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| 1 | +570. Managers with at Least 5 Direct Reports |
| 2 | +The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id. |
| 3 | +SQL Schema |
| 4 | +Pandas Schema |
| 5 | +Table: Employee |
| 6 | + |
| 7 | ++-------------+---------+ |
| 8 | +| Column Name | Type | |
| 9 | ++-------------+---------+ |
| 10 | +| id | int | |
| 11 | +| name | varchar | |
| 12 | +| department | varchar | |
| 13 | +| managerId | int | |
| 14 | ++-------------+---------+ |
| 15 | +id is the primary key (column with unique values) for this table. |
| 16 | +Each row of this table indicates the name of an employee, their department, and the id of their manager. |
| 17 | +If managerId is null, then the employee does not have a manager. |
| 18 | +No employee will be the manager of themself. |
| 19 | + |
| 20 | + |
| 21 | +Write a solution to find managers with at least five direct reports. |
| 22 | + |
| 23 | +Return the result table in any order. |
| 24 | + |
| 25 | +The result format is in the following example. |
| 26 | + |
| 27 | + |
| 28 | + |
| 29 | +Example 1: |
| 30 | + |
| 31 | +Input: |
| 32 | +Employee table: |
| 33 | ++-----+-------+------------+-----------+ |
| 34 | +| id | name | department | managerId | |
| 35 | ++-----+-------+------------+-----------+ |
| 36 | +| 101 | John | A | null | |
| 37 | +| 102 | Dan | A | 101 | |
| 38 | +| 103 | James | A | 101 | |
| 39 | +| 104 | Amy | A | 101 | |
| 40 | +| 105 | Anne | A | 101 | |
| 41 | +| 106 | Ron | B | 101 | |
| 42 | ++-----+-------+------------+-----------+ |
| 43 | +Output: |
| 44 | ++------+ |
| 45 | +| name | |
| 46 | ++------+ |
| 47 | +| John | |
| 48 | ++------+ |
| 49 | + |
| 50 | +Explanation: |
| 51 | +John is the only manager who has at least 5 direct reports. |
| 52 | + |
| 53 | +Example 2: |
| 54 | + |
| 55 | +Input: |
| 56 | +Employee table: |
| 57 | ++-----+-------+------------+-----------+ |
| 58 | +| id | name | department | managerId | |
| 59 | ++-----+-------+------------+-----------+ |
| 60 | + | 101 | John | A | null | |
| 61 | +| 102 | Dan | A | 101 | |
| 62 | + | 103 | James | A | 101 | |
| 63 | + | 104 | Amy | A | 101 | |
| 64 | + |
| 65 | ++-----+-------+------------+-----------+ |
| 66 | +Output: |
| 67 | ++-----------+ |
| 68 | + |
| 69 | +# Write your MySQL query statement below |
| 70 | + |
| 71 | +-- Solution 2: Using subquery for counting direct reports and join for filtering managers with at least 5 direct reports |
| 72 | +SELECT name |
| 73 | +FROM Employee |
| 74 | +WHERE id IN ( |
| 75 | + SELECT managerId |
| 76 | + FROM Employee |
| 77 | + GROUP BY managerId |
| 78 | + HAVING COUNT(*) >= 5 |
| 79 | +); |
| 80 | + |
| 81 | +-- Solution 3: Using window function for counting direct reports and filtering managers with at least 5 direct reports |
| 82 | + |
| 83 | +SELECT name |
| 84 | +FROM Employee |
| 85 | + JOIN ( |
| 86 | + SELECT managerId |
| 87 | + FROM Employee |
| 88 | + GROUP BY managerId |
| 89 | + HAVING COUNT(*) >= 5 |
| 90 | + ) AS managers |
| 91 | + ON Employee.id = managers.managerId |
| 92 | + |
| 93 | + GROUP BY name |
| 94 | + HAVING COUNT(*) >= 5; |
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