diff --git a/1378-replace-employee-id-with-the-unique-identifier/README.md b/1378-replace-employee-id-with-the-unique-identifier/README.md new file mode 100644 index 0000000..283a0d5 --- /dev/null +++ b/1378-replace-employee-id-with-the-unique-identifier/README.md @@ -0,0 +1,73 @@ +
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. ++