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Reducing Dishes

Difficulty: Hard

Problem Description

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

Key Insights

  • The order in which dishes are cooked affects the total like-time coefficient.
  • Discarding dishes with lower satisfaction levels can potentially maximize the total like-time coefficient.
  • Sorting the satisfaction levels in non-decreasing order allows for easier calculation of coefficients.
  • A greedy approach can be applied, starting from the dish with the highest satisfaction level and moving downwards.

Space and Time Complexity

Time Complexity: O(n log n) - due to sorting the satisfaction array. Space Complexity: O(1) - no additional space used beyond input storage.

Solution

To solve this problem, we can use a greedy approach combined with sorting. First, we sort the satisfaction array in non-decreasing order. Then, we iterate through the sorted array in reverse (from highest satisfaction to lowest) and calculate the like-time coefficients. We maintain a running total of the coefficients and a separate variable to keep track of the current sum of satisfaction levels. If this current sum becomes negative, we stop, as adding more dishes will decrease the total like-time coefficient. The final result will be the maximum sum calculated.

Code Solutions

def maxSatisfaction(satisfaction):
    satisfaction.sort()  # Sort the satisfaction levels
    max_sum = 0
    current_sum = 0
    total = 0
    
    for i in range(len(satisfaction) - 1, -1, -1):
        current_sum += satisfaction[i]  # Add satisfaction level
        if current_sum > 0:  # Only consider if the current sum is positive
            total += current_sum  # Update total like-time coefficient
        else:
            break  # Stop if current_sum is not positive
    
    return total
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