finish 181, 182

Author: BaffinLee

101-200/181. Employees Earning More Than Their Managers.md

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+# 181. Employees Earning More Than Their Managers
+
+- Difficulty: Easy.
+- Related Topics: .
+- Similar Questions: .
+
+## Problem
+
+The ```Employee``` table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.
+
+```
++----+-------+--------+-----------+
+| Id | Name  | Salary | ManagerId |
++----+-------+--------+-----------+
+| 1  | Joe   | 70000  | 3         |
+| 2  | Henry | 80000  | 4         |
+| 3  | Sam   | 60000  | NULL      |
+| 4  | Max   | 90000  | NULL      |
++----+-------+--------+-----------+
+```
+
+Given the ```Employee``` table, write a SQL query that finds out employees who earn more than their managers. For the above table, Joe is the only employee who earns more than his manager.
+
+```
++----------+
+| Employee |
++----------+
+| Joe      |
++----------+
+```
+
+## Solution 1
+
+```sql
+# Write your MySQL query statement below
+select
+  a.Name as Employee
+  from Employee a
+  left join Employee b
+  on a.ManagerId = b.Id
+  where a.Salary > b.Salary
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity :
+* Space complexity :
+
+## Solution 2
+
+```sql
+# Write your MySQL query statement below
+select
+  a.Name as Employee
+  from Employee a, Employee b
+  where a.ManagerId = b.Id
+    and a.Salary > b.Salary
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity :
+* Space complexity :

101-200/182. Duplicate Emails.md

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+# 182. Duplicate Emails
+
+- Difficulty: Easy.
+- Related Topics: .
+- Similar Questions: .
+
+## Problem
+
+Write a SQL query to find all duplicate emails in a table named ```Person```.
+
+```
++----+---------+
+| Id | Email   |
++----+---------+
+| 1  | [email protected] |
+| 2  | [email protected] |
+| 3  | [email protected] |
++----+---------+
+```
+
+For example, your query should return the following for the above table:
+
+```
++---------+
+| Email   |
++---------+
+| [email protected] |
++---------+
+```
+
+**Note**: All emails are in lowercase.
+
+## Solution 1
+
+```sql
+# Write your MySQL query statement below
+select Email
+  from Person
+  group by Email
+  having count(Id) > 1
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity :
+* Space complexity :
+
+## Solution 2
+
+```sql
+# Write your MySQL query statement below
+select
+  distinct a.Email as Email
+  from Person a, Person b
+  where a.Id <> b.Id
+    and a.Email = b.Email
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity :
+* Space complexity :