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This repository was archived by the owner on Sep 8, 2025. It is now read-only.

Commit 1173045

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department-highest-salary
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‎MySQL/department-highest-salary.sql‎

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# Time: O(n^2)
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# Space: O(n)
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SELECT d.Department AS Department, e.Name AS Employee, d.Salary AS Salary
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FROM (SELECT Department.Id AS DepartmentId, Department.Name AS Department, emp.Salary AS Salary
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FROM Department JOIN (SELECT DepartmentId, MAX(Salary) AS Salary FROM Employee GROUP BY DepartmentId) emp
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ON Department.Id = emp.DepartmentId) d
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JOIN Employee e
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ON e.DepartmentId = d.DepartmentId and e.Salary = d.Salary
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SELECT Department.Name AS Department, Employee.Name AS Employee, Employee.Salary AS Salary
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FROM Department JOIN Employee ON Employee.DepartmentId = Department.Id
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WHERE Employee.Salary IN (SELECT MAX(e.Salary) FROM Employee e WHERE e.DepartmentId = Employee.DepartmentId)
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SELECT DEPARTMENT,EMPLOYEE,SALARY FROM
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(SELECT D.NAME AS DEPARTMENT,E.NAME AS EMPLOYEE,E.SALARY,RANK()OVER(PARTITION BY D.ID ORDER BY SALARY DESC)
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AS RN FROM DEPARTMENT D JOIN EMPLOYEE E ON E.DEPARTMENTID = D.ID)TEMP WHERE RN=1;
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/** SOLUTION TWO **/
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SELECT
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D.name AS Department,rankedE.name AS Employee,rankedE.salary AS Salary
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FROM (
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SELECT
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RANK() OVER (
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PARTITION BY E.departmentId ORDER BY E.salary DESC) AS s_rank
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,id,name,salary,departmentId
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FROM Employee E
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) rankedE
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LEFT JOIN Department D
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ON rankedE.departmentId = D.id
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WHERE rankedE.s_rank = 1
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