Hand out bonuses to the top 2 most productive employees: the best one first, then the second best. You don't need any sorting. Find the best employee with the loop from v4, then find the best of the other employees with a second loop.
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];
Build:
Alice Chen: $37500 Mary Jane: $50250
Rules:
- Index
iin all three vectors refers to the same employee. - The best employee is the one with the highest productivity; the second best is the one with the highest productivity among all the other employees.
- 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 two lines, the best employee first:
name: $bonus.
Find the best employee first, exactly as in v4. Remember: strictly greater, so the first listed wins a tie.
For the second place, loop over all the indices and ignore the best one. The whole condition is
if i != bestIndex && (secondIndex == names.len() || productivities[i] > productivities[secondIndex]) {. "Nobody chosen yet" must also let an employee in.
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 mut bestIndex = 0;
for i in 1..names.len() {
if productivities[i] > productivities[bestIndex] {
bestIndex = i;
}
}
let mut secondIndex = names.len();
for i in 0..names.len() {
if i != bestIndex && (secondIndex == names.len() || productivities[i] > productivities[secondIndex]) {
secondIndex = i;
}
}
let bestBonus = salaries[bestIndex] * 15 / 100;
let secondBonus = salaries[secondIndex] * 15 / 100;
println!("{}: ${}", names[bestIndex], bestBonus);
println!("{}: ${}", names[secondIndex], secondBonus);
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