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3057. Employees Project Allocation

Description

Table: Project

+-------------+---------+
\| Column Name \| Type    \|
+-------------+---------+
\| project_id  \| int     \|
\| employee_id \| int     \|
\| workload    \| int     \|
+-------------+---------+
employee_id is the primary key (column with unique values) of this table.
employee_id is a foreign key (reference column) to Employee table.
Each row of this table indicates that the employee with employee_id is working on the project with project_id and the workload of the project.

Table: Employees

+------------------+---------+
\| Column Name      \| Type    \|
+------------------+---------+
\| employee_id      \| int     \|
\| name             \| varchar \|
\| team             \| varchar \|
+------------------+---------+
employee_id is the primary key (column with unique values) of this table.
Each row of this table contains information about one employee.

Write a solution to find the employees who are allocated to projects with a workload that exceeds the average workload of all employees for their respective teams

Return the result table ordered by employee_id, project_id in ascending order.

The result format is in the following example.

 

Example 1:

Input: 
Project table:
+-------------+-------------+----------+
\| project_id  \| employee_id \| workload \|
+-------------+-------------+----------+
\| 1           \| 1           \|  45      \|
\| 1           \| 2           \|  90      \| 
\| 2           \| 3           \|  12      \|
\| 2           \| 4           \|  68      \|
+-------------+-------------+----------+
Employees table:
+-------------+--------+------+
\| employee_id \| name   \| team \|
+-------------+--------+------+
\| 1           \| Khaled \| A    \|
\| 2           \| Ali    \| B    \|
\| 3           \| John   \| B    \|
\| 4           \| Doe    \| A    \|
+-------------+--------+------+
Output: 
+-------------+------------+---------------+------------------+
\| employee_id \| project_id \| employee_name \| project_workload \|
+-------------+------------+---------------+------------------+  
\| 2           \| 1          \| Ali           \| 90               \| 
\| 4           \| 2          \| Doe           \| 68               \| 
+-------------+------------+---------------+------------------+
Explanation: 
- Employee with ID 1 has a project workload of 45 and belongs to Team A, where the average workload is 56.50. Since his project workload does not exceed the team's average workload, he will be excluded.
- Employee with ID 2 has a project workload of 90 and belongs to Team B, where the average workload is 51.00. Since his project workload does exceed the team's average workload, he will be included.
- Employee with ID 3 has a project workload of 12 and belongs to Team B, where the average workload is 51.00. Since his project workload does not exceed the team's average workload, he will be excluded.
- Employee with ID 4 has a project workload of 68 and belongs to Team A, where the average workload is 56.50. Since his project workload does exceed the team's average workload, he will be included.
Result table orderd by employee_id, project_id in ascending order.

Solutions

Solution 1: Grouping Statistics + Equi-Join

First, we join the Project table and the Employees table based on employee_id, then group by team to calculate the average workload of each team, and record it in the temporary table T.

Then, we join the Project table and the Employees table again, and also join the T table, to find employees whose workload is greater than the average workload of the team. Finally, we sort by employee_id and project_id.

• Python
• SQL

• import pandas as pd
  
  
  def employees_with_above_avg_workload(
      project: pd.DataFrame, employees: pd.DataFrame
  ) -> pd.DataFrame:
      merged_df = pd.merge(project, employees, on="employee_id")
      avg_workload_per_team = merged_df.groupby("team")["workload"].mean().reset_index()
      merged_df = pd.merge(
          merged_df, avg_workload_per_team, on="team", suffixes=("", "_avg")
      )
      ans = merged_df[merged_df["workload"] > merged_df["workload_avg"]]
      ans = ans[["employee_id", "project_id", "name", "workload"]]
      ans = ans.rename(columns={"name": "employee_name", "workload": "project_workload"})
      return ans.sort_values(by=["employee_id", "project_id"])
  
  

• # Write your MySQL query statement below
  WITH
      T AS (
          SELECT team, AVG(workload) AS avg_workload
          FROM
              Project
              JOIN Employees USING (employee_id)
          GROUP BY 1
      )
  SELECT
      employee_id,
      project_id,
      name AS employee_name,
      workload AS project_workload
  FROM
      Project
      JOIN Employees USING (employee_id)
      JOIN T USING (team)
  WHERE workload > avg_workload
  ORDER BY 1, 2;
  
  

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