Reducing Dishes

Hard

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ArraysGreedy Algorithms

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.

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

Reducing Dishes

Hard

Asked by:

2 views

Topics:

ArraysGreedy Algorithms

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.

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

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the satisfaction ratings; can they be negative, zero, or only positive?
What is the maximum number of dishes (length of the input array) that I should expect?
If no combination of dishes yields a positive 'like-time coefficient' sum, what should I return?
Are there any duplicate satisfaction ratings in the input array?
Is the input array guaranteed to be non-empty?

Brute Force Solution

Approach

The brute force approach is like trying every single way you could cook the dishes. We will explore every possible subset of dishes, calculate the total 'satisfaction' for each, and then pick the best one we find. This guarantees we find the absolute best answer, but might take a very long time.

Here's how the algorithm would work step-by-step:

First, consider cooking only the first dish, and calculate the total satisfaction.
Next, consider cooking only the second dish, and calculate the total satisfaction.
Do this for each dish individually.
Then, consider cooking the first and second dishes together, and calculate the total satisfaction.
Consider cooking the first and third dishes together, and calculate the total satisfaction.
Do this for every possible pair of dishes.
Now, consider cooking the first, second, and third dishes together, and calculate the total satisfaction.
Continue this process, considering all possible groups of three dishes, then four dishes, and so on, until you consider cooking all the dishes together.
For each group of dishes you consider, make sure you calculate the satisfaction correctly, taking into account the order in which you cook them and the time it takes.
As you go, keep track of the highest total satisfaction you have found so far.
After you have tried every single possible combination of dishes, the highest total satisfaction you have found is the answer.

Code Implementation

def reducing_dishes_brute_force(satisfaction):
    number_of_dishes = len(satisfaction)
    maximum_satisfaction = 0

    # Iterate through all possible subsets of dishes
    for i in range(1 << number_of_dishes):
        current_dishes = []
        for j in range(number_of_dishes):
            # Check if j-th dish is included in the current subset
            if (i >> j) & 1:
                current_dishes.append(satisfaction[j])

        current_dishes.sort()
        current_satisfaction = 0
        time = 1

        # Calculate the total satisfaction for the current subset
        for dish_satisfaction in current_dishes:
            current_satisfaction += dish_satisfaction * time
            time += 1

        # Update the maximum satisfaction found so far
        maximum_satisfaction = max(maximum_satisfaction, current_satisfaction)

    return maximum_satisfaction

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(2^n) – The provided brute force solution explores every possible subset of the input dishes. With n dishes, there are 2^n possible subsets (each dish can either be included or excluded). For each subset, the algorithm calculates the total satisfaction, which takes O(n) time in the worst case to iterate through the dishes in the subset. Therefore, the overall time complexity is O(n * 2^n). However, since 2^n grows much faster than n, we can simplify the overall time complexity to O(2^n), as the exponential term dominates the runtime. We choose to represent the most significant term as 2^n as it reflects the exponential growth of subsets.

Space Complexity

O(1) – The brute force approach described only uses a variable to keep track of the highest total satisfaction found so far. It also iterates through all possible subsets of dishes without using any auxiliary data structures to store these subsets explicitly. Therefore, the space used is constant and does not depend on the number of dishes N. The auxiliary space complexity is O(1).

Optimal Solution

Approach

To maximize the total 'like-time coefficient,' we need to focus on dishes that contribute the most when multiplied by their preparation time. The core idea is to prepare the most likable dishes later so they are multiplied by the largest preparation times, and we can skip dishes that actively reduce our score.

Here's how the algorithm would work step-by-step:

First, sort the dishes by their 'like' values from least to most. This allows us to easily pick out the most likable dishes to cook last.
Now, consider the dishes one by one, starting from the most likable. Keep track of the total 'like-time coefficient' as we add each dish.
Before adding a dish, check if including it will increase the overall 'like-time coefficient.' If adding the dish decreases the coefficient, skip it. It's better to not cook it at all.
If adding the dish increases the coefficient (or keeps it the same), include it in our schedule and update the total coefficient.
Continue this process for all dishes. At the end, the total 'like-time coefficient' will be maximized because we only included dishes that contributed positively and cooked the most likable dishes at the end.

Code Implementation

def reducing_dishes(satisfaction_levels):
    satisfaction_levels.sort()
    max_like_time_coefficient = 0
    current_like_time_coefficient = 0
    sum_of_satisfaction_levels = 0

    # Iterate from right to left (most to least negative).
    for i in range(len(satisfaction_levels) - 1, -1, -1):

        # Calculate coefficient if current dish is included.
        new_like_time_coefficient = current_like_time_coefficient + satisfaction_levels[i] + sum_of_satisfaction_levels

        # Check if adding current dish increases total coefficient.
        if new_like_time_coefficient > current_like_time_coefficient:
            current_like_time_coefficient = new_like_time_coefficient
            sum_of_satisfaction_levels += satisfaction_levels[i]
            max_like_time_coefficient = current_like_time_coefficient

        # If it decreases or stays the same,
        # check if the current max is still 0 and if current dish can become the max.
        elif max_like_time_coefficient == 0 and satisfaction_levels[i] > 0:
            max_like_time_coefficient = satisfaction_levels[i]

    return max_like_time_coefficient

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation is sorting the input array of size n, which takes O(n log n) time. After sorting, we iterate through the sorted array once, performing constant-time operations for each element to calculate the maximum like-time coefficient. Therefore, the iteration is O(n), which is dominated by the sorting complexity. Thus, the overall time complexity is O(n log n).

Space Complexity

O(1) – The provided algorithm primarily sorts the input in place and then iterates through it. The only extra space used is for a few integer variables to track the current index and the cumulative like-time coefficient. The amount of extra memory doesn't depend on the number of dishes (N), making the auxiliary space constant.

Edge Cases

Case	How to Handle
Null or empty satisfaction array	Return 0, as there are no dishes to prepare.
Satisfaction array with a single dish	Return the satisfaction value if it's positive, otherwise return 0.
All dishes have negative satisfaction	Return 0, as no dishes should be prepared to maximize the result.
Satisfaction values include zero	Zeros can be included or excluded without affecting the validity, so treat them like other numbers.
Large satisfaction values that could lead to integer overflow	Use a 64-bit integer data type to store intermediate and final results to prevent overflow.
Satisfaction array is already sorted in descending order	The dynamic programming or greedy approach should still work correctly without modification.
Satisfaction array is sorted in ascending order (worst case)	The algorithm should sort the array into descending order, requiring O(n log n) time.
Maximum input size constraints - large number of dishes	Ensure algorithm has optimal time complexity (e.g., O(n log n) for sorting and O(n) for the rest) to avoid exceeding time limit.