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Commitfbd06e7

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‎README.md

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@@ -642,6 +642,7 @@ Your ideas/fixes/algorithms are more than welcome!
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|577|[Employee Bonus](https://leetcode.com/problems/employee-bonus/)|[Solution](../master/database/_577.sql)||| Easy|
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|574|[Winning Candidate](https://leetcode.com/problems/winning-candidate/)|[Solution](../master/database/_574.sql)||| Medium|
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|570|[Managers with at Least 5 Direct Reports](https://leetcode.com/problems/managers-with-at-least-5-direct-reports/)|[Solution](../master/database/_570.sql)||| Medium|
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|569|[Median Employee Salary](https://leetcode.com/problems/median-employee-salary/)|[Solution](../master/database/_569.sql)||| Hard|
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|262|[Trips and Users](https://leetcode.com/problems/trips-and-users/)|[Solution](../master/database/_262.sql)||| Hard| Inner Join
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|197|[Rising Temperature](https://leetcode.com/problems/rising-temperature/)|[Solution](../master/database/_197.sql)| O(n^2)|O(n)| Easy|
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|196|[Delete Duplicate Emails](https://leetcode.com/problems/delete-duplicate-emails/)|[Solution](../master/database/_196.sql)| O(n^2)|O(n)| Easy|

‎database/_569.sql

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--569. Median Employee Salary
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-- The Employee table holds all employees. The employee table has three columns: Employee Id, Company Name, and Salary.
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--
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--+-----+------------+--------+
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--|Id | Company | Salary |
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--+-----+------------+--------+
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--|1 | A | 2341 |
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--|2 | A | 341 |
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--|3 | A | 15 |
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--|4 | A | 15314 |
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--|5 | A | 451 |
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--|6 | A | 513 |
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--|7 | B | 15 |
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--|8 | B | 13 |
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--|9 | B | 1154 |
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--|10 | B | 1345 |
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--|11 | B | 1221 |
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--|12 | B | 234 |
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--|13 | C | 2345 |
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--|14 | C | 2645 |
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--|15 | C | 2645 |
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--|16 | C | 2652 |
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--|17 | C | 65 |
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--+-----+------------+--------+
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--
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--Write a SQL query to find the median salary of each company. Bonus points if you can solve it without using any built-in SQL functions.
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--
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--+-----+------------+--------+
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--|Id | Company | Salary |
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--+-----+------------+--------+
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--|5 | A | 451 |
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--|6 | A | 513 |
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--|12 | B | 234 |
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--|9 | B | 1154 |
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--|14 | C | 2645 |
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--+-----+------------+--------+
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select Id, Company, Salaryfrom
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(
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selecte.Id,e.Salary,e.Company, if( @prev=e.Company , @Rank := @Rank+1, @Rank :=1)as rank, @prev :=e.Company
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from Employee e , (select @Rank :=0, @prev :=0)as temporder bye.Company,e.Salary,e.Id
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) Ranking
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INNER JOIN
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(
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selectcount(*)as totalcount, Companyas namefrom Employee e2group bye2.Company
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) companycount
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oncompanycount.name=Ranking.Company
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where Rank= floor((totalcount+1)/2)or Rank= floor((totalcount+2)/2)

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