LeetCode-in-Java

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.Arrays;
import java.util.PriorityQueue;

public class Solution {
    public int maxPerformance(int n, int[] speed, int[] efficiency, int k) {
        int[][] engineers = new int[n][2];
        for (int i = 0; i < n; i++) {
            engineers[i][0] = speed[i];
            engineers[i][1] = efficiency[i];
        }
        Arrays.sort(engineers, (engineer1, engineer2) -> engineer2[1] - engineer1[1]);
        long speedSum = 0;
        long maximumPerformance = 0;
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        for (int[] engineer : engineers) {
            if (minHeap.size() == k) {
                speedSum -= minHeap.poll();
            }
            speedSum += engineer[0];
            minHeap.offer(engineer[0]);
            maximumPerformance = Math.max(maximumPerformance, speedSum * engineer[1]);
        }
        return (int) (maximumPerformance % 1000_000_007);
    }
}

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