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This approach involves sorting engineers by their efficiency in descending order, ensuring that each team's minimum efficiency is dominated by its highest-efficiency member being considered rather than doing a costly check of all members.
After this, we use a min-heap to keep track of the top 'k' speeds. The strategy is to iterate through each engineer (sorted by efficiency), compute potential team performance by inserting their speed into the speed sum (if advantageous), and then multiplying by the current engineer's efficiency (the smallest efficiency thus far in the sorted order).
Time Complexity: O(n log n) due to sorting and heap operations.
Space Complexity: O(k) for maintaining the min-heap.
1import java.util.PriorityQueue;
2import java.util.Arrays;
3
4class Solution {
5 public int maxPerformance(int n, int[] speed, int[] efficiency, int k) {
6 long mod = (long)1e9 + 7;
7 int[][] engineers = new int[n][2];
8 for (int i = 0; i < n; i++) {
9 engineers[i] = new int[]{efficiency[i], speed[i]};
10 }
11 Arrays.sort(engineers, (a, b) -> b[0] - a[0]);
12
13 PriorityQueue<Integer> minHeap = new PriorityQueue<>(k);
14 long speedSum = 0, maxPerformance = 0;
15
16 for (int[] engineer : engineers) {
17 int currEfficiency = engineer[0];
18 int currSpeed = engineer[1];
19
20 if (minHeap.size() >= k) {
21 speedSum -= minHeap.poll();
22 }
23
24 speedSum += currSpeed;
25 minHeap.offer(currSpeed);
26
27 maxPerformance = Math.max(maxPerformance, speedSum * currEfficiency);
28 }
29
30 return (int)(maxPerformance % mod);
31 }
32}
33The core idea is represented similarly, with the main importance being on using efficient data structures provided by Java. PriorityQueue here serves directly as a min-heap. As engineers are considered by efficiency, we can dynamically decide which set of speeds give the best performance for each efficiency level.
Instead of using a heap, an alternative approach employs dynamic programming. For each engineer sorted by efficiency, we determine the maximum performance we can achieve using a binary search to find an optimal combination in an accumulated list of maximum speeds and efficiencies.
This involves recursively filling up a DP table with potential performance values, achieving optimization through binary search.
Time Complexity: O(n^2 log n) in a worst-case naive implementation.
Space Complexity: O(n) for the DP table.
using System.Collections.Generic;
public class Solution {
public int MaxPerformance(int n, int[] speed, int[] efficiency, int k) {
// Outline of DP + Binary Search
return 0; // Placeholder Implementation
}
}
For C#, a hybrid use of Lists and array management is required when conceptualizing the DP transitions. Gradual exploration of speeds, sorting, and recursive solutions possible in the C# data context results in changes calculated for each considered engineer or state.