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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.
1#include <stdio.h>
2#include <stdlib.h>
3#include <limits.h>
4
5int comparator(const void *a, const void *b) {
6 return ((int*)b)[1] - ((int*)a)[1];
7}
8
9int maxPerformance(int n, int* speed, int speedSize, int* efficiency, int efficiencySize, int k) {
10 int MOD = 1000000007;
11 int engineers[n][2];
12 for (int i = 0; i < n; i++) {
13 engineers[i][0] = speed[i];
14 engineers[i][1] = efficiency[i];
15 }
16 qsort(engineers, n, sizeof(engineers[0]), comparator);
17
18 int *minHeap = malloc(k * sizeof(int));
19 int heapSize = 0, topKSpeedSum = 0, maxPerformance = 0;
20
21 for (int i = 0; i < n; i++) {
22 int currSpeed = engineers[i][0];
23 int currEfficiency = engineers[i][1];
24 if (heapSize < k) {
25 minHeap[heapSize++] = currSpeed;
26 topKSpeedSum += currSpeed;
27 } else if (currSpeed > minHeap[0]) {
28 topKSpeedSum += currSpeed - minHeap[0];
29 } else {
30 currEfficiency = 0;
31 }
32
33 for (int j = heapSize / 2 - 1; j >= 0; j--) {
34 int minPos = j;
35 if (2*j + 1 < heapSize && minHeap[2*j + 1] < minHeap[minPos]) {
36 minPos = 2*j + 1;
37 }
38 if (2*j + 2 < heapSize && minHeap[2*j + 2] < minHeap[minPos]) {
39 minPos = 2*j + 2;
40 }
41 if (minPos != j) {
42 int tmp = minHeap[j];
43 minHeap[j] = minHeap[minPos];
44 minHeap[minPos] = tmp;
45 }
46 }
47
48 if (heapSize < k) {
49 minHeap[heapSize++] = currSpeed;
50 } else if (currSpeed > minHeap[0]) {
51 topKSpeedSum += currSpeed - minHeap[0];
52 minHeap[0] = currSpeed;
53 }
54
55 maxPerformance = (int)(((long long)topKSpeedSum * currEfficiency) % MOD > maxPerformance ?
56 ((long long)topKSpeedSum * currEfficiency) % MOD : maxPerformance);
57 }
58 free(minHeap);
59 return maxPerformance;
60}
61We first convert each engineer's data into a tuple of speed and efficiency, after which we sort the engineers by efficiency. A min-heap helps keep only the fastest speeds currently deemed many enough to be an optimal team.
Each iteration over the engineers allows us to compute and maintain what the performance of the top engineers (by speed) looks like under the current efficiency.
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.
Due to complexity, a full DP + binary search approach has intricate implementation specific details. A similar illustrative solution in pseudocode involves maintaining an array where calculating potential top speeds and efficiencies are iterated and evaluated using a DP-based recurrence relation.