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To maximize the like-time coefficient, we can sort the satisfaction levels in non-descending order and calculate the possible like-time coefficients by considering prefixes from the end of the sorted array. This allows us to selectively choose higher satisfaction dishes to maximize the total like-time coefficient.
Time Complexity: O(n log n) due to sorting.
Space Complexity: O(1) for using a constant amount of extra space.
1#include <stdio.h>
2#include <stdlib.h>
3
4int compare(const void* a, const void* b) {
5 return (*(int*)b - *(int*)a);
6}
7
8int maxSatisfaction(int* satisfaction, int size) {
9 qsort(satisfaction, size, sizeof(int), compare);
10 int total = 0, sum = 0;
11 for (int i = 0; i < size; ++i) {
12 sum += satisfaction[i];
13 if (sum > 0) {
14 total += sum;
15 } else {
16 break;
17 }
18 }
19 return total;
20}
21
22int main() {
23 int satisfaction[] = {-1, -8, 0, 5, -9};
24 int size = sizeof(satisfaction) / sizeof(satisfaction[0]);
25 printf("%d", maxSatisfaction(satisfaction, size));
26 return 0;
27}This C program sorts the satisfaction array in decreasing order and iteratively calculates the like-time coefficient. If adding the current satisfaction level to the sum results in a non-positive value, the computation stops as it indicates that further additions would only decrease the total result.
An alternative approach involves using dynamic programming to maximize the like-time coefficient. We maintain a DP table where `dp[i]` represents the maximum like-time coefficient attainable using the first `i` dishes. This approach computes overlap coefficients dynamically using state transitions calculated for each dish considered.
Time Complexity: O(n^2)
Space Complexity: O(n)
1#include <iostream>
2#include <vector>
#include <algorithm>
int maxSatisfaction(std::vector<int>& satisfaction) {
std::sort(satisfaction.begin(), satisfaction.end());
int n = satisfaction.size();
std::vector<int> dp(n + 1, 0);
for (int i = n - 1; i >= 0; i--) {
for (int t = 0; t <= i; t++) {
dp[t] = std::max(dp[t], dp[t + 1] + satisfaction[i] * (t + 1));
}
}
return dp[0];
}
int main() {
std::vector<int> satisfaction = {4, 3, 2};
std::cout << maxSatisfaction(satisfaction);
return 0;
}This C++ version applies dynamic programming to update the DP array by comparing the inclusion and exclusion of each dish, computing the optimal like-time coefficient.