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:
val names = Array("John Doe", "Mary Jane", "Bob Smith", "Alice Chen", "Carlos Diaz")
val productivities = Array(83, 133, 97, 150, 120)
val salaries = Array(135000, 335000, 210000, 250000, 180000)
val k = 3
Build:
Alice Chen: $37500 Mary Jane: $50250 Carlos Diaz: $27000
Rules:
- Index
iin all three arrays 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:
val taken = new Array[Boolean](names.size). 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.
val names = Array("John Doe", "Mary Jane", "Bob Smith", "Alice Chen", "Carlos Diaz")
val productivities = Array(83, 133, 97, 150, 120)
val salaries = Array(135000, 335000, 210000, 250000, 180000)
val k = 3
val taken = new Array[Boolean](names.size)
for (rank <- 0 until k) {
var bestIndex = -1
for (i <- 0 until names.size) {
if (!taken(i) && (bestIndex == -1 || productivities(i) > productivities(bestIndex))) {
bestIndex = i
}
}
if (bestIndex != -1) {
taken(bestIndex) = true
val bonus = salaries(bestIndex) * 15 / 100
println(names(bestIndex) + ": $" + bonus)
}
}
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