Formatted question description: https://leetcode.ca/all/1402.html

# 1402. Reducing Dishes (Hard)

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.

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 don't like the dishes. No dish is prepared.


Example 4:

Input: satisfaction = [-2,5,-1,0,3,-3]
Output: 35


Constraints:

• n == satisfaction.length
• 1 <= n <= 500
• -10^3 <= satisfaction[i] <= 10^3

Related Topics:
Dynamic Programming

## Solution 1. Brute Force

// OJ: https://leetcode.com/problems/reducing-dishes/

// Time: O(N^2)
// Space: O(1)
class Solution {
typedef long long LL;
public:
int maxSatisfaction(vector<int>& A) {
int N = A.size();
sort(A.begin(), A.end());
LL ans = 0;
for (int i = 0; i < N; ++i) {
LL sum = 0;
for (int j = i; j < N; ++j) sum += A[j] * (j - i + 1);
ans = max(ans, sum);
}
return ans;
}
};


## Solution 2. Prefix Sum

If array is sorted in descending order. The answer is one of the following:

• B[0] = A[0] * 1
• B[1] = A[0] * 2 + A[1] * 1
• B[2] = A[0] * 3 + A[1] * 2 + A[2] * 1

And we have

• B[0] = Sum(0)
• B[1] = B[0] + Sum(1)
• B[2] = B[1] + Sum(2)

Where Sum(i) = A[0] + ... + A[i], i.e. prefix sum of A.

• B[0] = Sum(0)
• B[1] = Sum(0) + Sum(1)
• B[2] = Sum(0) + Sum(1) + Sum(2)

And thus B is the prefix sum of prefix sum of A.

// OJ: https://leetcode.com/problems/reducing-dishes/

// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int maxSatisfaction(vector<int>& A) {
sort(begin(A), end(A), greater<int>());
partial_sum(begin(A), end(A), begin(A));
partial_sum(begin(A), end(A), begin(A));
return max(0, *max_element(begin(A), end(A)));
}
};


Java

• class Solution {
public int maxSatisfaction(int[] satisfaction) {
Arrays.sort(satisfaction);
int startIndex = -1;
int length = satisfaction.length;
for (int i = 0; i < length; i++) {
if (satisfaction[i] >= 0) {
startIndex = i;
break;
}
}
if (startIndex < 0)
return 0;
int sum = 0;
int totalSatisfaction = 0;
int time = 1;
for (int i = startIndex, j = 1; i < length; i++, j++) {
sum += satisfaction[i];
totalSatisfaction += satisfaction[i] * j;
}
int maxSatisfaction = totalSatisfaction;
for (int i = startIndex - 1; i >= 0; i--) {
sum += satisfaction[i];
if (sum < 0)
break;
totalSatisfaction += sum;
maxSatisfaction = Math.max(maxSatisfaction, totalSatisfaction);
}
return maxSatisfaction;
}
}

• // OJ: https://leetcode.com/problems/reducing-dishes/
// Time: O(N^2)
// Space: O(1)
class Solution {
typedef long long LL;
public:
int maxSatisfaction(vector<int>& A) {
int N = A.size();
sort(A.begin(), A.end());
LL ans = 0;
for (int i = 0; i < N; ++i) {
LL sum = 0;
for (int j = i; j < N; ++j) sum += A[j] * (j - i + 1);
ans = max(ans, sum);
}
return ans;
}
};

• print("Todo!")