Add problem 1378 from Leetcode

Author: absognety

LeetCode/SQL50/SQL1378.sql

@@ -0,0 +1,78 @@
+-- Table: Employees
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | id            | int     |
+-- | name          | varchar |
+-- +---------------+---------+
+-- id is the primary key (column with unique values) for this table.
+-- Each row of this table contains the id and the name of an employee in a company.
+ 
+
+-- Table: EmployeeUNI
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | id            | int     |
+-- | unique_id     | int     |
+-- +---------------+---------+
+-- (id, unique_id) is the primary key (combination of columns with unique values) for this table.
+-- Each row of this table contains the id and the corresponding unique id of an employee in the company.
+ 
+
+-- Write a solution to show the unique ID of each user, If a user does not have a unique ID replace just show null.
+
+-- Return the result table in any order.
+
+-- The result format is in the following example.
+
+ 
+
+-- Example 1:
+
+-- Input: 
+-- Employees table:
+-- +----+----------+
+-- | id | name     |
+-- +----+----------+
+-- | 1  | Alice    |
+-- | 7  | Bob      |
+-- | 11 | Meir     |
+-- | 90 | Winston  |
+-- | 3  | Jonathan |
+-- +----+----------+
+-- EmployeeUNI table:
+-- +----+-----------+
+-- | id | unique_id |
+-- +----+-----------+
+-- | 3  | 1         |
+-- | 11 | 2         |
+-- | 90 | 3         |
+-- +----+-----------+
+-- Output: 
+-- +-----------+----------+
+-- | unique_id | name     |
+-- +-----------+----------+
+-- | null      | Alice    |
+-- | null      | Bob      |
+-- | 2         | Meir     |
+-- | 3         | Winston  |
+-- | 1         | Jonathan |
+-- +-----------+----------+
+-- Explanation: 
+-- Alice and Bob do not have a unique ID, We will show null instead.
+-- The unique ID of Meir is 2.
+-- The unique ID of Winston is 3.
+-- The unique ID of Jonathan is 1.
+
+
+-- Solution (PostgreSQL):
+select case 
+       when EmployeeUNI.unique_id is null then null
+       else EmployeeUNI.unique_id
+       end as unique_id,
+       Employees.name
+from Employees 
+     left join EmployeeUNI on (Employees.id = EmployeeUNI.id);