Problem Statement

Table: CollaborationAccepted
Each row in this table records an instance where one employee invites another to collaborate on a project, and the invitation is accepted.

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| inviter_id     | int     |
| invitee_id     | int     |
| project_id     | int     |
| accept_date    | date    |
+----------------+---------+
The combination of (inviter_id, invitee_id, project_id) is the primary key for this table.
This table includes the ID of the inviter, the ID of the invitee, the project they are collaborating on, and the date the invitation was accepted.

Develop a solution to find the employee(s) who have collaborated on the most number of unique projects. In case of a tie, list all such employees.

Example

Input:

CollaborationAccepted table:
+------------+------------+------------+-------------+
| inviter_id | invitee_id | project_id | accept_date |
+------------+------------+------------+-------------+
| 1          | 2          | 100        | 2020-05-01  |
| 1          | 3          | 101        | 2020-05-02  |
| 2          | 3          | 102        | 2020-05-03  |
| 3          | 4          | 103        | 2020-05-04  |
| 2          | 4          | 104        | 2020-05-05  |
+------------+------------+------------+-------------+

Output:

+----+--------------+
| id | num_projects |
+----+--------------+
| 2  | 3            |
| 3  | 3            |
+----+--------------+

In this example, both employees 2 and 3 have collaborated on 3 unique projects, which is the highest among all employees.

Try It yourself

Database Exercise

Database Schema:

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Exercise Script:

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Available actions: Execute

Solution

To identify the employee(s) who have collaborated on the most number of unique projects, we analyze the CollaborationAccepted table. We can accomplishes this by leveraging Common Table Expressions (CTEs) and set operations to aggregate and determine the desired results. Below is a comprehensive breakdown of the approach, step-by-step execution, and detailed explanations for each line of the SQL query.

Approach Overview

1. Combine Collaborations: Merge the inviter and invitee collaborations to treat all collaborations uniformly.
2. Count Unique Projects per Employee: Calculate the number of distinct projects each employee has collaborated on.
3. Identify Maximum Project Count: Determine the highest number of unique projects any employee has collaborated on.
4. Select Top Collaborators: Retrieve the employee(s) whose project count matches the maximum identified.
5. Present the Final Results: Display the employee IDs alongside their respective number of unique projects.

SQL Query

WITH AllCollaborations AS (
    -- Combine inviter and invitee collaborations
    SELECT inviter_id AS id, project_id
    FROM CollaborationAccepted
    UNION
    SELECT invitee_id AS id, project_id
    FROM CollaborationAccepted
),
ProjectCounts AS (
    -- Count unique projects per employee
    SELECT id, COUNT(DISTINCT project_id) AS num_projects
    FROM AllCollaborations
    GROUP BY id
)
-- Select employees with the maximum number of projects
SELECT id, num_projects
FROM ProjectCounts
WHERE num_projects = (
    SELECT MAX(num_projects)
    FROM ProjectCounts
)
ORDER BY id;

Step-by-Step Approach

Step 1: Combine Collaborations (AllCollaborations)

Merge the inviter and invitee collaborations to create a unified view of all collaborations, treating both roles equally.

SQL Snippet:

WITH AllCollaborations AS (
    -- Combine inviter and invitee collaborations
    SELECT inviter_id AS id, project_id
    FROM CollaborationAccepted
    UNION
    SELECT invitee_id AS id, project_id
    FROM CollaborationAccepted
),

Explanation:

1. WITH AllCollaborations AS (

   • Initiates a Common Table Expression (CTE) named AllCollaborations. CTEs allow for temporary result sets that can be referenced within the main query.

2. SELECT inviter_id AS id, project_id

   • Selects the inviter_id from the CollaborationAccepted table and aliases it as id.
   • Retrieves the associated project_id for each invitation.

3. UNION

   • Combines the results of the first SELECT with the second SELECT.
   • The UNION operator ensures that duplicate records are eliminated, resulting in a list of unique (id, project_id) pairs.

4. SELECT invitee_id AS id, project_id

   • Selects the invitee_id from the CollaborationAccepted table and aliases it as id.
   • Retrieves the associated project_id for each accepted invitation.

Intermediate Output After Step 1 (AllCollaborations):

+----+------------+
| id | project_id |
+----+------------+
| 1  | 100        |
| 2  | 100        |
| 1  | 101        |
| 3  | 101        |
| 2  | 102        |
| 3  | 102        |
| 3  | 103        |
| 4  | 103        |
| 2  | 104        |
| 4  | 104        |
+----+------------+

Step 2: Count Unique Projects per Employee (ProjectCounts)

Calculate the number of distinct projects each employee has collaborated on.

SQL Snippet:

ProjectCounts AS (
    -- Count unique projects per employee
    SELECT id, COUNT(DISTINCT project_id) AS num_projects
    FROM AllCollaborations
    GROUP BY id
)

Explanation:

1. ProjectCounts AS (

   • Initiates a second CTE named ProjectCounts.

2. SELECT id, COUNT(DISTINCT project_id) AS num_projects

   • Selects the id (employee ID) from the AllCollaborations CTE.
   • Counts the number of distinct project_ids each employee has collaborated on.
   • Aliases the count as num_projects.

3. GROUP BY id

   • Groups the data by id to perform the aggregation for each employee.

Intermediate Output After Step 2 (ProjectCounts):

+----+--------------+
| id | num_projects |
+----+--------------+
| 1  | 2            |
| 2  | 3            |
| 3  | 3            |
| 4  | 2            |
+----+--------------+

Step 3: Identify Maximum Project Count

Determine the highest number of unique projects any employee has collaborated on.

SQL Snippet:

SELECT id, num_projects
FROM ProjectCounts
WHERE num_projects = (
    SELECT MAX(num_projects)
    FROM ProjectCounts
)

Explanation:

1. SELECT id, num_projects

   • Selects the id and num_projects columns from the ProjectCounts CTE.

2. WHERE num_projects = (

   • Introduces a condition to filter employees whose num_projects equals the maximum found.

3. SELECT MAX(num_projects)

   • Selects the highest value of num_projects across all employees.

4. ORDER BY id;

   • Orders the final result set by id in ascending order for clarity.

Final Output:

+----+--------------+
| id | num_projects |
+----+--------------+
| 2  | 3            |
| 3  | 3            |
+----+--------------+

Explanation of Output:

• Employee 2:

  • Collaborated on 3 unique projects (100, 102, 104)

• Employee 3:

  • Collaborated on 3 unique projects (101, 102, 103)