LESSON 192 OF 414 · NESTED LOOPS

Employee of the Year V6 - Project in Kotlin

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:

        val names = arrayOf("John Doe", "Mary Jane", "Bob Smith", "Alice Chen", "Carlos Diaz")
        val productivities = intArrayOf(83, 133, 97, 150, 120)
        val salaries = intArrayOf(135000, 335000, 210000, 250000, 180000)
        val k = 3

Build:

Alice Chen: $37500
Mary Jane: $50250
Carlos Diaz: $27000

Rules:

Index i in 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 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: val taken = BooleanArray(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 = arrayOf("John Doe", "Mary Jane", "Bob Smith", "Alice Chen", "Carlos Diaz")
        val productivities = intArrayOf(83, 133, 97, 150, 120)
        val salaries = intArrayOf(135000, 335000, 210000, 250000, 180000)
        val k = 3

        val taken = BooleanArray(names.size)
        for (rank in 0 until k) {
            var bestIndex = -1
            for (i in 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)
            }
        }

STEP 1 OF 9

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Employee of the Year V6 - Project in Kotlin

Employee Of The Year V6 in Kotlin