add 570

Author: fishercoder1534

README.md

@@ -610,6 +610,7 @@ Your ideas/fixes/algorithms are more than welcome!
 |580|[Count Student Number in Departments](https://leetcode.com/problems/count-student-number-in-departments/)|[Solution](../master/database/_580.sql) | || Medium | Left Join
 |577|[Employee Bonus](https://leetcode.com/problems/employee-bonus/)|[Solution](../master/database/_577.sql) | || Easy |
 |574|[Winning Candidate](https://leetcode.com/problems/winning-candidate/)|[Solution](../master/database/_574.sql) | || Medium |
+|570|[Managers with at Least 5 Direct Reports](https://leetcode.com/problems/managers-with-at-least-5-direct-reports/)|[Solution](../master/database/_570.sql) | || Medium |
 |197|[Rising Temperature](https://leetcode.com/problems/rising-temperature/)|[Solution](../master/database/_197.sql)| O(n^2)|O(n) | Easy|
 |196|[Delete Duplicate Emails](https://leetcode.com/problems/delete-duplicate-emails/)|[Solution](../master/database/_196.sql)| O(n^2)|O(n) | Easy|
 |184|[Department Highest Salary](https://leetcode.com/problems/department-highest-salary)|[Solution](../master/database/_184.sql)| O(n^2)|O(n) | Medium|

database/_570.sql

@@ -0,0 +1,26 @@
+--570. Managers with at Least 5 Direct Reports
+--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 	  |Department |ManagerId |
+--+------+----------+-----------+----------+
+--|101   |John 	  |A 	      |null      |
+--|102   |Dan 	  |A 	      |101       |
+--|103   |James 	  |A 	      |101       |
+--|104   |Amy 	  |A 	      |101       |
+--|105   |Anne 	  |A 	      |101       |
+--|106   |Ron 	  |B 	      |101       |
+--+------+----------+-----------+----------+
+--Given the Employee table, write a SQL query that finds out managers with at least 5 direct report. For the above table, your SQL query should return:
+--
+--+-------+
+--| Name  |
+--+-------+
+--| John  |
+--+-------+
+--Note:
+--No one would report to himself.
+
+select Name from Employee as Name where Id in
+    (select ManagerId from Employee group by ManagerId having count(*) >= 5)
+;