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 <= 105
• speed.length == n
• efficiency.length == n
• 1 <= speed[i] <= 105
• 1 <= 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()
    }
}

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