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‎6-Subqueries/1321_Restaurant_Growth.sql

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-- Source: https://leetcode.com/problems/restaurant-growth/description/?envType=study-plan-v2&envId=top-sql-50
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-- Table: Customer
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-- +---------------+---------+
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-- | Column Name | Type |
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-- +---------------+---------+
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-- | customer_id | int |
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-- | name | varchar |
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-- | visited_on | date |
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-- | amount | int |
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-- +---------------+---------+
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-- In SQL,(customer_id, visited_on) is the primary key for this table.
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-- This table contains data about customer transactions in a restaurant.
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-- visited_on is the date on which the customer with ID (customer_id) has visited the restaurant.
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-- amount is the total paid by a customer.
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-- You are the restaurant owner and you want to analyze a possible expansion (there will be at least one customer every day).
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-- Compute the moving average of how much the customer paid in a seven days window (i.e., current day + 6 days before). average_amount should be rounded to two decimal places.
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-- Return the result table ordered by visited_on in ascending order.
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-- The result format is in the following example.
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-- Example 1:
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-- Input:
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-- Customer table:
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-- +-------------+--------------+--------------+-------------+
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-- | customer_id | name | visited_on | amount |
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-- +-------------+--------------+--------------+-------------+
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-- | 1 | Jhon | 2019-01-01 | 100 |
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-- | 2 | Daniel | 2019-01-02 | 110 |
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-- | 3 | Jade | 2019-01-03 | 120 |
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-- | 4 | Khaled | 2019-01-04 | 130 |
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-- | 5 | Winston | 2019-01-05 | 110 |
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-- | 6 | Elvis | 2019-01-06 | 140 |
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-- | 7 | Anna | 2019-01-07 | 150 |
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-- | 8 | Maria | 2019-01-08 | 80 |
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-- | 9 | Jaze | 2019-01-09 | 110 |
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-- | 1 | Jhon | 2019-01-10 | 130 |
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-- | 3 | Jade | 2019-01-10 | 150 |
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-- +-------------+--------------+--------------+-------------+
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-- Output:
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-- +--------------+--------------+----------------+
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-- | visited_on | amount | average_amount |
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-- +--------------+--------------+----------------+
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-- | 2019-01-07 | 860 | 122.86 |
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-- | 2019-01-08 | 840 | 120 |
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-- | 2019-01-09 | 840 | 120 |
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-- | 2019-01-10 | 1000 | 142.86 |
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-- +--------------+--------------+----------------+
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-- Explanation:
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-- 1st moving average from 2019-01-01 to 2019-01-07 has an average_amount of (100 + 110 + 120 + 130 + 110 + 140 + 150)/7 = 122.86
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-- 2nd moving average from 2019-01-02 to 2019-01-08 has an average_amount of (110 + 120 + 130 + 110 + 140 + 150 + 80)/7 = 120
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-- 3rd moving average from 2019-01-03 to 2019-01-09 has an average_amount of (120 + 130 + 110 + 140 + 150 + 80 + 110)/7 = 120
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-- 4th moving average from 2019-01-04 to 2019-01-10 has an average_amount of (130 + 110 + 140 + 150 + 80 + 110 + 130 + 150)/7 = 142.86
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------------------------------------------------------------------------------
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-- SQL Schema
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Create table If Not Exists Customer (customer_id int, name varchar(20), visited_on date, amount int)
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Truncate table Customer
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insert into Customer (customer_id, name, visited_on, amount) values ('1', 'Jhon', '2019-01-01', '100')
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insert into Customer (customer_id, name, visited_on, amount) values ('2', 'Daniel', '2019-01-02', '110')
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insert into Customer (customer_id, name, visited_on, amount) values ('3', 'Jade', '2019-01-03', '120')
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insert into Customer (customer_id, name, visited_on, amount) values ('4', 'Khaled', '2019-01-04', '130')
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insert into Customer (customer_id, name, visited_on, amount) values ('5', 'Winston', '2019-01-05', '110')
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insert into Customer (customer_id, name, visited_on, amount) values ('6', 'Elvis', '2019-01-06', '140')
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insert into Customer (customer_id, name, visited_on, amount) values ('7', 'Anna', '2019-01-07', '150')
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insert into Customer (customer_id, name, visited_on, amount) values ('8', 'Maria', '2019-01-08', '80')
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insert into Customer (customer_id, name, visited_on, amount) values ('9', 'Jaze', '2019-01-09', '110')
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insert into Customer (customer_id, name, visited_on, amount) values ('1', 'Jhon', '2019-01-10', '130')
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insert into Customer (customer_id, name, visited_on, amount) values ('3', 'Jade', '2019-01-10', '150')
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-- MS SQL Server Code
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WITH
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amount_per_day AS (
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SELECT
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visited_on
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, SUM(SUM(amount)) OVER(ORDER BY visited_on
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ROWS BETWEEN 6 PRECEDING AND CURRENT ROW) AS amount
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, ROUND(AVG(SUM(amount)*1.0) OVER(ORDER BY visited_on
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ROWS BETWEEN 6 PRECEDING AND CURRENT ROW),2) AS average_amount
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FROM Customer
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GROUP BY visited_on
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),
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starting_date AS (
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SELECT DATEADD(day, 6, earliest_date) AS starting_date
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FROM
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( SELECT MIN(visited_on) AS earliest_date
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FROM amount_per_day
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) AS a
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)
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SELECT visited_on, amount, average_amount
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FROM amount_per_day
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CROSS JOIN starting_date
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WHERE visited_on >= starting_date
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ORDER BY visited_on;

‎6-Subqueries/185_Department_Top_Three_Salaries.sql

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-- Source: https://leetcode.com/problems/department-top-three-salaries/description/?envType=study-plan-v2&envId=top-sql-50
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-- Table: Employee
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-- +--------------+---------+
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-- | Column Name | Type |
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-- +--------------+---------+
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-- | id | int |
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-- | name | varchar |
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-- | salary | int |
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-- | departmentId | int |
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-- +--------------+---------+
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-- id is the primary key (column with unique values) for this table.
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-- departmentId is a foreign key (reference column) of the ID from the Department table.
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-- Each row of this table indicates the ID, name, and salary of an employee. It also contains the ID of their department.
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-- Table: Department
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-- +-------------+---------+
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-- | Column Name | Type |
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-- +-------------+---------+
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-- | id | int |
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-- | name | varchar |
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-- +-------------+---------+
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-- id is the primary key (column with unique values) for this table.
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-- Each row of this table indicates the ID of a department and its name.
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-- A company's executives are interested in seeing who earns the most money in each of the company's departments. A high earner in a department is an employee who has a salary in the top three unique salaries for that department.
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-- Write a solution to find the employees who are high earners in each of the departments.
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-- Return the result table in any order.
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-- The result format is in the following example.
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-- Example 1:
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-- Input:
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-- Employee table:
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-- +----+-------+--------+--------------+
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-- | id | name | salary | departmentId |
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-- +----+-------+--------+--------------+
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-- | 1 | Joe | 85000 | 1 |
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-- | 2 | Henry | 80000 | 2 |
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-- | 3 | Sam | 60000 | 2 |
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-- | 4 | Max | 90000 | 1 |
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-- | 5 | Janet | 69000 | 1 |
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-- | 6 | Randy | 85000 | 1 |
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-- | 7 | Will | 70000 | 1 |
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-- +----+-------+--------+--------------+
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-- Department table:
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-- +----+-------+
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-- | id | name |
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-- +----+-------+
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-- | 1 | IT |
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-- | 2 | Sales |
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-- +----+-------+
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-- Output:
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-- +------------+----------+--------+
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-- | Department | Employee | Salary |
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-- +------------+----------+--------+
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-- | IT | Max | 90000 |
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-- | IT | Joe | 85000 |
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-- | IT | Randy | 85000 |
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-- | IT | Will | 70000 |
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-- | Sales | Henry | 80000 |
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-- | Sales | Sam | 60000 |
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-- +------------+----------+--------+
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-- Explanation:
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-- In the IT department:
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-- - Max earns the highest unique salary
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-- - Both Randy and Joe earn the second-highest unique salary
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-- - Will earns the third-highest unique salary
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-- In the Sales department:
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-- - Henry earns the highest salary
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-- - Sam earns the second-highest salary
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-- - There is no third-highest salary as there are only two employees
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------------------------------------------------------------------------------
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-- SQL Schema
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Create table If Not Exists Employee (id int, name varchar(255), salary int, departmentId int)
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Create table If Not Exists Department (id int, name varchar(255))
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Truncate table Employee
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insert into Employee (id, name, salary, departmentId) values ('1', 'Joe', '85000', '1')
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insert into Employee (id, name, salary, departmentId) values ('2', 'Henry', '80000', '2')
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insert into Employee (id, name, salary, departmentId) values ('3', 'Sam', '60000', '2')
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insert into Employee (id, name, salary, departmentId) values ('4', 'Max', '90000', '1')
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insert into Employee (id, name, salary, departmentId) values ('5', 'Janet', '69000', '1')
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insert into Employee (id, name, salary, departmentId) values ('6', 'Randy', '85000', '1')
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insert into Employee (id, name, salary, departmentId) values ('7', 'Will', '70000', '1')
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Truncate table Department
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insert into Department (id, name) values ('1', 'IT')
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insert into Department (id, name) values ('2', 'Sales')
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-- MS SQL Server Code
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SELECT d.name AS Department, e.name AS Employee, e.salary AS Salary
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FROM Employee e
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JOIN Department d
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ON e.departmentId = d.id
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JOIN (
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SELECT *, DENSE_RANK() OVER(PARTITION BY departmentId ORDER BY salary DESC) AS 'rank'
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FROM Employee
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) AS t
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ON t.id = e.id
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WHERE t.rank <= 3
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ORDER BY 1, 3 DESC

‎6-Subqueries/1978_Employees_Whose_Manager_Left_the_Company.sql

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-- Source: https://leetcode.com/problems/employees-whose-manager-left-the-company/description/?envType=study-plan-v2&envId=top-sql-50
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-- Table: Employees
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-- +-------------+----------+
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-- | Column Name | Type |
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-- +-------------+----------+
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-- | employee_id | int |
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-- | name | varchar |
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-- | manager_id | int |
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-- | salary | int |
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-- +-------------+----------+
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-- In SQL, employee_id is the primary key for this table.
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-- This table contains information about the employees, their salary, and the ID of their manager. Some employees do not have a manager (manager_id is null).
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-- Find the IDs of the employees whose salary is strictly less than $30000 and whose manager left the company. When a manager leaves the company, their information is deleted from the Employees table, but the reports still have their manager_id set to the manager that left.
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-- Return the result table ordered by employee_id.
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-- The result format is in the following example.
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-- Example 1:
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-- Input:
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-- Employees table:
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-- +-------------+-----------+------------+--------+
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-- | employee_id | name | manager_id | salary |
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-- +-------------+-----------+------------+--------+
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-- | 3 | Mila | 9 | 60301 |
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-- | 12 | Antonella | null | 31000 |
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-- | 13 | Emery | null | 67084 |
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-- | 1 | Kalel | 11 | 21241 |
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-- | 9 | Mikaela | null | 50937 |
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-- | 11 | Joziah | 6 | 28485 |
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-- +-------------+-----------+------------+--------+
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-- Output:
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-- +-------------+
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-- | employee_id |
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-- +-------------+
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-- | 11 |
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-- +-------------+
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-- Explanation:
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-- The employees with a salary less than $30000 are 1 (Kalel) and 11 (Joziah).
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-- Kalel's manager is employee 11, who is still in the company (Joziah).
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-- Joziah's manager is employee 6, who left the company because there is no row for employee 6 as it was deleted.
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------------------------------------------------------------------------------
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-- SQL Schema
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Create table If Not Exists Employees (employee_id int, name varchar(20), manager_id int, salary int)
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Truncate table Employees
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insert into Employees (employee_id, name, manager_id, salary) values ('3', 'Mila', '9', '60301')
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insert into Employees (employee_id, name, manager_id, salary) values ('12', 'Antonella', 'None', '31000')
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insert into Employees (employee_id, name, manager_id, salary) values ('13', 'Emery', 'None', '67084')
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insert into Employees (employee_id, name, manager_id, salary) values ('1', 'Kalel', '11', '21241')
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insert into Employees (employee_id, name, manager_id, salary) values ('9', 'Mikaela', 'None', '50937')
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insert into Employees (employee_id, name, manager_id, salary) values ('11', 'Joziah', '6', '28485')
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-- MS SQL Server Code
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SELECT DISTINCT e.employee_id
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FROM Employees e
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WHERE e.manager_id IS NOT NULL
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AND e.salary < 30000
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AND NOT EXISTS
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(
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SELECT employee_id
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FROM Employees
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WHERE employee_id = e.manager_id
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)
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ORDER BY employee_id;

‎README.md

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### Subqueries
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| # | Problem | Level | Status |
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| ---- | ------------------------------------------------ | ------ | ------ |
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| 1978 | [Employees Whose Manager Left the Company]() | Easy | Solved |
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| 626 | [Exchange Seats]() | Medium | - |
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| 1341 | [Movie Rating]() | Medium | - |
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| 1321 | [Restaurant Growth]() | Medium | Solved |
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| 602 | [Friend Requests II: Who Has the Most Friends]() | Medium | - |
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| 585 | [Investments in 2016]() | Medium | - |
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| 185 | [Department Top Three Salaries]() | Hard | Solved |
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| # | Problem | Level | Status |
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| ---- | ----------------------------------------------------------------------------------------------------------- | ------ | ------ |
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| 1978 | [Employees Whose Manager Left the Company](/6-Subqueries/1978_Employees_Whose_Manager_Left_the_Company.sql) | Easy | Solved |
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| 626 | Exchange Seats | Medium | - |
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| 1341 | Movie Rating | Medium | - |
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| 1321 | [Restaurant Growth](/6-Subqueries/1321_Restaurant_Growth.sql) | Medium | Solved |
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| 602 | Friend Requests II: Who Has the Most Friends | Medium | - |
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| 585 | Investments in 2016 | Medium | - |
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| 185 | [Department Top Three Salaries](/6-Subqueries/185_Department_Top_Three_Salaries.sql) | Hard | Solved |
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### Advanced String Functions / Regex Clause
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| # | Problem | Level | Status |
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| ---- | ----------------------------------------- | ------ | ------ |
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| 1667 | [Fix Names in a Table]() | Easy | - |
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| 1527 | [Patients with a Condition]() | Easy | - |
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| 196 | [Delete Duplicate Emails]() | Easy | Solved |
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| 176 | [Second Highest Salary]() | Medium | Solved |
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| 1484 | [Group Sold Products By The Date]() | Easy | - |
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| 1327 | [List the Products Ordered in a Period]() | Easy | - |
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| 1517 | [Find Users With Valid E-mails]() | Easy | Solved |
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| # | Problem | Level | Status |
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| ---- | ------------------------------------- | ------ | ------ |
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| 1667 | Fix Names in a Table | Easy | - |
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| 1527 | Patients with a Condition | Easy | - |
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| 196 | [Delete Duplicate Emails]() | Easy | Solved |
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| 176 | [Second Highest Salary]() | Medium | Solved |
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| 1484 | Group Sold Products By The Date | Easy | - |
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| 1327 | List the Products Ordered in a Period | Easy | - |
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| 1517 | [Find Users With Valid E-mails]() | Easy | Solved |

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