Find N’th Highest Salary Employee in Each Department

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1. Group employees by department
   • Collectors.groupingBy
2. For each department:

   • Sort employees by salary in descending order

     • sorted(Comparator.comparing(Employee::getSalary).reversed())
   • Skip the highest salary employee skip
   • Pick the next one (2nd highest) findFirst
3. Store result as:
   • Map<Department, Optional<Employee>> OR
   • Map<Department, Employee> (if data is guaranteed)

2nd Highest Salary Employee in Each Department

  public static Map<String, Optional<Employee>> getSecondHighestSalaryEmployeeByDepartment(List<Employee> employees) {

        return employees.stream().collect(Collectors.groupingBy(Employee::getDepartment,
                Collectors.collectingAndThen(Collectors.toList(),list -> list.stream().
                        sorted(Comparator.comparing(Employee::getSalary).reversed())
                        .skip(1) 
                        .findFirst())
        ));
    }

First Highest Salary Employee in Each Department

    public static Map<String, Optional<Employee>> getFirstHighestSalaryEmployeeByDepartment(List<Employee> employees) {

        return employees.stream().collect(Collectors.groupingBy(Employee::getDepartment,
                Collectors.collectingAndThen(Collectors.toList(),list ->
                        list.stream().max(Comparator.comparing(Employee::getSalary)))
        ));
    }

Least Salary Employee in Each Department

    public static Map<String, Optional<Employee>> getLeastSalaryEmployeeByDepartment(List<Employee> employees) {

        return employees.stream().collect(Collectors.groupingBy(Employee::getDepartment,
                Collectors.collectingAndThen(Collectors.toList(),list ->
                        list.stream().min(Comparator.comparing(Employee::getSalary)))
        ));
    }