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LESSON 193 OF 415 · NESTED LOOPS

Employee of the Year V6 - Project in Swift

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:

        let names = ["John Doe", "Mary Jane", "Bob Smith", "Alice Chen", "Carlos Diaz"]
        let productivities = [83, 133, 97, 150, 120]
        let salaries = [135000, 335000, 210000, 250000, 180000]
        let 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: var taken = Array(repeating: false, count: names.count). 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.

        let names = ["John Doe", "Mary Jane", "Bob Smith", "Alice Chen", "Carlos Diaz"]
        let productivities = [83, 133, 97, 150, 120]
        let salaries = [135000, 335000, 210000, 250000, 180000]
        let k = 3

        var taken = Array(repeating: false, count: names.count)
        for rank in 0..<k {
            var bestIndex = -1
            for i in 0..<names.count {
                if !taken[i] && (bestIndex == -1 || productivities[i] > productivities[bestIndex]) {
                    bestIndex = i
                }
            }
            if bestIndex != -1 {
                taken[bestIndex] = true
                let bonus = salaries[bestIndex] * 15 / 100
                print(names[bestIndex] + ": $" + String(bonus))
            }
        }

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    Employee Of The Year V6 in Swift

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