Problem
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 do not like the dishes. No dish is prepared.
Constraints:
n == satisfaction.length1 <= n <= 500-1000 <= satisfaction[i] <= 1000
Solution
class Solution {
public int maxSatisfaction(int[] satisfaction) {
Arrays.sort(satisfaction);
int sum = 0;
int mulSum = 0;
for (int i = 0; i < satisfaction.length; i++) {
sum += satisfaction[i];
mulSum += (i + 1) * satisfaction[i];
}
int maxVal = Math.max(0, mulSum);
for (int j : satisfaction) {
mulSum -= sum;
sum -= j;
maxVal = Math.max(maxVal, mulSum);
}
return maxVal;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).