Population Density Difference in SQL | HackerRank Programming Solutions | HackerRank SQL Solutions

Hello Programmers/Coders, Today we are going to share solutions of Programming problems of HackerRank of Programming Language SQL. At Each Problem with Successful submission with all Test Cases Passed, you will get an score or marks. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for Population Density Difference 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.

Use “Ctrl+F” To Find Any Questions Answer. & For Mobile User, You Just Need To Click On Three dots In Your Browser & You Will Get A “Find” Option There. Use These Option to Get Any Random Questions Answer.

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 Problem – Population Density Difference SQl- Hacker Rank Solution

Population Density Difference SQl- Hacker Rank Solution

Problem:

Query the difference between the maximum and minimum populations in CITY.

Input Format

The CITY table is described as follows: 

Population Density Difference SQl- Hacker Rank Solution

select max(Population)-min(Population) from City;