Create a New Column

Create a New Column - Problem

Database Easy

You are given a DataFrame employees with the following structure:

Column Name	Type
name	object
salary	int

A company plans to provide its employees with a bonus.

Write a solution to create a new column named bonus that contains the doubled values of the salary column.

Input & Output

Example 1 — Basic Employee Bonus

$ Input: employees = [{"name": "Alice", "salary": 50000}, {"name": "Bob", "salary": 60000}]

› Output: [{"name": "Alice", "salary": 50000, "bonus": 100000}, {"name": "Bob", "salary": 60000, "bonus": 120000}]

💡 Note: Alice's bonus = 50000 * 2 = 100000, Bob's bonus = 60000 * 2 = 120000

Example 2 — Single Employee

$ Input: employees = [{"name": "Charlie", "salary": 75000}]

› Output: [{"name": "Charlie", "salary": 75000, "bonus": 150000}]

💡 Note: Charlie's bonus = 75000 * 2 = 150000

Example 3 — Multiple Employees

$ Input: employees = [{"name": "David", "salary": 45000}, {"name": "Eva", "salary": 55000}, {"name": "Frank", "salary": 65000}]

› Output: [{"name": "David", "salary": 45000, "bonus": 90000}, {"name": "Eva", "salary": 55000, "bonus": 110000}, {"name": "Frank", "salary": 65000, "bonus": 130000}]

💡 Note: Each employee gets bonus = salary * 2: David: 90000, Eva: 110000, Frank: 130000

Constraints

1 ≤ employees.length ≤ 10⁴
1 ≤ salary ≤ 10⁶
name contains only alphabetic characters

Visualization

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Asked in

N Netflix 15 A Airbnb 12

The key insight is to use DataFrame vectorized operations for efficient column manipulation. The optimal approach uses employees['bonus'] = employees['salary'] * 2 to create the new column in one line. Time: O(n), Space: O(1)

Common Approaches

✓ Row-by-Row Iteration

⏱️ Time: O(n) Space: O(1)

Loop through each row of the DataFrame, access the salary value, multiply by 2, and assign to a new bonus column. This approach processes one employee at a time.

Vectorized Column Operation

⏱️ Time: O(n) Space: O(1)

Apply multiplication operation directly to the entire salary column at once. This leverages pandas' optimized vectorized operations for better performance.

Row-by-Row Iteration — Algorithm Steps

Step 1: Initialize empty bonus column
Step 2: Loop through each row
Step 3: Calculate salary * 2 for each row
Step 4: Assign to bonus column

Visualization

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Step-by-Step Walkthrough

Initialize

Start with original DataFrame

Iterate

Loop through each row and calculate salary * 2

Result

DataFrame with new bonus column

Code -

solution.c — C

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

#define MAX_INPUT 10000
#define MAX_NAME 100

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

int parseEmployees(char* input, Employee* employees) {
    int count = 0;
    char* ptr = input;
    
    // Find the start of the array
    while (*ptr && *ptr != '[') ptr++;
    if (!*ptr) return 0;
    ptr++; // skip '['
    
    while (*ptr && *ptr != ']') {
        // Skip whitespace and commas
        while (*ptr && (*ptr == ' ' || *ptr == ',' || *ptr == '\n' || *ptr == '\t')) ptr++;
        
        if (*ptr == '}' || *ptr == ']') break;
        
        // Expect '{' to start object
        if (*ptr != '{') break;
        ptr++; // skip '{'
        
        Employee emp = {0};
        
        // Parse object fields
        while (*ptr && *ptr != '}') {
            // Skip whitespace and commas
            while (*ptr && (*ptr == ' ' || *ptr == ',' || *ptr == '\n' || *ptr == '\t')) ptr++;
            
            if (*ptr == '}') break;
            
            // Expect '"' to start field name
            if (*ptr != '"') break;
            ptr++; // skip '"'
            
            // Read field name
            char fieldName[50] = {0};
            int i = 0;
            while (*ptr && *ptr != '"' && i < 49) {
                fieldName[i++] = *ptr++;
            }
            if (*ptr != '"') break;
            ptr++; // skip '"'
            
            // Skip whitespace and ':'
            while (*ptr && (*ptr == ' ' || *ptr == ':' || *ptr == '\n' || *ptr == '\t')) ptr++;
            
            if (strcmp(fieldName, "name") == 0) {
                // Expect '"' to start string value
                if (*ptr != '"') break;
                ptr++; // skip '"'
                
                i = 0;
                while (*ptr && *ptr != '"' && i < MAX_NAME - 1) {
                    emp.name[i++] = *ptr++;
                }
                emp.name[i] = '\0';
                
                if (*ptr != '"') break;
                ptr++; // skip '"'
            }
            else if (strcmp(fieldName, "salary") == 0) {
                // Parse integer value
                emp.salary = 0;
                while (*ptr && isdigit(*ptr)) {
                    emp.salary = emp.salary * 10 + (*ptr - '0');
                    ptr++;
                }
                emp.bonus = emp.salary * 2;
            }
        }
        
        if (*ptr == '}') {
            ptr++; // skip '}'
            employees[count++] = emp;
        }
    }
    
    return count;
}

void printEmployees(Employee* employees, int count) {
    printf("[");
    for (int i = 0; i < count; i++) {
        if (i > 0) printf(", ");
        printf("{\"name\": \"%s\", \"salary\": %d, \"bonus\": %d}", 
               employees[i].name, employees[i].salary, employees[i].bonus);
    }
    printf("]");
}

int main() {
    static char input[MAX_INPUT];
    static Employee employees[100];
    
    // Read all input
    int pos = 0;
    int c;
    while ((c = getchar()) != EOF && pos < MAX_INPUT - 1) {
        input[pos++] = c;
    }
    input[pos] = '\0';
    
    // Find the JSON array in the input
    char* jsonStart = strstr(input, "[");
    if (!jsonStart) {
        printf("[]");
        return 0;
    }
    
    int empCount = parseEmployees(jsonStart, employees);
    printEmployees(employees, empCount);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Must visit each of the n rows once to calculate bonus

✓ Linear Growth

Space Complexity

O(1)

Only creates one new column, no additional data structures

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Row-by-Row Iteration — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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