feat(sql): solutions

Author: hongbo-miao

SQL/0175. Combine Two Tables.sql

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+-- Table: Person
+--
+-- +-------------+---------+
+-- | Column Name | Type    |
+-- +-------------+---------+
+-- | personId    | int     |
+-- | lastName    | varchar |
+-- | firstName   | varchar |
+-- +-------------+---------+
+-- personId is the primary key column for this table.
+-- This table contains information about the ID of some persons and their first and last names.
+--
+-- Table: Address
+--
+-- +-------------+---------+
+-- | Column Name | Type    |
+-- +-------------+---------+
+-- | addressId   | int     |
+-- | personId    | int     |
+-- | city        | varchar |
+-- | state       | varchar |
+-- +-------------+---------+
+-- addressId is the primary key column for this table.
+-- Each row of this table contains information about the city and state of one person with ID = PersonId.
+--
+-- Write an SQL query to report the first name, last name, city, and state of each person in the Person table. If the address of a personId is not present in the Address table, report null instead.
+-- Return the result table in any order.
+-- The query result format is in the following example.
+--
+-- Example 1:
+--
+-- Input:
+-- Person table:
+-- +----------+----------+-----------+
+-- | personId | lastName | firstName |
+-- +----------+----------+-----------+
+-- | 1        | Wang     | Allen     |
+-- | 2        | Alice    | Bob       |
+-- +----------+----------+-----------+
+-- Address table:
+-- +-----------+----------+---------------+------------+
+-- | addressId | personId | city          | state      |
+-- +-----------+----------+---------------+------------+
+-- | 1         | 2        | New York City | New York   |
+-- | 2         | 3        | Leetcode      | California |
+-- +-----------+----------+---------------+------------+
+-- Output:
+-- +-----------+----------+---------------+----------+
+-- | firstName | lastName | city          | state    |
+-- +-----------+----------+---------------+----------+
+-- | Allen     | Wang     | Null          | Null     |
+-- | Bob       | Alice    | New York City | New York |
+-- +-----------+----------+---------------+----------+
+-- Explanation:
+-- There is no address in the address table for the personId = 1 so we return null in their city and state.
+-- addressId = 1 contains information about the address of personId = 2.
+
+SELECT Person.FirstName, Person.LastName, Address.City, Address.State
+FROM Person
+         LEFT JOIN Address
+                   ON Person.PersonId = Address.PersonId;

SQL/0181. Employees Earning More Than Their Managers.sql

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+-- Table: Employee
+--
+-- +-------------+---------+
+-- | Column Name | Type    |
+-- +-------------+---------+
+-- | id          | int     |
+-- | name        | varchar |
+-- | salary      | int     |
+-- | managerId   | int     |
+-- +-------------+---------+
+-- id is the primary key column for this table.
+-- Each row of this table indicates the ID of an employee, their name, salary, and the ID of their manager.
+--
+--
+-- Write an SQL query to find the employees who earn more than their managers.
+--
+-- Return the result table in any order.
+--
+-- The query result format is in the following example.
+--
+--
+--
+-- Example 1:
+--
+-- Input:
+-- Employee table:
+-- +----+-------+--------+-----------+
+-- | id | name  | salary | managerId |
+-- +----+-------+--------+-----------+
+-- | 1  | Joe   | 70000  | 3         |
+-- | 2  | Henry | 80000  | 4         |
+-- | 3  | Sam   | 60000  | Null      |
+-- | 4  | Max   | 90000  | Null      |
+-- +----+-------+--------+-----------+
+-- Output:
+-- +----------+
+-- | Employee |
+-- +----------+
+-- | Joe      |
+-- +----------+
+-- Explanation: Joe is the only employee who earns more than his manager.
+
+SELECT a.Name AS 'Employee'
+FROM Employee AS a,
+     Employee AS b
+WHERE a.ManagerId = b.Id
+  AND a.Salary > b.Salary;