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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.
1import java.util.Arrays;
2
3public class ReducingDishes {
4 public static int maxSatisfaction(int[] satisfaction) {
5 Arrays.sort(satisfaction);
6 int total = 0, sum = 0;
7 for (int i = satisfaction.length - 1; i >= 0; i--) {
8 sum += satisfaction[i];
9 if (sum > 0) {
10 total += sum;
11 } else {
12 break;
13 }
14 }
15 return total;
16 }
17
18 public static void main(String[] args) {
19 int[] satisfaction = {-1, -8, 0, 5, -9};
20 System.out.println(maxSatisfaction(satisfaction));
21 }
22}This Java solution sorts the satisfaction array in ascending order, then effectively handles calculations from the end of the array to maximize the like-time coefficient.
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)
1import
In this Java implementation, dynamic programming is used. A 1D DP array computes the best attainable like-time coefficient by considering each dish and its contribution to the past prepared dishes.