Finding the Second Highest Salary from Repeating Values

In this article, we will explore a common problem in data analysis: finding the second highest value in a dataset when there are repeating values. This problem can be solved using various techniques, including sorting and ranking.

We will start by examining the given query and identifying its strengths and weaknesses. Then, we will discuss alternative approaches to solving this problem, including using window functions like dense_rank().

Understanding the Problem

The problem is asking us to find the second highest salary from a table that contains multiple employees with the same salary. The table has three columns: emp_name, emp_id, and salary. We will assume that the salaries are integers and that the employees have unique names.

The given query uses the ROW_NUMBER() function to assign a rank to each employee based on their salary in descending order. However, this approach has some limitations. For example, it only works correctly when there are no ties for the same rank.

Examining the Given Query

Let’s examine the given query and identify its strengths and weaknesses:

-- Create a table with three columns: emp_name, emp_id, and salary
CREATE TABLE sals (emp_name VARCHAR(10), emp_id INTEGER, salary INTEGER);

-- Insert some data into the table
INSERT INTO sals VALUES ('rr', 12, 100000);
INSERT INTO sals VALUES ('ed', 13, 100000);
INSERT INTO sals VALUES ('ty', 14, 100000);
INSERT INTO sals VALUES ('we', 15, 80000);
INSERT INTO sals VALUES ('wse', 16, 80000);
INSERT INTO sals VALUES ('wa', 17, 74000);

-- Query the table and assign a rank to each employee based on their salary
SELECT emp_name,
       emp_id,
       salary,
       ROW_NUMBER() OVER (ORDER BY salary DESC) AS rrnk
FROM (
  SELECT emp_name,
         emp_id,
         salary,
         ROW_NUMBER() OVER (PARTITION BY salary ORDER BY salary DESC) AS rnk
  FROM sals
) x
WHERE x.rnk = 2 AND rrnk = 2;

The query has two main issues:

1. The WHERE clause is using both the rrnk and rnk columns, which are derived from different queries. This can lead to incorrect results if the subquery returns a different number of rows for each salary.
2. The query only returns employees who have a rank of 2 in both the main query and the subquery. However, this may not be the second highest salary overall.

Alternative Approaches

Let’s discuss alternative approaches to solving this problem:

Using dense_rank()

One common approach to solving this problem is to use the dense_rank() function instead of ROW_NUMBER(). The dense_rank() function assigns a rank to each row based on the values in the specified column, and it uses the previous value if there are ties.

Here’s an example query that uses dense_rank():

SELECT emp_name,
       emp_id,
       salary,
       dense_rank() OVER (ORDER BY salary DESC) AS dr
FROM sals;

This query assigns a rank to each employee based on their salary in descending order. If two employees have the same salary, they will both receive the same rank.

To find the second highest salary, we can use the following query:

SELECT emp_name,
       emp_id,
       salary
FROM (
  SELECT emp_name,
         emp_id,
         salary,
         dense_rank() OVER (ORDER BY salary DESC) AS dr
  FROM sals
) x
WHERE dr = 2;

This query uses the dense_rank() function to assign a rank to each employee based on their salary. Then, it selects only the rows where the rank is equal to 2.

Using a Window Function with MAX()

Another approach to solving this problem is to use a window function with the MAX() aggregation function.

Here’s an example query that uses a window function with MAX():

SELECT emp_name,
       emp_id,
       salary,
       MAX(salary) OVER (ORDER BY salary DESC) AS max_salary
FROM sals;

This query assigns a value to each row based on the maximum salary seen so far. If two employees have the same salary, they will both receive the same value.

To find the second highest salary, we can use the following query:

SELECT emp_name,
       emp_id,
       salary
FROM (
  SELECT emp_name,
         emp_id,
         salary,
         MAX(salary) OVER (ORDER BY salary DESC) AS max_salary
  FROM sals
) x
WHERE max_salary = (SELECT MAX(max_salary) FROM (
  SELECT MAX(salary) OVER (ORDER BY salary DESC) AS max_salary
  FROM sals
));

This query uses the MAX() aggregation function to find the maximum salary seen so far. Then, it selects only the rows where the value is equal to the maximum salary.

Conclusion

In this article, we have explored a common problem in data analysis: finding the second highest value in a dataset when there are repeating values. We discussed three alternative approaches to solving this problem:

• Using dense_rank() and selecting only the rows with a rank of 2
• Using a window function with MAX() aggregation function
• Using both dense_rank() and MAX() aggregation functions

We hope that these examples have provided a clear understanding of how to solve this problem using various techniques.

Last modified on 2024-07-22