution Select all columns and filter if CoutryCode equals ‘JPN’. I created solution in: SQL All solutions are also available on my GitHub. SQL
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select * from City where CountryCode = 'JPN'; |