# New Companies in SQL | HackerRank Solution

Hello coders, today we are going to solve New Companies HackerRank Solution in SQL.

## Problem

Amber’s conglomerate corporation just acquired some new companies. Each of the companies follows this hierarchy:

Given the table schemas below, write a query to print the company_codefounder name, total number of lead managers, total number of senior managers, total number of managers, and total number of employees. Order your output by ascending company_code.

Note:

• The tables may contain duplicate records.
• The company_code is string, so the sorting should not be numeric. For example, if the company_codes are C_1C_2, and C_10, then the ascending company_codes will be C_1C_10, and C_2.

## Input Format

The following tables contain company data:

• Company: The company_code is the code of the company and founder is the founder of the company.
• Lead_Manager: The lead_manager_code is the code of the lead manager, and the company_code is the code of the working company.
• Senior_Manager: The senior_manager_code is the code of the senior manager, the lead_manager_code is the code of its lead manager, and the company_code is the code of the working company.
• Manager: The manager_code is the code of the manager, the senior_manager_code is the code of its senior manager, the lead_manager_code is the code of its lead manager, and the company_code is the code of the working company.
• Employee: The employee_code is the code of the employee, the manager_code is the code of its manager, the senior_manager_code is the code of its senior manager, the lead_manager_code is the code of its lead manager, and the company_code is the code of the working company.

Sample Input

Sample Output

``````C1 Monika 1 2 1 2
C2 Samantha 1 1 2 2``````

## Explanation

In company C1, the only lead manager is LM1. There are two senior managers, SM1 and SM2, under LM1. There is one manager, M1, under senior manager SM1. There are two employees, E1 and E2, under manager M1.

In company C2, the only lead manager is LM2. There is one senior manager, SM3, under LM2. There are two managers, M2 and M3, under senior manager SM3. There is one employee, E3, under manager M2, and another employee, E4, under manager, M3.

## Solution – New Companies in SQL

### MySQL

```select c.company_code, c.founder, count(distinct lm.lead_manager_code), count(distinct sm.senior_manager_code), count(distinct m.manager_code), count(distinct e.employee_code) from Company c, Lead_Manager lm, Senior_Manager sm, Manager m, Employee e
where c.company_code = lm.company_code and lm.lead_manager_code = sm.lead_manager_code and sm.senior_manager_code = m.senior_manager_code and m.manager_code = e.manager_code group by c.company_code, c.founder
order by c.company_code```

Disclaimer: The above Problem (New Companies) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.

### 3 thoughts on “New Companies in SQL | HackerRank Solution”

1. thanks for the answer but explanation in video will be better

2. select Company.company_code, Company.founder , count(distinct Lead_Manager.lead_manager_code),count(distinct Senior_Manager.senior_manager_code), count(distinct Manager.manager_code), count(distinct Employee.employee_code ) from Company
inner join Senior_Manager on Lead_Manager.company_code = Senior_Manager.company_code
inner join Manager on Senior_Manager.company_code = Manager.company_code
inner join Employee on Manager.company_code = Employee.company_code
group by Company.company_code,Company.founder order by Company.company_code;

3. select c.company_code,c.founder,