Problem Description

You are given a DataFrame called employees that contains two columns:

• name: stores employee names as strings (object type)
• salary: stores employee salaries as integers

The task is to give all employees a pay raise by doubling their current salaries. You need to modify the salary column in-place by multiplying each salary value by 2.

For example, if an employee currently has a salary of 50000, after the modification their salary should become 100000.

The solution demonstrates a straightforward approach using pandas DataFrame operations. The code employees['salary'] *= 2 directly multiplies all values in the salary column by 2, which is equivalent to employees['salary'] = employees['salary'] * 2. This operation modifies the DataFrame in-place and returns the updated DataFrame with all salaries doubled.

Intuition

When we need to modify all values in a DataFrame column by applying the same operation, pandas provides vectorized operations that work on entire columns at once. Instead of iterating through each row individually (which would be inefficient), we can leverage pandas' built-in broadcasting capability.

The key insight is that pandas columns behave like arrays, allowing us to perform arithmetic operations directly on them. When we write employees['salary'] *= 2, pandas automatically applies this multiplication to every element in the salary column. This is much more efficient than a loop-based approach because pandas uses optimized C implementations under the hood.

We arrive at this solution by recognizing that:

1. The operation is uniform across all rows (every salary gets multiplied by 2)
2. Pandas supports in-place modifications using compound assignment operators (*=)
3. Column-wise operations are the most natural and efficient way to handle such transformations in pandas

This approach follows the principle of working with data at the column level rather than the row level whenever possible, which is a fundamental best practice in pandas programming.

Solution Approach

The implementation uses a direct column manipulation approach in pandas. Here's how the solution works:

1. Access the salary column: We use bracket notation employees['salary'] to access the entire salary column of the DataFrame.

2. Apply in-place multiplication: The compound assignment operator *= is used to multiply all values in the column by 2. This operation:

   • Takes each value in the salary column
   • Multiplies it by 2
   • Stores the result back in the same location

3. Return the modified DataFrame: After the in-place modification, we return the employees DataFrame which now contains the updated salary values.

The complete implementation is just one line of actual logic:

employees['salary'] *= 2

This is equivalent to writing:

employees['salary'] = employees['salary'] * 2

The advantage of using *= is that it clearly indicates we're modifying the data in-place rather than creating a new column. The operation is vectorized, meaning pandas applies the multiplication to all elements in the column simultaneously using optimized numpy operations underneath, making it extremely efficient even for large datasets.

The time complexity is O(n) where n is the number of employees, as we need to update each salary value once. The space complexity is O(1) for the operation itself since we're modifying the values in-place without creating additional data structures.

Example Walkthrough

Let's walk through a concrete example to illustrate how the solution works.

Suppose we have the following DataFrame with 3 employees:

Initial DataFrame:
   name     salary
0  Alice    50000
1  Bob      75000
2  Charlie  60000

Step 1: Access the salary column When we write employees['salary'], we're accessing the entire salary column as a Series:

[50000, 75000, 60000]

Step 2: Apply multiplication by 2 The operation employees['salary'] *= 2 multiplies each value in the column by 2:

• Alice's salary: 50000 × 2 = 100000
• Bob's salary: 75000 × 2 = 150000
• Charlie's salary: 60000 × 2 = 120000

This happens in a single vectorized operation - pandas doesn't loop through each value individually but instead applies the multiplication to the entire column at once.

Step 3: In-place update The *= operator updates the original DataFrame directly, replacing the old salary values with the new doubled values:

Final DataFrame:
   name     salary
0  Alice    100000
1  Bob      150000
2  Charlie  120000

The key point is that this transformation happens efficiently in one operation rather than requiring us to iterate through each row. The DataFrame is modified in-place, meaning we don't create a copy of the data - we directly update the existing salary column values.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n) where n is the number of rows in the DataFrame. The operation employees['salary'] *= 2 iterates through each element in the salary column exactly once to multiply it by 2.

Space Complexity: O(1) for the additional space used by the algorithm. The multiplication operation is performed in-place on the existing DataFrame column, modifying the values directly without creating a new copy of the data. The only space used is for temporary variables during the multiplication operation, which is constant regardless of input size.

Note: The total space occupied by the DataFrame itself is O(n), but this is not counted as additional space since it's the input that already exists in memory.

Common Pitfalls

1. Type Mismatch Issues

One common pitfall occurs when the salary column contains non-numeric data types or mixed types (e.g., strings that look like numbers, NaN values, or object dtype instead of int/float). This can happen if the data was imported incorrectly or contains formatting characters like currency symbols.

Problem Example:

# If salary column contains strings like "50000" or "$50,000"
employees['salary'] *= 2  # This will cause a TypeError

Solution:

def modifySalaryColumn(employees: pd.DataFrame) -> pd.DataFrame:
    # Ensure salary column is numeric type
    employees['salary'] = pd.to_numeric(employees['salary'], errors='coerce')
    employees['salary'] *= 2
    return employees

2. Handling Missing Values (NaN)

If the salary column contains NaN or None values, the multiplication will maintain these as NaN, which might not be the desired behavior in all business contexts.

Problem Example:

# If an employee has NaN salary, it remains NaN after multiplication
# NaN * 2 = NaN

Solution:

def modifySalaryColumn(employees: pd.DataFrame) -> pd.DataFrame:
    # Option 1: Fill NaN with 0 before multiplication
    employees['salary'] = employees['salary'].fillna(0)
    employees['salary'] *= 2
  
    # Option 2: Only multiply non-null values
    # employees.loc[employees['salary'].notna(), 'salary'] *= 2
  
    return employees

3. Integer Overflow for Large Salaries

While less common with pandas/numpy's int64 default, extremely large salary values could theoretically cause overflow issues, especially if the DataFrame was created with smaller integer types (int32, int16).

Solution:

def modifySalaryColumn(employees: pd.DataFrame) -> pd.DataFrame:
    # Convert to float64 to handle larger numbers
    employees['salary'] = employees['salary'].astype('float64')
    employees['salary'] *= 2
    return employees

4. Unintended Reference Modification

If the function caller expects the original DataFrame to remain unchanged, the in-place modification could cause unexpected side effects since DataFrames are mutable objects passed by reference.

Solution:

def modifySalaryColumn(employees: pd.DataFrame) -> pd.DataFrame:
    # Create a copy to avoid modifying the original
    employees_copy = employees.copy()
    employees_copy['salary'] *= 2
    return employees_copy