Hello coders, today we are going to solve Average Population of Each Continent HackerRank Solution in SQL.
Problem
Given the CITY and COUNTRY tables, query the names of all the continents (COUNTRY.Continent) and their respective average city populations (CITY.Population) rounded down to the nearest integer.
Note: CITY.CountryCode and COUNTRY.Code are matching key columns.
Input Format
The CITY and COUNTRY tables are described as follows:
| Field | Type |
|---|---|
| ID | NUMBER |
| NAME | VARCHAR2(17) |
| COUNTRYCODE | VARCHAR2(3) |
| DISTRICT | VARCHAR2(20) |
| POPULATION | NUMBER |
| Field | Type |
|---|---|
| CODE | VARCHAR2(3) |
| NAME | VARCHAR2(44) |
| CONTINENT | VARCHAR2(13) |
| REGION | VARCHAR2(25) |
| SURFACEAREA | NUMBER |
| INDEPYEAR | VARCHAR2(5) |
| POPULATION | NUMBER |
| LIFEEXPECTANCY | VARCHAR2(4) |
| GNP | NUMBER |
| GNPOLD | VARCHAR2(9) |
| LOCALNAME | VARCHAR2(44) |
| GOVERNMENTFORM | VARCHAR2(44) |
| HEADOFSTATE | VARCHAR2(32) |
| CAPITAL | VARCHAR2(4) |
| CODE2 | VARCHAR2(2) |
Solution – Average Population of Each Continent in SQL
MySQL
select country.continent, floor(avg(city.population)) from country join city on city.countrycode = country.code group by country.continent;
Disclaimer: The above Problem (Average Population of Each Continent) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.