LeetCode 2884 - Modify Columns

LeetCode Problem 2884

Difficulty: 🟢 Easy
Topics: —

Solution

Problem Understanding

The problem provides a Pandas DataFrame named employees with two columns:

Column	Type
name	object
salary	int

Each row represents one employee and their current salary. The task is to modify the salary column so that every salary value becomes twice its original amount.

In other words, for every row in the DataFrame, we must replace:

$salary = salary \times 2$

The updated DataFrame should then be returned.

This is a straightforward DataFrame transformation problem. We are not creating new rows, filtering data, or aggregating values. Instead, we are performing an in place column update where every element in the salary column is multiplied by 2.

The input represents tabular employee data, and the expected output is the same table structure with updated salary values.

The problem guarantees that:

• The salary column contains integers.
• The DataFrame already exists and is properly formatted.
• Every row has a valid salary value.

Since this is an Easy level Pandas problem, the main focus is understanding how vectorized column operations work in Pandas.

An important edge case is an empty DataFrame. If there are no employees, the solution should still work correctly and simply return the empty DataFrame unchanged. Another potential edge case is salaries equal to 0, which should remain 0 after multiplication. Large salary values are also safe because Python integers can grow dynamically without overflow issues.

Approaches

Brute Force Approach

A brute force solution would iterate through every row individually and manually update each salary value one by one.

For example, we could use a loop over the DataFrame indices, retrieve the current salary, multiply it by 2, and assign the updated value back into the DataFrame.

This approach works because every employee record is visited exactly once, ensuring that every salary gets doubled correctly.

However, row by row iteration is inefficient in Pandas. Pandas is designed for vectorized operations, and manual loops introduce unnecessary overhead. While this approach is still linear in time complexity, it does not take advantage of optimized internal implementations.

Optimal Approach

The optimal solution uses a vectorized column operation:

employees["salary"] *= 2

This applies the multiplication to the entire column at once.

The key insight is that Pandas supports element wise arithmetic directly on Series objects. The salary column is internally represented as a vectorized data structure, so multiplying it by 2 automatically updates every element efficiently.

This approach is cleaner, more concise, and significantly more idiomatic for Pandas based problems.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n)	O(1)	Iterates row by row and updates salaries manually
Optimal	O(n)	O(1)	Uses vectorized Pandas column multiplication

Algorithm Walkthrough

1. Access the salary column from the employees DataFrame.
2. Multiply the entire column by 2. Pandas automatically applies this operation element wise to every row.
3. Store the updated values back into the same salary column.
4. Return the modified DataFrame.

Why it works

The algorithm works because Pandas vectorized arithmetic guarantees that the multiplication operation is applied independently to every element in the column. Since every salary value is transformed from x to 2 * x, the resulting DataFrame correctly represents doubled salaries for all employees.

Python Solution

import pandas as pd

def modifySalaryColumn(employees: pd.DataFrame) -> pd.DataFrame:
    employees["salary"] *= 2
    return employees

The implementation directly follows the optimal algorithm.

First, the salary column is selected using employees["salary"]. In Pandas, this returns a Series containing all salary values.

The statement:

employees["salary"] *= 2

performs an in place vectorized multiplication. Every value in the Series is doubled automatically.

Finally, the modified DataFrame is returned.

This solution is concise because Pandas internally handles iteration and element wise arithmetic efficiently.

Go Solution

package main

type Employee struct {
	Name   string
	Salary int
}

func modifySalaryColumn(employees []Employee) []Employee {
	for i := 0; i < len(employees); i++ {
		employees[i].Salary *= 2
	}

	return employees
}

Go does not have a built in DataFrame structure like Pandas, so we model the data using a struct and a slice of employees.

The algorithm iterates through the slice and doubles each employee's salary in place.

Unlike Python, Go integers have fixed sizes. However, for the constraints typically used in this problem, integer overflow is not a concern.

Slices in Go behave similarly to references to underlying arrays, so modifying elements inside the slice updates the original data structure directly.

Worked Examples

Example 1

Initial DataFrame:

name	salary
Jack	19666
Piper	74754
Mia	62509
Ulysses	54866

The algorithm applies:

$newSalary = oldSalary \times 2$

Step by step transformation:

Employee	Original Salary	Updated Salary
Jack	19666	39332
Piper	74754	149508
Mia	62509	125018
Ulysses	54866	109732

Final DataFrame:

name	salary
Jack	39332
Piper	149508
Mia	125018
Ulysses	109732

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Every salary value is processed once
Space	O(1)	Modification occurs in place

The algorithm must touch every salary at least once to update it, so the runtime is linear in the number of employees.

The space complexity is constant because no additional data structures proportional to the input size are created. The DataFrame is modified directly.

Test Cases

import pandas as pd

def modifySalaryColumn(employees: pd.DataFrame) -> pd.DataFrame:
    employees["salary"] *= 2
    return employees

# Standard example from the problem statement
df1 = pd.DataFrame({
    "name": ["Jack", "Piper", "Mia", "Ulysses"],
    "salary": [19666, 74754, 62509, 54866]
})

result1 = modifySalaryColumn(df1)

assert result1["salary"].tolist() == [39332, 149508, 125018, 109732]  # normal case

# Single employee
df2 = pd.DataFrame({
    "name": ["Alice"],
    "salary": [50000]
})

result2 = modifySalaryColumn(df2)

assert result2["salary"].tolist() == [100000]  # single row case

# Salary equal to zero
df3 = pd.DataFrame({
    "name": ["Bob"],
    "salary": [0]
})

result3 = modifySalaryColumn(df3)

assert result3["salary"].tolist() == [0]  # zero salary remains zero

# Empty DataFrame
df4 = pd.DataFrame({
    "name": [],
    "salary": []
})

result4 = modifySalaryColumn(df4)

assert result4.empty  # empty input case

# Large salary values
df5 = pd.DataFrame({
    "name": ["Rich"],
    "salary": [10**9]
})

result5 = modifySalaryColumn(df5)

assert result5["salary"].tolist() == [2 * 10**9]  # large integer handling

Test	Why
Standard multi employee input	Verifies normal functionality
Single employee	Confirms algorithm works with minimal non empty input
Zero salary	Ensures multiplication handles zero correctly
Empty DataFrame	Validates robustness for no rows
Large salary values	Confirms large integers are processed properly

Edge Cases

One important edge case is an empty DataFrame. A naive implementation that assumes at least one row exists could fail when accessing elements directly. The vectorized Pandas solution handles this naturally because multiplying an empty column simply produces another empty column.

Another edge case is salaries equal to zero. Some implementations may accidentally introduce issues if they rely on truthy or falsy checks before updating values. Since the algorithm directly multiplies every value by 2, zero remains zero correctly.

Large salary values are also important to consider. In some programming languages, integer overflow could occur when doubling very large numbers. Python handles arbitrarily large integers automatically, so the implementation remains correct even for extremely large salary values.

A final edge case is a DataFrame containing only one employee. Some incorrect implementations accidentally rely on iteration patterns that assume multiple rows exist. The vectorized operation works identically regardless of the number of rows.