Add: Weather Type in Each Country

Author: openset

problems/weather-type-in-each-country/README.md

@@ -11,4 +11,86 @@
 
 ## [1294. Weather Type in Each Country (Easy)](https://leetcode.com/problems/weather-type-in-each-country "")
 
+<p>Table: <code>Countries</code></p>
+<pre>
++---------------+---------+
+| Column Name   | Type    |
++---------------+---------+
+| country_id    | int     |
+| country_name  | varchar |
++---------------+---------+
+country_id is the primary key for this table.
+Each row of this table contains the ID and the name of one country.
+</pre>
+ 
+<p>Table: <code>Weather</code></p>
+<pre>
++---------------+---------+
+| Column Name   | Type    |
++---------------+---------+
+| country_id    | int     |
+| weather_state | varchar |
+| day           | date    |
++---------------+---------+
+(country_id, day) is the primary key for this table.
+Each row of this table indicates the weather state in a country for one day.
+</pre>
+ 
+Write an SQL query to find the type of weather in each country for November 2019.
 
+The type of weather is Cold if the average weather_state is less than or equal 15, Hot if the average weather_state is greater than or equal 25 and Warm otherwise.
+
+Return result table in any order.
+
+The query result format is in the following example:
+<pre>
+Countries table:
++------------+--------------+
+| country_id | country_name |
++------------+--------------+
+| 2          | USA          |
+| 3          | Australia    |
+| 7          | Peru         |
+| 5          | China        |
+| 8          | Morocco      |
+| 9          | Spain        |
++------------+--------------+
+Weather table:
++------------+---------------+------------+
+| country_id | weather_state | day        |
++------------+---------------+------------+
+| 2          | 15            | 2019-11-01 |
+| 2          | 12            | 2019-10-28 |
+| 2          | 12            | 2019-10-27 |
+| 3          | -2            | 2019-11-10 |
+| 3          | 0             | 2019-11-11 |
+| 3          | 3             | 2019-11-12 |
+| 5          | 16            | 2019-11-07 |
+| 5          | 18            | 2019-11-09 |
+| 5          | 21            | 2019-11-23 |
+| 7          | 25            | 2019-11-28 |
+| 7          | 22            | 2019-12-01 |
+| 7          | 20            | 2019-12-02 |
+| 8          | 25            | 2019-11-05 |
+| 8          | 27            | 2019-11-15 |
+| 8          | 31            | 2019-11-25 |
+| 9          | 7             | 2019-10-23 |
+| 9          | 3             | 2019-12-23 |
++------------+---------------+------------+
+Result table:
++--------------+--------------+
+| country_name | weather_type |
++--------------+--------------+
+| USA          | Cold         |
+| Austraila    | Cold         |
+| Peru         | Hot          |
+| China        | Warm         |
+| Morocco      | Hot          |
++--------------+--------------+
+Average weather_state in USA in November is (15) / 1 = 15 so weather type is Cold.
+Average weather_state in Austraila in November is (-2 + 0 + 3) / 3 = 0.333 so weather type is Cold.
+Average weather_state in Peru in November is (25) / 1 = 25 so weather type is Hot.
+Average weather_state in China in November is (16 + 18 + 21) / 3 = 18.333 so weather type is Warm.
+Average weather_state in Morocco in November is (25 + 27 + 31) / 3 = 27.667 so weather type is Hot.
+We know nothing about average weather_state in Spain in November so we don't include it in the result table. 
+</pre>

problems/weather-type-in-each-country/weather_type_in_each_country.sql

@@ -0,0 +1,11 @@
+# Write your MySQL query statement below
+
+SELECT country_name,
+       CASE
+           WHEN avg(weather_state) <= 15 THEN "Cold"
+           WHEN avg(weather_state) >= 25 THEN "Hot"
+           ELSE "Warm" END AS weather_type
+FROM Countries
+         INNER JOIN Weather ON Countries.country_id = Weather.country_id
+WHERE LEFT(DAY, 7) = '2019-11'
+GROUP BY country_name