Write a SQL query to get the second highest salary from the table above. Also write a query to find the nth highest salary in SQL, where n can be any number.
select MAX(Salary) from Employee;
Figuring out the answer to find the 2nd highest salary
What if we try to exclude the highest salary value from the result set returned by the SQL that we run? If we remove the highest salary from a group of salary values, then we will have a new group of values whose highest salary is actually the 2nd highest in the original Employee table.
So, if we can somehow select the highest value from a result set that excludes the highest value, then we would actually be selecting the 2nd highest salary value. Think about that carefully and see if you can come up with the actual SQL yourself before you read the answer that we provide below. Here is a small hint to help you get started: you will have to use the “NOT IN” SQL operator.
Solution to finding the 2nd highest salary in SQL
Now, here is what the SQL will look like:
SELECT MAX(Salary) FROM Employee
WHERE Salary NOT IN (SELECT MAX(Salary) FROM Employee )
Running the SQL above would return us “450″, which is of course the 2nd highest salary in the Employee table.
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An explanation of the solution
The SQL above first finds the highest salary value in the Employee table using “(select MAX(Salary) from Employee)”. Then, adding the “WHERE Salary NOT IN” in front basically creates a new set of Salary values that does not include the highest Salary value. For instance, if the highest salary in the Employee table is 200,000 then that value will be excluded from the results using the “NOT IN” operator, and all values except for 200,000 will be retained in the results.
This now means that the highest value in this new result set will actually be the 2nd highest value in the Employee table. So, we then select the max Salary from the new result set, and that gives us 2nd highest Salary in the Employee table. And that is how the query above works.
An alternative solution using the not equals SQL operator
We can actually use the not equals operator – the “<>” – instead of the NOT IN operator as an alternative solution to this problem. This is what the SQL would look like:
select MAX(Salary) from Employee
WHERE Salary <> (select MAX(Salary) from Employee )
How would you write a SQL query to find the Nth highest salary?
What we did above was write a query to find the 2nd highest Salary value in the Employee table. But, another commonly asked interview question is how can we use SQL to find the Nth highest salary, where N can be any number whether it’s the 3rd highest, 4th highest, 5th highest, 10th highest, etc? This is also an interesting question – try to come up with an answer yourself before reading the one below to see what you come up with.
The answer and explanation to finding the nth highest salary in SQL
Here we will present one possible answer to finding the nth highest salary first, and the explanation of that answer after since it’s actually easier to understand that way. Note that the first answer we present is actually not optimal from a performance standpoint since it uses a subquery, but we think that it will be interesting for you to learn about because you might just learn something new about SQL. If you want to see the more optimal solutions first, you can skip down to the sections that says “Find the nth highest salary without a subquery” instead.
The SQL below will give you the correct answer – but you will have to plug in an actual value for N of course. This SQL to find the Nth highest salary should work in SQL Server, MySQL, DB2, Oracle, Teradata, and almost any other RDBMS:
SELECT * /*This is the outer query part */
FROM Employee Emp1
WHERE (N-1) = ( /* Subquery starts here */
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > Emp1.Salary)
How does the query above work?
The query above can be quite confusing if you have not seen anything like it before – pay special attention to the fact that “Emp1″ appears in both the subquery (also known as an inner query) and the “outer” query. The outer query is just the part of the query that is not the subquery/inner query – both parts of the query are clearly labeled in the comments.
The subquery is a correlated subquery
The subquery in the SQL above is actually a specific type of subquery known as a correlated subquery. The reason it is called a correlated subquery is because the the subquery uses a value from the outer query in it’s WHERE clause. In this case that value is the Emp1 table alias as we pointed out earlier. A normal subquery can be run independently of the outer query, but a correlated subquery can NOT be run independently of the outer query. If you want to read more about the differences between correlated and uncorrelated subqueries you can go here: Correlated vs Uncorrelated Subqueries.
The most important thing to understand in the query above is that the subquery is evaluated each and every time a row is processed by the outer query. In other words, the inner query can not be processed independently of the outer query since the inner query uses the Emp1 value as well.
Finding nth highest salary example and explanation
Let’s step through an actual example to see how the query above will actually execute step by step. Suppose we are looking for the 2nd highest Salary value in our table above, so our N is 2. This means that the query will look like this:
SELECT *
FROM Employee Emp1
WHERE (1) = (
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > Emp1.Salary)
You can probably see that Emp1 and Emp2 are just aliases for the same Employee table – it’s like we just created 2 separate clones of the Employee table and gave them different names.
Understanding and visualizing how the query above works
Let’s assume that we are using this data:
Employee Employee ID Salary
3 200
4 800
7 450
For the sake of our explanation, let’s assume that N is 2 – so the query is trying to find the 2nd highest salary in the Employee table. The first thing that the query above does is process the very first row of the Employee table, which has an alias of Emp1.
The salary in the first row of the Employee table is 200. Because the subquery is correlated to the outer query through the alias Emp1, it means that when the first row is processed, the query will essentially look like this – note that all we did is replace Emp1.Salary with the value of 200:
SELECT *
FROM Employee Emp1
WHERE (1) = (
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > 200)
So, what exactly is happening when that first row is processed? Well, if you pay special attention to the subquery you will notice that it’s basically searching for the count of salary entries in the Employee table that are greater than 200. Basically, the subquery is trying to find how many salary entries are greater than 200. Then, that count of salary entries is checked to see if it equals 1 in the outer query, and if so then everything from that particular row in Emp1 will be returned.
Note that Emp1 and Emp2 are both aliases for the same table – Employee. Emp2 is only being used in the subquery to compare all the salary values to the current salary value chosen in Emp1. This allows us to find the number of salary entries (the count) that are greater than 200. And if this number is equal to N-1 (which is 1 in our case) then we know that we have a winner – and that we have found our answer.
But, it’s clear that the subquery will return a 2 when Emp1.Salary is 200, because there are clearly 2 salaries greater than 200 in the Employee table. And since 2 is not equal to 1, the salary of 200 will clearly not be returned.
So, what happens next? Well, the SQL processor will move on to the next row which is 800, and the resulting query looks like this:
SELECT *
FROM Employee Emp1
WHERE (1) = (
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > 800)
Since there are no salaries greater than 800, the query will move on to the last row and will of course find the answer as 450. This is because 800 is greater than 450, and the count will be 1. More precisely, the entire row with the desired salary would be returned, and this is what it would look like:
Employee ID Salary
3 200
4 800
7 450
It’s also worth pointing out that the reason DISTINCT is used in the query above is because there may be duplicate salary values in the table. In that scenario, we only want to count repeated salaries just once, which is exactly why we use the DISTINCT operator.
select MAX(Salary) from Employee;
Figuring out the answer to find the 2nd highest salary
What if we try to exclude the highest salary value from the result set returned by the SQL that we run? If we remove the highest salary from a group of salary values, then we will have a new group of values whose highest salary is actually the 2nd highest in the original Employee table.
So, if we can somehow select the highest value from a result set that excludes the highest value, then we would actually be selecting the 2nd highest salary value. Think about that carefully and see if you can come up with the actual SQL yourself before you read the answer that we provide below. Here is a small hint to help you get started: you will have to use the “NOT IN” SQL operator.
Solution to finding the 2nd highest salary in SQL
Now, here is what the SQL will look like:
SELECT MAX(Salary) FROM Employee
WHERE Salary NOT IN (SELECT MAX(Salary) FROM Employee )
Running the SQL above would return us “450″, which is of course the 2nd highest salary in the Employee table.
Subscribe to our newsletter for more free interview questions.
An explanation of the solution
The SQL above first finds the highest salary value in the Employee table using “(select MAX(Salary) from Employee)”. Then, adding the “WHERE Salary NOT IN” in front basically creates a new set of Salary values that does not include the highest Salary value. For instance, if the highest salary in the Employee table is 200,000 then that value will be excluded from the results using the “NOT IN” operator, and all values except for 200,000 will be retained in the results.
This now means that the highest value in this new result set will actually be the 2nd highest value in the Employee table. So, we then select the max Salary from the new result set, and that gives us 2nd highest Salary in the Employee table. And that is how the query above works.
An alternative solution using the not equals SQL operator
We can actually use the not equals operator – the “<>” – instead of the NOT IN operator as an alternative solution to this problem. This is what the SQL would look like:
select MAX(Salary) from Employee
WHERE Salary <> (select MAX(Salary) from Employee )
How would you write a SQL query to find the Nth highest salary?
What we did above was write a query to find the 2nd highest Salary value in the Employee table. But, another commonly asked interview question is how can we use SQL to find the Nth highest salary, where N can be any number whether it’s the 3rd highest, 4th highest, 5th highest, 10th highest, etc? This is also an interesting question – try to come up with an answer yourself before reading the one below to see what you come up with.
The answer and explanation to finding the nth highest salary in SQL
Here we will present one possible answer to finding the nth highest salary first, and the explanation of that answer after since it’s actually easier to understand that way. Note that the first answer we present is actually not optimal from a performance standpoint since it uses a subquery, but we think that it will be interesting for you to learn about because you might just learn something new about SQL. If you want to see the more optimal solutions first, you can skip down to the sections that says “Find the nth highest salary without a subquery” instead.
The SQL below will give you the correct answer – but you will have to plug in an actual value for N of course. This SQL to find the Nth highest salary should work in SQL Server, MySQL, DB2, Oracle, Teradata, and almost any other RDBMS:
SELECT * /*This is the outer query part */
FROM Employee Emp1
WHERE (N-1) = ( /* Subquery starts here */
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > Emp1.Salary)
How does the query above work?
The query above can be quite confusing if you have not seen anything like it before – pay special attention to the fact that “Emp1″ appears in both the subquery (also known as an inner query) and the “outer” query. The outer query is just the part of the query that is not the subquery/inner query – both parts of the query are clearly labeled in the comments.
The subquery is a correlated subquery
The subquery in the SQL above is actually a specific type of subquery known as a correlated subquery. The reason it is called a correlated subquery is because the the subquery uses a value from the outer query in it’s WHERE clause. In this case that value is the Emp1 table alias as we pointed out earlier. A normal subquery can be run independently of the outer query, but a correlated subquery can NOT be run independently of the outer query. If you want to read more about the differences between correlated and uncorrelated subqueries you can go here: Correlated vs Uncorrelated Subqueries.
The most important thing to understand in the query above is that the subquery is evaluated each and every time a row is processed by the outer query. In other words, the inner query can not be processed independently of the outer query since the inner query uses the Emp1 value as well.
Finding nth highest salary example and explanation
Let’s step through an actual example to see how the query above will actually execute step by step. Suppose we are looking for the 2nd highest Salary value in our table above, so our N is 2. This means that the query will look like this:
SELECT *
FROM Employee Emp1
WHERE (1) = (
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > Emp1.Salary)
You can probably see that Emp1 and Emp2 are just aliases for the same Employee table – it’s like we just created 2 separate clones of the Employee table and gave them different names.
Understanding and visualizing how the query above works
Let’s assume that we are using this data:
Employee Employee ID Salary
3 200
4 800
7 450
For the sake of our explanation, let’s assume that N is 2 – so the query is trying to find the 2nd highest salary in the Employee table. The first thing that the query above does is process the very first row of the Employee table, which has an alias of Emp1.
The salary in the first row of the Employee table is 200. Because the subquery is correlated to the outer query through the alias Emp1, it means that when the first row is processed, the query will essentially look like this – note that all we did is replace Emp1.Salary with the value of 200:
SELECT *
FROM Employee Emp1
WHERE (1) = (
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > 200)
So, what exactly is happening when that first row is processed? Well, if you pay special attention to the subquery you will notice that it’s basically searching for the count of salary entries in the Employee table that are greater than 200. Basically, the subquery is trying to find how many salary entries are greater than 200. Then, that count of salary entries is checked to see if it equals 1 in the outer query, and if so then everything from that particular row in Emp1 will be returned.
Note that Emp1 and Emp2 are both aliases for the same table – Employee. Emp2 is only being used in the subquery to compare all the salary values to the current salary value chosen in Emp1. This allows us to find the number of salary entries (the count) that are greater than 200. And if this number is equal to N-1 (which is 1 in our case) then we know that we have a winner – and that we have found our answer.
But, it’s clear that the subquery will return a 2 when Emp1.Salary is 200, because there are clearly 2 salaries greater than 200 in the Employee table. And since 2 is not equal to 1, the salary of 200 will clearly not be returned.
So, what happens next? Well, the SQL processor will move on to the next row which is 800, and the resulting query looks like this:
SELECT *
FROM Employee Emp1
WHERE (1) = (
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > 800)
Since there are no salaries greater than 800, the query will move on to the last row and will of course find the answer as 450. This is because 800 is greater than 450, and the count will be 1. More precisely, the entire row with the desired salary would be returned, and this is what it would look like:
Employee ID Salary
3 200
4 800
7 450
It’s also worth pointing out that the reason DISTINCT is used in the query above is because there may be duplicate salary values in the table. In that scenario, we only want to count repeated salaries just once, which is exactly why we use the DISTINCT operator.
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