Create a New Column - Practice Coding Problems

Create a New Column - Problem

Database Easy

🏢 Employee Bonus Calculator

A company wants to reward its hardworking employees with a special bonus! You're given a DataFrame containing employee information with their names and current salaries.

Your task: Create a new column called bonus that contains double the salary for each employee.

Input Format:

You'll receive a pandas DataFrame employees with the following structure:

Column Name	Type
name	object
salary	int

Output:

Return the same DataFrame with an additional bonus column where each bonus equals 2 × salary.

Example: If Alice earns $50,000, her bonus should be $100,000!

Input & Output

example_1.py — Basic Employee Bonus

$ Input: employees = pd.DataFrame({ 'name': ['Alice', 'Bob'], 'salary': [50000, 60000] })

› Output: name salary bonus 0 Alice 50000 100000 1 Bob 60000 120000

💡 Note: Alice earns $50,000 so her bonus is $50,000 × 2 = $100,000. Bob earns $60,000 so his bonus is $60,000 × 2 = $120,000.

example_2.py — Multiple Employees

$ Input: employees = pd.DataFrame({ 'name': ['Charlie', 'Diana', 'Eve'], 'salary': [75000, 45000, 90000] })

› Output: name salary bonus 0 Charlie 75000 150000 1 Diana 45000 90000 2 Eve 90000 180000

💡 Note: Each employee's bonus is calculated as double their salary: Charlie gets $150,000, Diana gets $90,000, and Eve gets $180,000.

example_3.py — Single Employee Edge Case

$ Input: employees = pd.DataFrame({ 'name': ['John'], 'salary': [100000] })

› Output: name salary bonus 0 John 100000 200000

💡 Note: Even with just one employee, the same logic applies: John's bonus of $200,000 is double his $100,000 salary.

Visualization

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Understanding the Visualization

Original DataFrame

Start with employee data containing names and salaries

Column Selection

Select the salary column for transformation

Vectorized Operation

Apply multiplication by 2 to all values simultaneously

New Column Creation

Assign the results to a new 'bonus' column

Key Takeaway

🎯 Key Insight: Pandas vectorized operations allow us to transform entire columns with mathematical operations in a single, efficient step - just like applying formulas in spreadsheets!

Time & Space Complexity

Time Complexity

⏱️

O(n)

Vectorized operations still process n elements, but with much lower constant factors due to optimized C implementation

✓ Linear Growth

Space Complexity

O(n)

We create a new column to store n bonus values

⚡ Linearithmic Space

Constraints

1 ≤ employees.shape[0] ≤ 10⁴
1 ≤ employees['salary'][i] ≤ 10⁶
All salary values are positive integers
Employee names are non-empty strings

Asked in

N Netflix 35 ♪ Spotify 28 🏠 Airbnb 22 U Uber 18

The optimal solution uses pandas vectorized operations to create the bonus column in a single line: employees['bonus'] = employees['salary'] * 2. This approach is efficient, readable, and leverages pandas' optimized internal implementation for maximum performance.

Common Approaches

Approach	Time	Space	Notes
✓ Iterative Row-by-Row Processing	O(n)	O(n)	Loop through each row and manually calculate bonus for each employee
Vectorized Column Operations (Optimal)	O(n)	O(n)	Use pandas vectorization to calculate all bonuses simultaneously

Iterative Row-by-Row Processing — Algorithm Steps

Initialize an empty list to store bonus values
Iterate through each row in the DataFrame
For each row, extract the salary value
Calculate bonus = salary × 2
Add the bonus to our list
Assign the bonus list as a new column to the DataFrame

Visualization

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Step-by-Step Walkthrough

Start with DataFrame

We have a DataFrame with name and salary columns

Process Row 1

Calculate Alice's bonus: 50000 × 2 = 100000

Process Row 2

Calculate Bob's bonus: 60000 × 2 = 120000

Process Row 3

Calculate Charlie's bonus: 75000 × 2 = 150000

Add Bonus Column

Assign all calculated bonuses to the new column

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct {
    char name[50];
    int salary;
    int bonus;
} Employee;

void createBonusColumn(Employee* employees, int count) {
    // Iterate through each employee
    for (int i = 0; i < count; i++) {
        // Calculate bonus as double the salary
        employees[i].bonus = employees[i].salary * 2;
    }
}

int main() {
    Employee employees[3] = {
        {"Alice", 50000, 0},
        {"Bob", 60000, 0},
        {"Charlie", 75000, 0}
    };
    
    createBonusColumn(employees, 3);
    
    for (int i = 0; i < 3; i++) {
        printf("%s: $%d -> Bonus: $%d\n", 
               employees[i].name, employees[i].salary, employees[i].bonus);
    }
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We iterate through each of the n employees once to calculate their bonus

✓ Linear Growth

Space Complexity

O(n)

We create a new list/column to store n bonus values

⚡ Linearithmic Space

Constraints

1 ≤ employees.shape[0] ≤ 10⁴
1 ≤ employees['salary'][i] ≤ 10⁶
All salary values are positive integers
Employee names are non-empty strings

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🏢 Employee Bonus Calculator

Input Format:

Output:

Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Iterative Row-by-Row Processing — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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