LESSON 154 OF 409 · LOOPING OVER ARRAY INDICES

Employee of the Year V4 - Project in C++

Practice

A whole team is competing. Use a for loop over the vectors to find the most productive employee, then print their 15% bonus.

You start with:

vector<string> names = {"John Doe", "Mary Jane", "Bob Smith", "Alice Chen", "Carlos Diaz"};
vector<int> productivities = {83, 133, 97, 150, 120};
vector<int> salaries = {135000, 335000, 210000, 250000, 180000};

Build:

Alice Chen
Bonus: $37500

Rules:

Index i in all three vectors refers to the same employee.
The winner is whichever employee has the highest productivity. If two employees share the highest productivity, the one listed first wins.
The winner's bonus is 15% of their salary, rounded down to a whole number.
Print the winner's name on the first line, then "Bonus: $<amount>" on the second.

Start with int bestIndex = 0 and loop over the other indices with for (int i = 1; i < (int)names.size(); i++). Inside the loop, update the best index only when productivities[i] > productivities[bestIndex]: a strict comparison keeps the first employee on a tie.

Dividing two int values drops the fraction, so salaries[bestIndex] * 15 / 100 is already rounded down: int bonus = salaries[bestIndex] * 15 / 100;

Chain the text and the number with <<: cout << "Bonus: $" << bonus << endl;

vector<string> names = {"John Doe", "Mary Jane", "Bob Smith", "Alice Chen", "Carlos Diaz"};
vector<int> productivities = {83, 133, 97, 150, 120};
vector<int> salaries = {135000, 335000, 210000, 250000, 180000};

int bestIndex = 0;
for (int i = 1; i < (int)names.size(); i++) {
    if (productivities[i] > productivities[bestIndex]) {
        bestIndex = i;
    }
}
int bonus = salaries[bestIndex] * 15 / 100;
cout << names[bestIndex] << endl;
cout << "Bonus: $" << bonus << endl;

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Employee of the Year V4 - Project in C++

Employee Of The Year V4 in C++