LeetCode in Kotlin
1383. Maximum Performance of a Team
Hard
You are given two integers n and k and two integer arrays speed and efficiency both of length n. There are n engineers numbered from 1 to n. speed[i] and efficiency[i] represent the speed and efficiency of the ith engineer respectively.
Choose at most k different engineers out of the n engineers to form a team with the maximum performance.
The performance of a team is the sum of their engineers’ speeds multiplied by the minimum efficiency among their engineers.
Return the maximum performance of this team. Since the answer can be a huge number, return it modulo 109 + 7.
Example 1:
Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2
Output: 60
Explanation:
We have the maximum performance of the team by selecting engineer 2 (with speed=10 and efficiency=4) and engineer 5 (with speed=5 and efficiency=7). That is, performance = (10 + 5) * min(4, 7) = 60.
Example 2:
Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3
Output: 68
Explanation:
This is the same example as the first but k = 3. We can select engineer 1, engineer 2 and engineer 5 to get the maximum performance of the team. That is, performance = (2 + 10 + 5) * min(5, 4, 7) = 68.
Example 3:
Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4
Output: 72
Constraints:
1 <= k <= n <= 105speed.length == nefficiency.length == n1 <= speed[i] <= 1051 <= efficiency[i] <= 108
Solution
import java.util.PriorityQueue
class Solution {
fun maxPerformance(n: Int, speed: IntArray, efficiency: IntArray, k: Int): Int {
val engineers = Array(n) { IntArray(2) }
for (i in 0 until n) {
engineers[i][0] = speed[i]
engineers[i][1] = efficiency[i]
}
engineers.sortWith { engineer1: IntArray, engineer2: IntArray -> engineer2[1] - engineer1[1] }
var speedSum: Long = 0
var maximumPerformance: Long = 0
val minHeap = PriorityQueue<Int>()
for (engineer in engineers) {
if (minHeap.size == k) {
speedSum -= minHeap.poll().toLong()
}
speedSum += engineer[0].toLong()
minHeap.offer(engineer[0])
maximumPerformance = Math.max(maximumPerformance, speedSum * engineer[1])
}
return (maximumPerformance % 1000000007).toInt()
}
}