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

Python3MySQL

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

# 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

MySQL

# 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