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islam added aggregate functions #[6,8]
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‎Basic Aggregate Functions/06 -Monthly Transactions I/README.md‎

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# [Monthly Transactions I](https://leetcode.com/problems/monthly-transactions-i/description/?envType=study-plan-v2&envId=top-sql-50)
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### Problem Requirements:
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Find for each month and country, the number of transactions and their total amount, the number of approved transactions and their total amount.
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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>How can you do some things when you find given conditions?</p>
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</details>
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<details>
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<summary>Hint#2</summary>
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<p>Try it using <code>CASE</code> statement</p>
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</details>
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<details>
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<summary>Hint#3</summary>
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<p>
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SQL has an aggregation function called <code>COUNT(expression)</code> which count all the rows that satisfy a specified condition
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</p>
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</details>
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<details>
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<summary>Hint#4</summary>
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<p>
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SQL has an aggregation function called <code>SUM(expression)</code> which calculate the sum of values in a set
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</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>GROUP BY</code> clause to group a set of rows into a set of summary rows</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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<ul>
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<li>To count the total number of transactions we can use <code>COUNT()</code> function </li>
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<li>
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To count the total number of approved transactions we can use <code>COUNT()</code> function with the help of <code>CASE</code> statement as follows
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<br>
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WHEN state = approved THEN add <code>1</code> ELSE add <code>0</code>
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</li>
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<li>To sum the total amount we can use <code>SUM()</code> function </li>
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<li>
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To sum the total approved amount we can use <code>SUM()</code> function with the help of <code>CASE</code> statement as follows
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<br>
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WHEN state = approved THEN add <code>amount</code> ELSE add <code>0</code>
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</li>
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<li>
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All these functions should grouped by the combination of month and country with the <code>GROUP BY</code> clause.
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</li>
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</ul>
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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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DATE_FORMAT(trans_date, '%Y-%m') AS month,
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country,
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COUNT(state) AS trans_count,
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SUM(
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CASE WHEN state = 'approved' THEN 1 ELSE 0 END
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) AS approved_count,
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SUM(amount) AS trans_total_amount,
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SUM(
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CASE WHEN state = 'approved' THEN amount ELSE 0 END
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) AS approved_total_amount
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FROM
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Transactions
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GROUP BY
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month,
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country
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-- we can count the number of approved transactions in another way.
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-- COUNT(
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-- CASE WHEN state = 'approved' THEN 1 ELSE NULL END
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-- ) AS approved_count
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-- COUNT function will ignore NULL.
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```
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</details>

‎Basic Aggregate Functions/08 -Game Play Analysis IV/README.md‎

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# [Game Play Analysis IV](https://leetcode.com/problems/game-play-analysis-iv/description/?envType=study-plan-v2&envId=top-sql-50)
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### Problem Requirements:
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Find the number of players that logged in for at least two consecutive days starting from their first login date, then divide that number by the total number of players.
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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>
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What if you have another column that has the <code>first_login</code>,
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The problem will be very easy, isn't it?
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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>
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Try use <code>WITH</code> clause. It will help you to define a temporary data set. It will make your life easier.
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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>
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The SQL <code>MIN(expression)</code> function returns the minimum value in a set of values
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</p>
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</details>
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<details>
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<summary>Hint#4</summary>
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<p>
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Try use <code>window function</code>.
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</p>
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</details>
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<details>
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<summary>Hint#5</summary>
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<p>SQL has an aggregation function called <code>COUNT(expression)</code> which count all the rows that satisfy a specified condition</p>
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</details>
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<details>
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<summary>Hint#6</summary>
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<p>
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To subtract an interval e.g., a year, a month and a day to date, you can use the <code>DATE_SUB()</code> function.
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</p>
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</details>
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<details>
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<summary>Hint#7</summary>
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<p>SQL has a <code>ROUND(number, decimals)</code> function which rounds a number to a specified number of decimal places. </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 count the number of players who logged in the next day of their first login day.
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<br>
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we should determine the first login date for each player.
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Then when we find a player logged in the next day of this day we should put this player into account.
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<ul>
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<li>We can determine the first login for each player with the help of aggregate window function
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<code>MIN(expression) OVER </code>
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. Aggregate window function will work the same as the normal aggregate function but it doesn't cause rows to become grouped into a single output row.
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</li>
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<li>To be able to use our original table and the new column we can use <code> WITH</code> clause , it will help us to define the temporary data set and use them in the query.
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</li>
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<li>
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We will filter this table according to our condition <code>(event_date and first_login are consecutive days)</code>.
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<br>
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We can use <code>DATE_SUB()</code> function to subtract a given duration from <code>event_date</code> As follows<br> <code>DATE_SUB(event_date, INTERVAL 1 DAY) = first_login</code>
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</li>
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<li>
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Finally, divide the player that satisfies the condition by the total number of player that we can calculate as a <code>subquery</code> then round the result using <code>ROUND()</code>function to round the result.
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</li>
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</ul>
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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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WITH temp AS(
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SELECT
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*,
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MIN(event_date) OVER (Partition BY player_id) AS first_login
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-- over means that the function before it is a window function
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FROM
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Activity
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)
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SELECT
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ROUND(
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(
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SELECT
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COUNT(player_id)
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FROM
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temp
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WHERE
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DATE_SUB(event_date, INTERVAL 1 DAY) = first_login
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) / (
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SELECT
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COUNT(DISTINCT player_id)
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FROM
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Activity
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),
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2
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) AS fraction
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```
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</details>

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