LESSON 187 OF 409 · NESTED LOOPS

Employee of the Year V6 - Project in C++

Practice

Generalize the previous version: give bonuses to the top K most productive employees, where k is given as a variable. Still no sorting: repeat the search for the best employee k times, and remember who already won so nobody wins twice.

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};
int k = 3;

Build:

Alice Chen: $37500
Mary Jane: $50250
Carlos Diaz: $27000

Rules:

Index i in all three vectors refers to the same employee.
Round 1 finds the employee with the highest productivity, round 2 the best among the employees who have not won yet, and so on, for k rounds.
If two employees have the same productivity, the one listed first wins.
Each bonus is 15% of the salary, rounded down (with integer division).
Print one line per winner, best first: name: $bonus.
If k is larger than the number of employees, everybody wins once and you print one line per employee: when nobody is left, a round prints nothing.
Your code must keep working when k changes to 1, 2, 5 or 7.

Use a list of booleans: vector<bool> taken(names.size(), false);. A round finds the best employee among those who are not taken yet, then marks that employee as taken.

The outer loop runs k times and the inner loop is the search from v4 with one more condition: if (!taken[i] && (bestIndex == -1 || productivities[i] > productivities[bestIndex])) {. If nobody is found in a round, skip the rest of that round.

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 k = 3;

vector<bool> taken(names.size(), false);
for (int rank = 0; rank < k; rank++) {
    int bestIndex = -1;
    for (int i = 0; i < (int)names.size(); i++) {
        if (!taken[i] && (bestIndex == -1 || productivities[i] > productivities[bestIndex])) {
            bestIndex = i;
        }
    }
    if (bestIndex != -1) {
        taken[bestIndex] = true;
        int bonus = salaries[bestIndex] * 15 / 100;
        cout << names[bestIndex] << ": $" << bonus << endl;
    }
}

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Employee of the Year V6 - Project in C++

Employee Of The Year V6 in C++