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176. Second Highest Salary

Description

Table: Employee

+-------------+------+
| Column Name | Type |
+-------------+------+
| id          | int  |
| salary      | int  |
+-------------+------+
id is the primary key (column with unique values) for this table.
Each row of this table contains information about the salary of an employee.

 

Write a solution to find the second highest distinct salary from the Employee table. If there is no second highest salary, return null (return None in Pandas).

The result format is in the following example.

 

Example 1:

Input: 
Employee table:
+----+--------+
| id | salary |
+----+--------+
| 1  | 100    |
| 2  | 200    |
| 3  | 300    |
+----+--------+
Output: 
+---------------------+
| SecondHighestSalary |
+---------------------+
| 200                 |
+---------------------+

Example 2:

Input: 
Employee table:
+----+--------+
| id | salary |
+----+--------+
| 1  | 100    |
+----+--------+
Output: 
+---------------------+
| SecondHighestSalary |
+---------------------+
| null                |
+---------------------+

Solutions

Solution 1: Use Sub Query and LIMIT

Thinking

The second highest salary is the second value of the distinct salaries in descending order, or \(\textit{NULL}\) if fewer than two exist. \(\textit{ORDER BY}\) plus \(\textit{LIMIT}\,1\,\textit{OFFSET}\,1\) picks that slot; an outer query keeps a single \(\textit{NULL}\) row when the offset is empty.

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import pandas as pd


def second_highest_salary(employee: pd.DataFrame) -> pd.DataFrame:
    # Drop any duplicate salary values to avoid counting duplicates as separate salary ranks
    unique_salaries = employee["salary"].drop_duplicates()

    # Sort the unique salaries in descending order and get the second highest salary
    second_highest = (
        unique_salaries.nlargest(2).iloc[-1] if len(unique_salaries) >= 2 else None
    )

    # If the second highest salary doesn't exist (e.g., there are fewer than two unique salaries), return None
    if second_highest is None:
        return pd.DataFrame({"SecondHighestSalary": [None]})

    # Create a DataFrame with the second highest salary
    result_df = pd.DataFrame({"SecondHighestSalary": [second_highest]})

    return result_df

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# Write your MySQL query statement below
SELECT
    (
        SELECT DISTINCT salary
        FROM Employee
        ORDER BY salary DESC
        LIMIT 1, 1
    ) AS SecondHighestSalary;

Solution 2: Use MAX() function

Thinking

Solution 1 needs a sort and an offset. The maximum among salaries strictly below the global maximum is the second highest; \(\textit{MAX}\) on an empty set is already \(\textit{NULL}\), so no \(\textit{LIMIT}\) is required.

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# Write your MySQL query statement below
SELECT MAX(salary) AS SecondHighestSalary
FROM Employee
WHERE salary < (SELECT MAX(salary) FROM Employee);

Solution 3: Use IFNULL() and window function

Thinking

After deduplicating, \(\textit{DENSE_RANK}\) numbers salaries descending and we keep rank \(2\). Ties at the top do not consume the second place, which matches “\(k\)-th highest” and extends cleanly.

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# Write your MySQL query statement below
WITH T AS (SELECT salary, DENSE_RANK() OVER (ORDER BY salary DESC) AS rk FROM Employee)
SELECT (SELECT DISTINCT salary FROM T WHERE rk = 2) AS SecondHighestSalary;

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