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| 1 | +class Solution { |
| 2 | +public: |
| 3 | + // Recursive function to solve the problem |
| 4 | + // satisfaction: vector containing satisfaction values |
| 5 | + // index: current index in the satisfaction array |
| 6 | + // time: current time multiplier for calculating the "like-time coefficient" |
| 7 | + int solve(vector<int>& satisfaction, int index, int time) { |
| 8 | + int n = satisfaction.size(); |
| 9 | + |
| 10 | + // Base case: if we've processed all dishes, return 0 |
| 11 | + if (index == n) return 0; |
| 12 | + |
| 13 | + // Option 1: Include the current dish |
| 14 | + // Add the "like-time coefficient" for the current dish and move to the next dish |
| 15 | + int include = satisfaction[index] * time + solve(satisfaction, index + 1, time + 1); |
| 16 | + |
| 17 | + // Option 2: Exclude the current dish |
| 18 | + // Skip the current dish and keep the same time multiplier |
| 19 | + int exclude = 0 + solve(satisfaction, index + 1, time); |
| 20 | + |
| 21 | + // Return the maximum of the two options |
| 22 | + return max(include, exclude); |
| 23 | + } |
| 24 | + |
| 25 | + // Main function to calculate the maximum "like-time coefficient" |
| 26 | + int maxSatisfaction(vector<int>& satisfaction) { |
| 27 | + // Step 1: Sort the satisfaction array in ascending order |
| 28 | + // This ensures we process the least satisfying dishes first, enabling us to skip them if needed |
| 29 | + sort(satisfaction.begin(), satisfaction.end()); |
| 30 | + |
| 31 | + // Step 2: Call the recursive function starting with index 0 and time 1 |
| 32 | + return solve(satisfaction, 0, 1); |
| 33 | + } |
| 34 | +}; |
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