Reducing Dishes

Problem Statement - link #

A chef has collected data on the satisfaction level of his n dishes. Chef can cook any dish in 1 unit of time.

Like-time coefficient of a dish is defined as the time taken to cook that dish including previous dishes multiplied by its satisfaction level i.e. time[i] * satisfaction[i].

Return the maximum sum of like-time coefficient that the chef can obtain after dishes preparation.

Dishes can be prepared in any order and the chef can discard some dishes to get this maximum value.

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
• -10^3 <= satisfaction[i] <= 10^3

Solutions #

class Solution {
public:
    int maxSatisfaction(vector<int>& satisfaction) {
        sort(satisfaction.begin(), satisfaction.end(), greater<int>());
        int prevSum = 0, res = 0;
        for(int cur: satisfaction) {
            prevSum += cur;
            if(prevSum < 0) {
                break;
            }
            res += prevSum;
        }
        return res;
    }
};