A whole team is competing. Use a for loop over the slices to find the most productive employee, then print their 15% bonus.
You start with:
names := []string{"John Doe", "Mary Jane", "Bob Smith", "Alice Chen", "Carlos Diaz"}
productivities := []int{83, 133, 97, 150, 120}
salaries := []int{135000, 335000, 210000, 250000, 180000}
Build:
Alice Chen Bonus: $37500
Rules:
- Index
iin all three slices 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
bestIndex := 0 and loop over the other indices with for i := 1; i < len(names); 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 whole numbers (
int) drops the fraction, so salaries[bestIndex] * 15 / 100 is already rounded down: bonus := salaries[bestIndex] * 15 / 100
Join text and a number with
strconv.Itoa: fmt.Println("Bonus: $" + strconv.Itoa(bonus))
names := []string{"John Doe", "Mary Jane", "Bob Smith", "Alice Chen", "Carlos Diaz"}
productivities := []int{83, 133, 97, 150, 120}
salaries := []int{135000, 335000, 210000, 250000, 180000}
bestIndex := 0
for i := 1; i < len(names); i++ {
if productivities[i] > productivities[bestIndex] {
bestIndex = i
}
}
bonus := salaries[bestIndex] * 15 / 100
fmt.Println(names[bestIndex])
fmt.Println("Bonus: $" + strconv.Itoa(bonus))
Free plan: 30 tutor messages. Premium is unlimited.
Saved
Ctrl + Enter to check
