Reducing Dishes

Problem Statement

In this problem, a chef has collected satisfaction levels for different dishes ranging from negative values, which imply dissatisfaction, to positive values indicating satisfaction. The goal is to maximize the sum of what's termed as the "like-time coefficient" for any subset of these dishes. The "like-time coefficient" for a dish is defined as the product of its satisfaction level and the time taken to cook it — including the cooking time of all previously cooked dishes. Each dish requires 1 unit of time to cook.

The chef can choose to cook any subset of dishes in any order, potentially even deciding not to cook some dishes (particularly those with negative satisfaction levels) in pursuit of achieving the highest sum of "like-time coefficients." Understanding the optimal sequence and selection of dishes to maximize this sum, given the variability in satisfaction levels, forms the crux of the problem.

Examples

Example 1

Input:

satisfaction = [-1,-8,0,5,-9]

Output:

14

Explanation:

After Removing the second and last dish, the maximum total like-time coefficient will be equal to (-1*1 + 0*2 + 5*3 = 14).
Each dish is prepared in one unit of time.

Example 2

Input:

satisfaction = [4,3,2]

Output:

20

Explanation:

Dishes can be prepared in any order, (2*1 + 3*2 + 4*3 = 20)

Example 3

Input:

satisfaction = [-1,-4,-5]

Output:

0

Explanation:

People do not like the dishes. No dish is prepared.

Constraints

• n == satisfaction.length
• 1 <= n <= 500
• -1000 <= satisfaction[i] <= 1000

Approach and Intuition

Given the requirements and the constraints, the problem can be approached efficiently:

1. Sort the Dishes by Satisfaction: This will allow us to consider potentially optimal sequences starting from the highest satisfaction to the lowest, facilitating the calculation by adding dishes that contribute most to the sum first.

2. Calculate Like-Time Coefficients: Start by calculating the total like-time coefficient by assuming all dishes are cooked in the sorted order.

3. Optimization by Exclusion: Begin to experiment by excluding dishes with the least satisfaction level and calculate the total like-time coefficient again. The idea is to identify a point where excluding any further dishes does not increase the overall sum or begins to decrease it. This involves:

   • Adjusting the cooking time for each subsequent dish whenever a dish is excluded.
   • Comparing the current sum with the maximum observed sum to decide whether to exclude a dish permanently from the optimal subset.

4. Systematic Exclusion: Use a loop to systematically exclude the lowest satisfaction dishes, recalculate the total, and compare with the maximum recorded total. The loop continues until all possibilities (each subset configuration) are explored or until the sum starts decreasing, indicating that further exclusions will not yield a better result.

• Edge Cases Handling: The edge cases such as all dishes having negative satisfaction levels where the best option would be not to cook any dish at all needs to be considered to immediately return a sum of zero.

This approach systematically evaluates each subset of dishes for their contribution to the maximal like-time coefficient and uses basic optimization techniques to efficiently reach a solution, especially given the potential constraints on the number of dishes and their satisfaction levels.

Solutions

class Solution {
public:
    int maxSatisfaction(vector<int>& feelings) {
        sort(feelings.begin(), feelings.end());
            
        int optimalHappiness = 0;
        int cumulativeSum = 0;
        for (int i = feelings.size() - 1; i >= 0 && cumulativeSum + feelings[i] > 0; i--) {
            cumulativeSum += feelings[i];
            optimalHappiness += cumulativeSum;
        }
            
        return optimalHappiness;
    }
};

class Solution {
    public int calculateMaxSatisfaction(int[] dishes) {
        Arrays.sort(dishes);
    
        int totalMaxSatisfaction = 0;
        int cumulativeSum = 0;
        for (int index = dishes.length - 1; index >= 0 && cumulativeSum + dishes[index] > 0; index--) {
            cumulativeSum += dishes[index];
            totalMaxSatisfaction += cumulativeSum;
        }
        return totalMaxSatisfaction;
    }
}