Java 8 stream Interview Questions & Answers

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Table of Contents

No.	Questions
1	Find list of employees whose name starts with alphabet A
2	Group The employees By Department Names
3	Find the total count of employees using stream
4	Find the max age of employees
5	Find all department names
6	Find the count of employee in each department
7	Find the list of employees whose age is less than 30
8	Find the list of employees whose age is in between 26 and 31
9	Find the average age of male and female employee
10	Find the department who is having maximum number of employee
11	Find the Employee who stays in Delhi and sort them by their names
12	Find the average salary in all departments
13	Find the highest salary in each department
14	Find the list of employee and sort them by their salary
15	Find the employee who has second highest salary

1. Find list of employees whose name starts with alphabet A

   List<Employee> aEmp = empList.stream()
     	.filter(emp -> emp.getName().startsWith("A"))
     				.collect(Collectors.toList());

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2. Group The employees By Department Names

      Map<String, List<Employee>> deptMap = empList.stream()
    .collect(Collectors.groupingBy(emp -> emp.getDepartmentName()));

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3. Find the total count of employees using stream

      long empCount = empList.stream().count();

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4. Find the max age of employees

      int maxAge = empList.stream().mapToInt(emp -> emp.getAge()).max().getAsInt();

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5. Find all department names

      List<String> deptNamesList = empList.stream()
    .map(emp -> emp.getDepartmentName()).collect(Collectors.toList());

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6. Find the count of employee in each department

      Map<String, Long> deptCountMap = empList.stream().collect(Collectors.groupingBy(Employee::getDepartmentName, Collectors.counting()));

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7. Find the list of employees whose age is less than 30

      List<Employee> ageList = empList.stream().filter(emp -> emp.getAge() < 30).collect(Collectors.toList());

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8. Find the list of employees whose age is in between 26 and 31

      List<Employee> ageBetween26And30 = employees.stream().filter(emp -> emp.getAge() < 30 && emp.getAge() > 26).collect(Collectors.toList());

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9. Find the average age of male and female employee

      Map<String, Double> avgAgeMap = empList.stream().collect(Collectors.groupingBy(Employee::getGender, Collectors.averagingInt(Employee::getAge)));

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10. Find the department who is having maximum number of employee

   Map.Entry<String, Long> deptMaxCount = empList.stream().collect(Collectors.groupingBy(
 			Employee::getDepartmentName, Collectors.counting()))
 			.entrySet().stream().max(Map.Entry.comparingByValue()).get();

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11. Find the Employee who stays in Delhi and sort them by their names

   List<Employee> employees = empList.stream().filter(emp -> emp.getAddress().equals("Delhi")).sorted(Comparator.comparing(Employee::getName)).collect(Collectors.toList());

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12. Find the average salary in all departments

   Map<String, Double> deptAvgSalary = empList.stream().collect(Collectors.groupingBy(Employee::getDepartmentName, Collectors.averagingDouble(Employee::getSalary)));

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13. Find the highest salary in each department

   Map<String, Optional<Employee>> highestSalForEachDedpt = employees.stream().collect(Collectors.groupingBy(Employee::getDepartNames, Collectors.minBy(Comparator.comparing(Employee::getSalary))));

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14. Find the list of employee and sort them by their salary

   List<Employee> emps = empList.stream().sorted(Comparator.comparing(Employee::getSalary)).collect(Collectors.toList());

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15. Find the employee who has second lowest salary

   Employee emp = empList.stream().sorted(Comparator.comparing(Employee::getSalary)).skip(1).findFirst().get();

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Good luck with your interview 😊

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