Exchange Seats and Employees Whose Manager Left the Company SQL MySQL leetcode Solution using mod
Query Structure and Explanation
Main Query:
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select employee_id from Employees
This part of the query specifies that we want to select the employee_id from the Employees table.
Where Clause:
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where salary < 30000
This condition filters the employees to include only those whose salary is less than 30,000.
Subquery in the Where Clause:
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and manager_id not in (
select employee_id from Employees
)
This part adds another condition: it checks that the manager_id of the employee is not in the list of employee_ids from the Employees table.
Order By Clause:
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order by employee_id;
This clause sorts the resulting list of employee_ids in ascending order.
Detailed Explanation
Subquery:
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select employee_id from Employees
The subquery fetches all the employee_ids from the Employees table.
Main Query Condition with Subquery:
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and manager_id not in (
select employee_id from Employees
)
The main query uses the results of the subquery to filter employees further. It ensures that the manager_id of the employees selected by the main query is not any of the employee_ids in the Employees table.
Purpose of the Query
The entire query aims to select the employee_ids of employees who:
Have a salary of less than 30,000.
Do not have a manager who is also an employee in the Employees table.
Use Case Example
This kind of query might be useful in scenarios where:
You are interested in finding employees who have lower salaries and whose managers are not part of the Employees table, which might indicate that their managers are external contractors or part of a different department not listed in the same table.
By combining these conditions, the query ensures that the result is a list of employee_ids meeting both criteria, and the results are neatly ordered by employee_id.
Challenge: Swapping Seats of Consecutive Students
Difficulty: Medium
Data:
A table named Seat stores student information:
id (int): Unique identifier for each student (primary key, continuous increment).
student (varchar): The student's name.
Objective:
Swap the seat IDs of every two consecutive students in the Seat table. If there's an odd number of students, the last student's seat ID remains unchanged.
Output:
A table with the same columns (id and student), but with seat IDs swapped for consecutive students, ordered by id in ascending order.
Approach:
We can achieve this using a window function like LAG() to access the previous student's ID within the query. Here's how:
Use a CASE expression to determine the swapped seat ID for each student.
If the current student's ID is even, the swapped ID is the LAG(id, 1) (previous student's ID).
If the current student's ID is odd and there's a next student (LAG(id, 1) IS NOT NULL), the swapped ID is the LAG(id, 1). Otherwise, the original ID remains unchanged (handling the odd number of students case).
Select the student name and the swapped (or original) ID for each row.
Order the result by id in ascending order.
Benefits of Window Functions:
This approach avoids complex joins or self-joins and achieves the desired result within a single query. It improves code readability by keeping the logic focused on manipulating the data within the window function clause.
Important Note:
The solution assumes students are assigned seat IDs based on a continuous increment starting from 1. If the seat IDs are not consecutive, modifications might be needed to handle the swapping logic.
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