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Added Count Mutual friends
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‎Count Mutual Friends/README.md

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# Count Mutual Friends
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![Star Badge](https://img.shields.io/static/v1?label=%F0%9F%8C%9F&message=If%20Useful&style=style=flat&color=BC4E99)
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[![View Main Folder](https://img.shields.io/badge/View-Main_Folder-971901?)](https://github.com/thecoddiwompler/SQL-Practice-Questions/tree/main)
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[![View Repositories](https://img.shields.io/badge/View-My_Repositories-blue?logo=GitHub)](https://github.com/thecoddiwompler?tab=repositories)
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[![View My Profile](https://img.shields.io/badge/View-My_Profile-green?logo=GitHub)](https://github.com/thecoddiwompler)
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---
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## 🛠️ Problem Statement
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<b>Table Name : Friends</b>
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| Column Name |Type |
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| ------------- | ------------- |
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| Friend1 | varchar(10) |
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| Friend2 | varchar(10) |
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</br>
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In this table, all pair of friends are given. For the given friends, Write a SQL Query to find the no of mutual friends.
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Return the result table in any order.
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The query result format is in the following example:
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<details>
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<summary>
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Input
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</summary>
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<br>
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<b> Table Name: Friends </b>
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<br>
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| Friend1 | Friend2 |
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|---------|---------|
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| Jason | Mary |
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| Mike | Mary |
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| Mike | Jason |
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| Susan | Jason |
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| John | Mary |
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| Susan | Mary |
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</details>
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<details>
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<summary>
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Output
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</summary>
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<br>
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| Friend1 | Friend2 | no_of_mutual_friends |
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|---------|---------|----------------------|
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| Jason | Mary | 2|
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| Mike | Mary |0|
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| Mike | Jason |1|
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| Susan | Jason |1|
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| John | Mary |1|
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| Susan | Mary |1|
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</details>
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---

‎Count Mutual Friends/schema.sql

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CREATE TABLE Friends (Friend1 VARCHAR(10), Friend2 VARCHAR(10));
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INSERT INTO
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Friends
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VALUES
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('Jason', 'Mary'),
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('Mike', 'Mary'),
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('Mike', 'Jason'),
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('Susan', 'Jason'),
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('John', 'Mary'),
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('Susan', 'Mary');
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COMMIT;

‎Count Mutual Friends/solution.sql

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WITH cte AS (
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SELECT
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friend1,
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friend2
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FROM
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friends
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UNION
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SELECT
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friend2,
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friend1
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FROM
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friends
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),
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friend1_friend AS (
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SELECT
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a.friend1,
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a.friend2,
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b.friend2 friend1_friend
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FROM
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friends a
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LEFT OUTER JOIN cte b ON a.friend1 = b.friend1
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AND a.friend2 <> b.friend2
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),
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friend2_friend AS (
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SELECT
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a.friend1,
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a.friend2,
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b.friend2 friend2_friend
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FROM
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friends a
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INNER JOIN cte b ON a.friend2 = b.friend1
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WHERE
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a.friend1 <> b.friend2
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)
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SELECT
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a.friend1,
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a.friend2,
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count(*) filter(
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WHERE
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a.friend1_friend IS NOT NULL
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) no_of_mutual_friends
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FROM
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friend1_friend a
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LEFT OUTER JOIN friend2_friend b ON a.friend1 = b.friend1
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AND a.friend2 = b.friend2
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AND a.friend1_friend = b.friend2_friend
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GROUP BY
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a.friend1,
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a.friend2;

‎iqsql.md

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3. Child - Parent - Grandparent Hierarchy
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4. Clocked Hours
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5. Company Structure [Extra Difficult]
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6. Department Top 3 Salary
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7. Employees Check-in Details
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8. Employees Hiring [Difficult]
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9. Employees Log [Extra Difficult]
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10. Final Destination
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11. Find Cumulative Salary of an Employee
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12. Highest-Grossing Items
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13. Increasing Sales Revenue
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14. Last Person to Fit in the Bus
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15. Manager with at least 5 direct reportees
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16. Mismatched IDs
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17. Missing Invoices
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18. Odd and Even Measurements
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19. Onboarded Cities
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20. Qualifying Criteria
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21. Travel Analytics
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6. Count Mutual Friends [Difficult]
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7. Department Top 3 Salary
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8. Employees Check-in Details
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9. Employees Hiring [Difficult]
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10. Employees Log [Extra Difficult]
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11. Final Destination
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12. Find Cumulative Salary of an Employee
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13. Highest-Grossing Items
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14. Increasing Sales Revenue
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15. Last Person to Fit in the Bus
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16. Manager with at least 5 direct reportees
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17. Mismatched IDs
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18. Missing Invoices
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19. Odd and Even Measurements
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20. Onboarded Cities
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21. Qualifying Criteria
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22. Travel Analytics

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