LESSON 195 OF 414 · NESTED LOOPS

Employee of the Year V6 - Project in Rust

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:

let names = vec!["John Doe", "Mary Jane", "Bob Smith", "Alice Chen", "Carlos Diaz"];
let productivities = vec![83, 133, 97, 150, 120];
let salaries = vec![135000, 335000, 210000, 250000, 180000];
let 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: let mut taken = vec![false; names.len()];. 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 == names.len() || productivities[i] > productivities[bestIndex]) {. If nobody is found in a round, skip the rest of that round.

let names = vec!["John Doe", "Mary Jane", "Bob Smith", "Alice Chen", "Carlos Diaz"];
let productivities = vec![83, 133, 97, 150, 120];
let salaries = vec![135000, 335000, 210000, 250000, 180000];
let k = 3;

let mut taken = vec![false; names.len()];
for _rank in 0..k {
    let mut bestIndex = names.len();
    for i in 0..names.len() {
        if !taken[i] && (bestIndex == names.len() || productivities[i] > productivities[bestIndex]) {
            bestIndex = i;
        }
    }
    if bestIndex != names.len() {
        taken[bestIndex] = true;
        let bonus = salaries[bestIndex] * 15 / 100;
        println!("{}: ${}", names[bestIndex], bonus);
    }
}

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Employee of the Year V6 - Project in Rust

Employee Of The Year V6 in Rust