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:
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}
k := 3
Build:
Alice Chen: $37500 Mary Jane: $50250 Carlos Diaz: $27000
Rules:
- Index
iin all three slices 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
krounds. - 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
kis 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
kchanges to 1, 2, 5 or 7.
Use a list of booleans:
taken := make([]bool, len(names)). 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.
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}
k := 3
taken := make([]bool, len(names))
for rank := 0; rank < k; rank++ {
bestIndex := -1
for i := 0; i < len(names); i++ {
if !taken[i] && (bestIndex == -1 || productivities[i] > productivities[bestIndex]) {
bestIndex = i
}
}
if bestIndex != -1 {
taken[bestIndex] = true
bonus := salaries[bestIndex] * 15 / 100
fmt.Println(names[bestIndex] + ": $" + strconv.Itoa(bonus))
}
}
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