New Companies in SQL | HackerRank Programming Solutions | HackerRank SQL Solutions

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In this post, you will find the solution for New Companies in SQL-HackerRank Problem. We are providing the correct and tested solutions of coding problems present on HackerRank. If you are not able to solve any problem, then you can take help from our Blog/website.

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Introduction To SQL

SQL stands for Structured Query Language. SQL is used to create, remove, alter the database and database objects in a database management system and to store, retrieve, update the data in a database. SQL is a standard language for creating, accessing, manipulating database management system. SQL works for all modern relational database management systems, like SQL Server, Oracle, MySQL, etc.

  • It is a standard language for Relational Database System. It enables a user to create, read, update and delete relational databases and tables.
  • All the RDBMS like MySQL, Informix, Oracle, MS Access and SQL Server use SQL as their standard database language.
  • SQL allows users to query the database in a number of ways, using English-like statements.

Link for the ProblemNew Companies SQL – Hacker Rank Solution

New Companies SQL – Hacker Rank Solution

Problem:

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

1458531031 249df3ae87 ScreenShot2016 03 21at8.59.56AM

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. 
  • 1458531125 deb0a57ae1 ScreenShot2016 03 21at8.50.04AM
  • Lead_Manager: The lead_manager_code is the code of the lead manager, and the company_code is the code of the working company. 
  • 1458534960 2c6d764e3c ScreenShot2016 03 21at8.50.12AM
  • 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. 
  • 1458534973 6548194998 ScreenShot2016 03 21at8.50.21AM
  • 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.
  •  1458534988 7fc0af46ce ScreenShot2016 03 21at8.50.29AM
  • 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.
  •  1458535002 d47f63cbb4 ScreenShot2016 03 21at8.50.41AM
New Companies

Sample Input

Company Table: 1458535049 2a207c44b3 ScreenShot2016 03 21at8.50.52AM Lead_Manager Table: 1458535073 919107f639 ScreenShot2016 03 21at8.51.03AM Senior_Manager Table: 1458535111 b1c48335b3 ScreenShot2016 03 21at8.51.15AM Manager Table: 1458535122 888f4bf340 ScreenShot2016 03 21at8.51.26AM Employee Table: 1458535134 878767e0d9 ScreenShot2016 03 21at8.51.52AM

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.

New Companies SQL – Hacker Rank Solution
SELECT C.COMPANY_CODE,
       C.FOUNDER,
  (SELECT COUNT(DISTINCT LEAD_MANAGER_CODE)
   FROM LEAD_MANAGER L
   WHERE L.COMPANY_CODE = C.COMPANY_CODE),
  (SELECT COUNT(DISTINCT SENIOR_MANAGER_CODE)
   FROM SENIOR_MANAGER S
   WHERE S.COMPANY_CODE = C.COMPANY_CODE),
  (SELECT COUNT(DISTINCT MANAGER_CODE)
   FROM MANAGER M
   WHERE M.COMPANY_CODE = C.COMPANY_CODE),
  (SELECT COUNT(DISTINCT EMPLOYEE_CODE)
   FROM EMPLOYEE E
   WHERE E.COMPANY_CODE = C.COMPANY_CODE)
FROM COMPANY C
ORDER BY C.COMPANY_CODE ASC;

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