Employee Bonus — LeetCode 577 Python Solution

The problem

Report the name and bonus of every employee whose bonus is less than 1000, and include the employees who have no bonus record at all, showing a null bonus for them. Employee has empId (int, unique), name (varchar), supervisor (int) and salary (int); Bonus has empId (int, primary key referencing an employee) and bonus (int). Return name and bonus.

Example

Input

Employee(empId, name, supervisor, salary) = (1, 'Ana', null, 5000), (2, 'Ben', 1, 3000), (3, 'Cleo', 1, 3500), (4, 'Dev', 1, 2500); Bonus(empId, bonus) = (2, 500), (3, 2000)

Output

('Ana', null), ('Ben', 500), ('Dev', null)

Explanation

Ben is under 1000, Ana and Dev have no bonus row at all, and Cleo's 2000 is excluded.

Python solution

import pandas as pd

def employee_bonus(employee: pd.DataFrame, bonus: pd.DataFrame) -> pd.DataFrame:
    df = employee.merge(bonus, on='empId', how='left')
    res = df[(df['bonus'].isna()) | (df['bonus'] < 1000)]
    return res[['name', 'bonus']]

Complexity

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