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‎Basic Aggregate Functions/01- Not Boring Movies/README.md‎

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# [Not Boring Movies](https://leetcode.com/problems/not-boring-movies/description/?envType=study-plan-v2&envId=top-sql-50)
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### Problem Requirements:
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Report the movies with an odd-numbered ID and a description that is not <code>"boring"</code>.
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Return the result table ordered by <code>rating</code> in <strong> descending order</strong>.
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<details>
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<summary style="font-size:1.3rem;"> <strong>Hints</strong> </summary>
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<details>
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<summary>Hint#1</summary>
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<p>SQL has a <code>MOD(num1, num2)</code> function that returns the value of <code>num1 % num2</code> (with <code>%</code> beging the modulo operator) </p>
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</details>
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<details>
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<summary>Hint#2</summary>
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<p>SQL has a <code>NOT</code> operator </p>
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</details>
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</details>
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<details>
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<summary style="font-size:1.3rem;"> <strong>Explanation</strong> </summary>
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In order to solve this problem we need to use the <code>mod</code> function to get the odd numbered IDs and the <code>not</code> operator to get the movies with description not equal to <code>"boring"</code>.
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Then, we can combine the two conditions using the <code>and</code> operator.
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Finally, we can order the result by <code>rating</code> in descending order using <code>desc</code> keyword.
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</details>
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<details>
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<summary style="font-size:1.3rem"><strong> SQL Solution</strong> </summary>
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```sql
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select id, movie, description, rating from Cinema
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where mod(id, 2)=1 and not description = "boring"
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order by rating desc;
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```
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</details>

‎Basic Aggregate Functions/03- Project Employees I/README.md‎

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# [Project Employees I](https://leetcode.com/problems/project-employees-i/description/?envType=study-plan-v2&envId=top-sql-50)
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### Problem Requirements:
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Report the <code>average</code> experience years of all the employees for each project, <strong> rounded to 2 digits</strong>.
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Return the result table in any order.
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<details>
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<summary style="font-size:1.3rem;"> <strong>Hints</strong> </summary>
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<details>
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<summary>Hint#1</summary>
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<p>When we have more than one tables we must think about joins first</p>
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</details>
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<details>
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<summary>Hint#2</summary>
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<p>SQL has a function <code>AVG(column_name)</code> returns the average value of that column </p>
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</details>
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<details>
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<summary>Hint#3</summary>
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<p>Here we are asked to find the average for each project, we must think about <strong>Grouping</strong> </p>
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</details>
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</details>
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<details>
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<summary style="font-size:1.3rem;"> <strong>Explanation</strong> </summary>
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We want to find the average experience years for each project, so we must join the two tables on the <code>employee_id</code> column, then we must group the result by the <code>project_id</code> column, and finally we must find the average of the <code>experience_years</code> column.
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When selecting the average of a column we must use the <code>AVG(column_name)</code> function, and to round the result to 2 digits we must use the <code>ROUND()</code> function, then we use <code>AS</code> to give the column an alias (<code>average_years</code>)
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</details>
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<details>
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<summary style="font-size:1.3rem"><strong> SQL Solution</strong> </summary>
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```sql
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select p.project_id, round(avg(e.experience_years),2) as average_years
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from Project p
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inner join Employee e on e.employee_id = p.employee_id
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group by p.project_id;
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```
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</details>

‎Basic Aggregate Functions/05- Queries Quality and Percentage/README.md‎

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# [Queries Quality and Percentage](https://leetcode.com/problems/queries-quality-and-percentage/description/?envType=study-plan-v2&envId=top-sql-50)
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### Problem Requirements:
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We define query <code>quality</code> as:
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The average of the ratio between query rating and its position.
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We also define <code>poor_query_percentage</code> as:
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The percentage of all queries with rating less than 3.
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Find each <code>query_name</code>, the <code>quality</code> and <code>poor_query_percentage</code>.
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Both <code>quality</code> and <code>poor_query_percentage</code> should be rounded to <code>2</code> decimal places.
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Return the result table in any order.
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<!-- <details open>
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<summary style="font-size:1.5rem;"> <strong>Solution#1</strong> </summary> -->
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<details>
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<summary style="font-size:1.3rem;"> <strong>Hints</strong> </summary>
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<details>
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<summary>Hint#1</summary>
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<p>How SQL <code>AVG</code> function works ? <br>
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<code>AVG(column_name) = SUM(column_name) / COUNT(column_name)</code>
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</p>
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</details>
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<details>
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<summary>Hint#2</summary>
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<p>SQL has something called <code>CASE</code> Expression <br>
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<code>CASE WHEN condition THEN value1 ELSE value2 END</code>
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</p>
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</details>
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<details>
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<summary>Hint#3</summary>
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<p>Try combining <code>AVG</code> and <code>CASE</code> together</p>
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</details>
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<details>
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<summary>Hint#5</summary>
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<p> Use Grouping </p>
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</details>
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<details>
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<summary>Hint#4</summary>
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<p> Note that <code>query_name</code> may be <code>NULL</code> </p>
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</details>
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</details>
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<details>
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<summary style="font-size:1.3rem;"> <strong>Explanation</strong> </summary>
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This problem is a bit tricky, we need to calculate the <code>quality</code> and <code>poor_query_percentage</code> for each <code>query_name</code>.
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First we must think about Grouping the data by <code>query_name</code>, then we can calculate the <code>quality</code> and <code>poor_query_percentage</code> for each group using <code>AVG</code> function.
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The <code>quality</code> is the average of the ratio between query rating and its position <code>q1.rating / q1.position</code>, so we can use <code>AVG</code> as following: <code>AVG(q1.rating / q1.position)</code>
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The <code>poor_query_percentage</code> is the percentage of all queries with rating less than <code>3</code>, so how do we get the avarge of this condition ? we can use <code>CASE</code> expression to check if the rating is less than <code>3</code>, then we can calculate the average of this condition, do not forget to multiply the avarage by <code>100</code> to get the percentage.
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Do not forget to round the results to <code>2</code> decimal places.
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Also do not forget to filter out the <code>NULL</code> values in <code>query_name</code> column.
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</details>
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<details>
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<summary style="font-size:1.3rem"><strong> SQL Solution</strong> </summary>
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```sql
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select
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q1.query_name as query_name,
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round(avg(q1.rating / q1.position), 2) as quality,
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round(100 * avg(case when q1.rating < 3 then 1 else 0 end), 2) as poor_query_percentage
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from Queries q1
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where q1.query_name is not null
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group by query_name;
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```
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</details>
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<!-- </details> -->

‎Basic Aggregate Functions/07- Immediate Food Delivery II/README.md‎

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# [Immediate Food Delivery II](https://leetcode.com/problems/immediate-food-delivery-ii/description/?envType=study-plan-v2&envId=top-sql-50)
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### Problem Requirements:
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If the customer's <strong> preferred delivery date </strong> is the same as the <strong>order date</strong>, then the order is called <strong>immediate</strong>; otherwise, it is called scheduled.
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The first order of a customer is the order with the <strong>earliest order date</strong> that the customer made. It is <strong>guaranteed</strong> that a customer has precisely one first order.
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Find the <strong>percentage of immediate orders in the first orders of all customers</strong>, rounded to <strong>2</strong> decimal places.
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<details>
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<summary style="font-size:1.3rem;"> <strong>Hints</strong> </summary>
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<details>
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<summary>Hint#1</summary>
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<p>Immediate order occurres when <code>order_date = customer_pref_delivery_date</code></p>
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</details>
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<details>
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<summary>Hint#2</summary>
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<p>Use <code>CASE</code> Expression and the <code>SUM</code> function to find the number of Immediate orders.
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What will the whole sample sapce of orders be ?</p>
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</details>
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<details>
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<summary>Hint#3</summary>
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<p>The sample space is the number of all orders in the table we can get it using <code>COUNT(*)</code> </p>
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</details>
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<details>
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<summary>Hint#4</summary>
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<p>How would you find the <code>MIN</code> <strong>order_date</strong> for each customer ? </p>
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</details>
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<details>
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<summary>Hint#5</summary>
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<p>Use <code>WHERE IN</code> clause (subquering) </p>
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</details>
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</details>
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<details>
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<summary style="font-size:1.3rem;"> <strong>Explanation</strong> </summary>
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This problem is somewhat tough, we need to calculate the percentage of immediate orders in the first orders of all customers.
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First we need to find a way to find the immediate orders. But, before that we must also find a way to find the first order of each customer.
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To find the first order of each customer we will be using <code>WHERE IN</code> clause.
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The <code>WHERE IN</code> clause is used to filter the result of a subquery. It extracts the rows from the outer query and then compares it with the result of the subquery. It looks like this:
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```sql
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WHERE (col1,col2) IN (
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SELECT SOME_FUNCTION(col1), SOME_FUNCTION(col2)
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FROM tableX
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WHERE condition
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);
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```
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Now we need to find the number of immediate orders, we can do that using <code>CASE</code> expression and <code>SUM</code> function.
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Since we did filter the Delivery table to have only the <code>customer_id</code> with the <strong>first order made </strong> we can now check the use <code>SUM</code> funtion on the <code>CASE</code> condition to find the number of <code>order_date = customer_pref_delivery_date</code> inside the <code>CASE</code> expression to find the count of immediate orders.
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We can divide the Summetion by the <code>COUNT(\*)</code> then multiply by <code>100</code> to get the percentage. Of course do not forget to <strong>round</strong> to <strong>2</strong> decimal places.
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</details>
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<details>
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<summary style="font-size:1.3rem"><strong> SQL Solution</strong> </summary>
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```sql
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SELECT ROUND(100 * SUM(CASE WHEN order_date = customer_pref_delivery_date THEN 1 ELSE 0 END) / COUNT(*), 2) AS immediate_percentage
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FROM Delivery
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WHERE (customer_id,order_date) IN (
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SELECT customer_id, MIN(order_date) AS od
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FROM Delivery
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GROUP BY customer_id
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);
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```
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</details>

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