Show the top k brands by rating. This is v3's search, repeated k times: find the best brand, mark it as taken, then search again among the brands that are not taken yet, until you have k brands in rank order.
You start with:
val brands = arrayOf("TechPro", "MegaCorp", "Quantum", "NeoByte", "Zenith")
val ratings = intArrayOf(4, 5, 3, 5, 4)
val k = 3
Build:
1. MegaCorp (5) 2. NeoByte (5) 3. TechPro (4)
Rules:
- Repeat v3's search
ktimes. Each round looks only at brands that are not taken yet and picks the one with the highest rating, then marks it as taken (one true/false flag per brand). - Ties go to the brand listed first: keep the first brand you find and replace it only on a strictly higher rating.
kis never larger than the number of brands.- Each line is
"rank. Brand (rating)", starting at rank 1. - Print the lines, one per line, with
println(...).
Start with the two declarations
val used = BooleanArray(ratings.size) and val lines = mutableListOf<String>(). A flag tells whether that position is already taken.
Each round is v3's search: nothing found yet, scan every position, skip taken ones, keep the first candidate and replace it only on a strictly higher rating. Then mark the winner as taken and add its line.
val brands = arrayOf("TechPro", "MegaCorp", "Quantum", "NeoByte", "Zenith")
val ratings = intArrayOf(4, 5, 3, 5, 4)
val k = 3
val used = BooleanArray(ratings.size)
val lines = mutableListOf<String>()
for (rank in 1..k) {
var bestIndex = -1
for (i in ratings.indices) {
if (!used[i]) {
if (bestIndex == -1 || ratings[i] > ratings[bestIndex]) {
bestIndex = i
}
}
}
used[bestIndex] = true
lines.add("$rank. ${brands[bestIndex]} (${ratings[bestIndex]})")
}
println(lines.joinToString("\n"))
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