How to display the presence and absence of nth-highest group-wise values in SQL

A step-by-step breakdown of the query.
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While skimming through SQL to prepare for interviews, I often come across this question: Find the employee with the highest or (second-highest) salary by joining a table containing employee information with another that contains department information. This raises a further question: What about finding the employee who earns the nth-highest salary department-wide?

Now I want to pose a more complex scenario: What will happen when a department doesn't have an employee earning the nth-highest salary? For example, a department with only two employees will not have an employee earning the third-highest salary.

Here's my approach to this question:

Create department and employee tables

I create a table that includes fields such as dept_id and dept_name.

CREATE TABLE department (
    dept_id int,
    dept_name varchar(60)
);

Now I insert various departments into the new table.

INSERT INTO department (dept_id,dept_name)
VALUES (780,'HR');
INSERT INTO department (dept_id,dept_name)
VALUES (781,'Marketing');
INSERT INTO department (dept_id,dept_name)
VALUES (782,'Sales');
INSERT INTO department (dept_id,dept_name)
VALUES (783,'Web Dev');
A table showing the data from the earlier code snippets with the columns "Department ID" and "Department Name"

Figure 1. The department table (Mohammed Kamil Khan, CC BY-SA 4.0)

Next, I create another table incorporating the fields first_name, last_name, dept_id, and salary.

CREATE TABLE employee (
    first_name varchar(100),
    last_name varchar(100),
    dept_id int,
    salary int
);

Then I insert values into the table:

INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Sam','Burton',781,80000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Peter','Mellark',780,90000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Happy','Hogan',782,110000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Steve','Palmer',782,120000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Christopher','Walker',783,140000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Richard','Freeman',781,85000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Alex','Wilson',782,115000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Harry','Simmons',781,90000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Thomas','Henderson',780,95000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Ronald','Thompson',783,130000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('James','Martin',783,135000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Laurent','Fisher',780,100000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Tom','Brooks',780,85000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Tom','Bennington',783,140000);
A table showing data from the earlier code snippets with first name, last name, dept ID, and salary columns, ordered by department ID number

Figure 2. A table of employees ordered by department ID (Mohammed Kamil Khan, CC BY-SA 4.0)

I can infer the number of employees in each department using this table (department ID:number of employees):

  • 780:4
  • 781:3
  • 782:3
  • 783:4

If I want the view the second-highest-earning employees from different departments, along with their department's name (using DENSE_RANK), the table will be as follows:

A table with department ID, department name, first name, last name, and salary columns, listing the second-highest-earning employee in each of four departments, ordered from lowest to highest salary

Figure 3. The second-highest-earning employee in each department (Mohammed Kamil Khan, CC BY-SA 4.0)

If I apply the same query to find the fourth-highest-earning employees, the output will be only Tom Brooks of department 780 (HR), with a salary of $85,000.

The table listing fourth-highest-earning employees lists only one employee.

Figure 4. The fourth-highest-earning employee (Mohammed Kamil Khan, CC BY-SA 4.0)

Though department 783 (Web Dev) has four employees, two (James Martin and Ronald Thompson) will be classified as the third-highest-earning employees of that department, since the top two earners have the same salary.

Finding the nth highest

Now, to the main question: What if I want to display the dept_ID and dept_name with null values for employee-related fields for departments that do not have an nth-highest-earning employee?

The list of fourth-highest-earning employee by department, showing "null" in the first name, last name, and salary columns for departments that do not have a fourth-highest earner.

Figure 5. All departments listed, whether or not they have an nth-highest-earning employee (Mohammed Kamil Khan, CC BY-SA 4.0)

The table displayed in Figure 5 is what I am aiming to obtain when specific departments do not have an nth-highest-earning employee: The marketing, sales, and web dev departments are listed, but the name and salary fields contain a null value.

The ultimate query that helps obtain the table in Figure 5 is as follows:

SELECT * FROM (WITH null1 AS (select A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
from (SELECT * FROM (
SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
FROM employee INNER JOIN department
ON employee.dept_id=department.dept_id) AS k
WHERE rank1=4)A),
full1 AS (SELECT dept_id, dept_name FROM department WHERE dept_id NOT IN (SELECT dept_id FROM null1 WHERE dept_id IS NOT null)),
nulled AS(SELECT
CASE WHEN null1.dept_id IS NULL THEN full1.dept_id ELSE null1.dept_id END,
CASE WHEN null1.dept_name IS NULL THEN full1.dept_name ELSE null1.dept_name END,
first_name,last_name,salary
FROM null1 RIGHT JOIN full1 ON null1.dept_id=full1.dept_id)
SELECT * from null1
UNION
SELECT * FROM nulled
ORDER BY dept_id)
B;

Breakdown of the query

I will break down the query to make it less overwhelming.

Use DENSE_RANK() to display employee and department information (not involving null for the absence of the nth-highest-earning member):

SELECT * FROM (
  SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
   employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
   FROM employee INNER JOIN department
   ON employee.dept_id=department.dept_id) AS k
   WHERE rank1=4

Output:

A table of the fourth-highest earners showing only the department with a fourth-highest earner

Figure 6. The fourth-highest earner (Mohammed Kamil Khan, CC BY-SA 4.0)

Exclude the rank1 column from the table in Figure 6, which identifies only one employee with a fourth-highest salary, even though there are four employees in another department.

select A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
    from (SELECT * FROM (
  SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
   employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
   FROM employee INNER JOIN department
   ON employee.dept_id=department.dept_id) AS k
   WHERE rank1=4)A

Output:

The fourth-highest earner table (table six) without the rank 1 column

Figure 7. The fourth-highest earner table without the rank 1 column (Mohammed Kamil Khan, CC BY-SA 4.0)

Point out the departments from the department table that do not have an nth-highest-earning employee:

SELECT * FROM (WITH null1 AS (select A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
    from (SELECT * FROM (
  SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
   employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
   FROM employee INNER JOIN department
   ON employee.dept_id=department.dept_id) AS k
   WHERE rank1=4)A),
full1 AS (SELECT dept_id, dept_name FROM department WHERE dept_id NOT IN (SELECT dept_id FROM null1 WHERE dept_id IS NOT null))
SELECT * FROM full1)B

Output:

The full1 table listing the departments without a fourth-highest earner by department ID and name: marketing, sales, web dev

Figure 8. The full1 table listing the departments without a fourth-highest earner (Mohammed Kamil Khan, CC BY-SA 4.0)

Replace full1 in the last line of the above code with null1:

SELECT * FROM (WITH null1 AS (select A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
    from (SELECT * FROM (
  SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
   employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
   FROM employee INNER JOIN department
   ON employee.dept_id=department.dept_id) AS k
   WHERE rank1=4)A),
full1 AS (SELECT dept_id, dept_name FROM department WHERE dept_id NOT IN (SELECT dept_id FROM null1 WHERE dept_id IS NOT null))
SELECT * FROM null1)B
The null1 table listing all departments, with null values for those without a fourth-highest earner

Figure 9. The null1 table listing all departments, with null values for those without a fourth-highest earner (Mohammed Kamil Khan, CC BY-SA 4.0)

Now, I fill the null values of the dept_id and dept_name fields in Figure 9 with the corresponding values from Figure 8.

SELECT * FROM (WITH null1 AS (select A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
    from (SELECT * FROM (
  SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
   employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
   FROM employee INNER JOIN department
   ON employee.dept_id=department.dept_id) AS k
   WHERE rank1=4)A),
full1 AS (SELECT dept_id, dept_name FROM department WHERE dept_id NOT IN (SELECT dept_id FROM null1 WHERE dept_id IS NOT null)),
nulled AS(SELECT
CASE WHEN null1.dept_id IS NULL THEN full1.dept_id ELSE null1.dept_id END,
CASE WHEN null1.dept_name IS NULL THEN full1.dept_name ELSE null1.dept_name END,
first_name,last_name,salary
FROM null1 RIGHT JOIN full1 ON null1.dept_id=full1.dept_id)
SELECT * from nulled) B;
The table with department id, department name, first name, last name, and salary columns, with null values in the name and salary columns

Figure 10. The result of the nulled query (Mohammed Kamil Khan, CC BY-SA 4.0)

The nulled query uses CASE WHEN on the nulls encountered in the dept_id and dept_name columns of the null1 table and replaces them with the corresponding values in the full1 table. Now all I need to do is apply UNION to the tables obtained in Figure 7 and Figure 10. This can be accomplished by declaring the last query in the previous code using WITH and then UNION-izing it with null1.

SELECT * FROM (WITH null1 AS (select A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
from (SELECT * FROM (
SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
FROM employee INNER JOIN department
ON employee.dept_id=department.dept_id) AS k
WHERE rank1=4)A),
full1 AS (SELECT dept_id, dept_name FROM department WHERE dept_id NOT IN (SELECT dept_id FROM null1 WHERE dept_id IS NOT null)),
nulled AS(SELECT
CASE WHEN null1.dept_id IS NULL THEN full1.dept_id ELSE null1.dept_id END,
CASE WHEN null1.dept_name IS NULL THEN full1.dept_name ELSE null1.dept_name END,
first_name,last_name,salary
FROM null1 RIGHT JOIN full1 ON null1.dept_id=full1.dept_id)
SELECT * from null1
UNION
SELECT * FROM nulled
ORDER BY dept_id)
B;
The complete table: department ID, department name, first name, last name, salary columns. The first row contains the information of the one fourth-highest earner, and the next three columns show the remaining departments, with ID, and null value in the other three columns.

Figure 11. The final result (Mohammed Kamil Khan, CC BY-SA 4.0)

Now I can infer from Figure 11 that marketing, sales, and web dev are the departments that do not have any employees earning the fourth-highest salary.

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