2884. Modify Columns

Easy
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Problem Description

In this problem, we have a pandas DataFrame named employees that contains two columns: name and salary. The name column is of type object, which typically means it contains string data, while the salary column contains integer values representing the salary of each employee in the company.

The task is to adjust the employee salaries by doubling each one. In other words, you must write a code that modifies the salary column by multiplying each salary by 2. This operation should update the original DataFrame in place.

The main objective is to return the updated DataFrame with the modified salaries.

Intuition

To solve this problem, we know that pandas DataFrames offer a very convenient way to perform operations on columns using simple arithmetic syntax. You can select a column and then apply an operation to each element (row) of the column efficiently.

Given that the problem statement requires the salary to be multiplied by 2, this can be accomplished by using the multiplication assignment operator (*=) in Python. This operator multiplies the left-hand operand (the salary column of the DataFrame) by the right-hand operand (the constant value 2) and assigns the result back to the left-hand operand.

The solution takes advantage of the fact that pandas will automatically apply the multiplication to each element in the column, thus doubling every employee's salary as required by the problem.

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1void search(Node root) {
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3  visit(root);
4  root.visited = true;
5  for (Node node in root.adjacent) {
6    if (!node.visited) {
7      search(node);
8    }
9  }
10}

Solution Approach

The solution approach for modifying the salary column in the employees DataFrame is quite straightforward and does not require complex algorithms or data structures. It simply leverages the features provided by the pandas library to manipulate DataFrame columns.

The steps for the implementation are as follows:

  1. The pandas library is imported with the alias pd.
  2. A function named modifySalaryColumn is defined, which takes a single parameter, employees. This parameter is expected to be a pandas DataFrame with a structure matching the one described in the problem (with name and salary columns).
  3. Within the function, the salary column of the DataFrame is accessed using bracket notation: employees['salary'].
  4. The multiplication assignment operator (*=) is applied to the salary column to multiply each value by 2. This operation is done in place, which means the existing salary column in the DataFrame is updated with the new values.
  5. Finally, the updated employees DataFrame is returned.

The solution does not make use of complex patterns, since it is able to fully utilize pandas' ability to perform vectorized operations across an entire column, applying the multiplication to each salary in one succinct line of code. This is one of the advantages of using libraries like pandas, designed for efficient data manipulation.

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Example Walkthrough

Let's walk through a small example to illustrate the solution approach:

Suppose we have the following employees DataFrame:

1   name  salary
20  Alice   50000
31   Bob    60000
42  Carol   55000

To adjust the salaries by doubling each one, we will apply the steps outlined in the solution approach:

  1. We start by importing pandas:
1import pandas as pd
  1. We define the modifySalaryColumn function that takes the employees DataFrame as a parameter:
1def modifySalaryColumn(employees):
  1. Inside the function, we access the salary column:
1    employees['salary']
  1. We use the multiplication assignment operator to double each salary:
1    employees['salary'] *= 2

After executing the above line of code, the salary column in the employees DataFrame is updated to:

1   name  salary
20  Alice  100000
31   Bob   120000
42  Carol  110000
  1. Lastly, we return the updated employees DataFrame:
1    return employees

Now, let's use this function with our example DataFrame:

1# Initial DataFrame
2employees = pd.DataFrame({
3    'name': ['Alice', 'Bob', 'Carol'],
4    'salary': [50000, 60000, 55000]
5})
6
7# Modify the salary column by doubling the salaries
8modified_employees = modifySalaryColumn(employees)
9
10# The resulting DataFrame
11print(modified_employees)

The output will be:

1   name  salary
20  Alice  100000
31   Bob   120000
42  Carol  110000

As demonstrated, each salary in the salary column has been successfully doubled, achieving the task of modifying the employee salaries in place. The modifySalaryColumn function is efficient and leverages the power of pandas for vectorized operations on DataFrame columns.

Solution Implementation

1import pandas as pd
2
3def modifySalaryColumn(employees: pd.DataFrame) -> pd.DataFrame:
4    # Double the values in the 'salary' column of the DataFrame
5    employees['salary'] = employees['salary'].apply(lambda x: x * 2)
6
7    # Return the modified DataFrame
8    return employees
9
1import java.util.List;
2import java.util.Map;
3
4public class SalaryModifier {
5
6    /**
7     * Doubles the salary for each employee in the list.
8     *
9     * @param employees A list of maps, where each map represents an employee with the 'salary' key.
10     * @return The list of employees with updated salaries.
11     */
12    public static List<Map<String, Object>> modifySalaryColumn(List<Map<String, Object>> employees) {
13        for (Map<String, Object> employee : employees) {
14            // Check if the employee map contains the 'salary' key
15            if (employee.containsKey("salary")) {
16                // Get the current salary
17                Object salaryObj = employee.get("salary");
18                if (salaryObj instanceof Number) {
19                    // Double the salary and update the employee map
20                    Number salary = (Number) salaryObj;
21                    employee.put("salary", salary.doubleValue() * 2);
22                }
23            }
24        }
25        // Return the modified list of employees with updated salaries
26        return employees;
27    }
28}
29
1#include <vector>
2#include <utility> // For std::pair
3#include <algorithm> // For std::transform
4#include <string>
5
6class Employee {
7public:
8    std::string name;
9    float salary;
10
11    Employee(std::string n, float s) : name(std::move(n)), salary(s) {}
12
13    // Function to double the salary
14    void doubleSalary() {
15        salary *= 2;
16    }
17};
18
19void modifySalaryColumn(std::vector<Employee>& employees) {
20    // Double the salaries of all employees in the vector
21    std::for_each(employees.begin(), employees.end(), [](Employee& e) {
22        e.doubleSalary();
23    });
24}
25
1// Define the employee type with at least a salary field
2type Employee = {
3    salary: number;
4    // ... other fields
5};
6
7/**
8 * Doubles the 'salary' property of each employee in an array
9 * @param employees - Array of employees to be modified
10 * @returns An array of employees with modified salaries
11 */
12function modifySalaryColumn(employees: Employee[]): Employee[] {
13    // Double the values in the 'salary' property of each employee
14    employees.forEach(employee => {
15        employee.salary *= 2;
16    });
17
18    // Return the modified array of employees
19    return employees;
20}
21```
22
23This function would be used with an array of objects, like so:
24
25```typescript
26// Example array of employees
27let employees: Employee[] = [
28    { salary: 50000 /*, other properties */ },
29    { salary: 60000 /*, other properties */ },
30    // ... other employees
31];
32
33// Doubling the salary of each employee
34employees = modifySalaryColumn(employees);
35
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What's the output of running the following function using input 56?

1KEYBOARD = {
2    '2': 'abc',
3    '3': 'def',
4    '4': 'ghi',
5    '5': 'jkl',
6    '6': 'mno',
7    '7': 'pqrs',
8    '8': 'tuv',
9    '9': 'wxyz',
10}
11
12def letter_combinations_of_phone_number(digits):
13    def dfs(path, res):
14        if len(path) == len(digits):
15            res.append(''.join(path))
16            return
17
18        next_number = digits[len(path)]
19        for letter in KEYBOARD[next_number]:
20            path.append(letter)
21            dfs(path, res)
22            path.pop()
23
24    res = []
25    dfs([], res)
26    return res
27
1private static final Map<Character, char[]> KEYBOARD = Map.of(
2    '2', "abc".toCharArray(),
3    '3', "def".toCharArray(),
4    '4', "ghi".toCharArray(),
5    '5', "jkl".toCharArray(),
6    '6', "mno".toCharArray(),
7    '7', "pqrs".toCharArray(),
8    '8', "tuv".toCharArray(),
9    '9', "wxyz".toCharArray()
10);
11
12public static List<String> letterCombinationsOfPhoneNumber(String digits) {
13    List<String> res = new ArrayList<>();
14    dfs(new StringBuilder(), res, digits.toCharArray());
15    return res;
16}
17
18private static void dfs(StringBuilder path, List<String> res, char[] digits) {
19    if (path.length() == digits.length) {
20        res.add(path.toString());
21        return;
22    }
23    char next_digit = digits[path.length()];
24    for (char letter : KEYBOARD.get(next_digit)) {
25        path.append(letter);
26        dfs(path, res, digits);
27        path.deleteCharAt(path.length() - 1);
28    }
29}
30
1const KEYBOARD = {
2    '2': 'abc',
3    '3': 'def',
4    '4': 'ghi',
5    '5': 'jkl',
6    '6': 'mno',
7    '7': 'pqrs',
8    '8': 'tuv',
9    '9': 'wxyz',
10}
11
12function letter_combinations_of_phone_number(digits) {
13    let res = [];
14    dfs(digits, [], res);
15    return res;
16}
17
18function dfs(digits, path, res) {
19    if (path.length === digits.length) {
20        res.push(path.join(''));
21        return;
22    }
23    let next_number = digits.charAt(path.length);
24    for (let letter of KEYBOARD[next_number]) {
25        path.push(letter);
26        dfs(digits, path, res);
27        path.pop();
28    }
29}
30

Time and Space Complexity

Time Complexity

The time complexity of the modifySalaryColumn function is O(n), where n is the number of elements in the 'salary' column of the employees DataFrame. This is because the function performs a single operation (multiplication by 2) on each element of the 'salary' column.

Space Complexity

The space complexity is O(1) as the modification is done in place and does not require additional space that grows with the input size. However, if the pandas DataFrame internally opts to create a copy during the operation due to its settings, the space complexity can be O(n) in practice.

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