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'; |
Problem Statement A description of the problem can be found on Hackerrank. Solution Select only Names and filter by CountryCode equals to ‘JPN’. I created solution in: SQL All solutions are also available on my GitHub. SQL
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select Name from City where CountryCode = 'JPN'; |
Problem Statement A description of the problem can be found on Hackerrank. Solution The difference between Revising the Select Query – 1 is that you select only name column and filter by same conditions, only the population number is higher. I created solution in: SQL All solutions are also available on my GitHub. SQL
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Select name from City where CountryCode='USA' and population > 120000; |
Problem Statement A description of the problem can be found on Hackerrank. Solution Select all columns and filter by CountryCode and Population. I created solution in: SQL All solutions are also available on my GitHub. SQL
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Select name from City where CountryCode='USA' and population > 120000; |
Problem Statement A description of the problem can be found on Hackerrank. Solution Simple. Just select all rows and columns. I created solution in: SQL All solutions are also available on my GitHub. SQL
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select * from City; |