#1076
Easy Database Project employees ii
Database
50.4% acceptance
Mar 31, 2026
207
63
No description available.
Solution
Pandas
Time O(1)
Space O(1)
# Table: Project
#
# +-------------+---------+
# | Column Name | Type |
# +-------------+---------+
# | project_id | int |
# | employee_id | int |
# +-------------+---------+
# (project_id, employee_id) is the primary key (combination of columns 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.
#
#
#
# Table: Employee
#
# +------------------+---------+
# | Column Name | Type |
# +------------------+---------+
# | employee_id | int |
# | name | varchar |
# | experience_years | int |
# +------------------+---------+
# 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 report all the projects that have the most employees.
#
# Return the result table in any order.
#
# The result format is in the following example.
#
# Example 1:
# Input:
# Project table:
# +-------------+-------------+
# | project_id | employee_id |
# +-------------+-------------+
# | 1 | 1 |
# | 1 | 2 |
# | 1 | 3 |
# | 2 | 1 |
# | 2 | 4 |
# +-------------+-------------+
# Employee table:
# +-------------+--------+------------------+
# | employee_id | name | experience_years |
# +-------------+--------+------------------+
# | 1 | Khaled | 3 |
# | 2 | Ali | 2 |
# | 3 | John | 1 |
# | 4 | Doe | 2 |
# +-------------+--------+------------------+
# Output:
# +-------------+
# | project_id |
# +-------------+
# | 1 |
# +-------------+
# Explanation: The first project has 3 employees while the second one has 2.
import pandas as pd
def project_employees_ii(project: pd.DataFrame, employee: pd.DataFrame) -> pd.DataFrame:
counts = project.groupby('project_id')['employee_id'].count()
max_count = counts.max()
result = counts[counts == max_count].reset_index()
return result[['project_id']]