How do you solve Reducing Dishes on LeetCode?

Reducing Dishes (LeetCode #1402) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting the satisfaction levels allows us to maximize the like-time coefficient effectively.

What is the brute force solution for Reducing Dishes?

The brute force approach for Reducing Dishes is Brute Force, with O(n² * 2^n) time complexity and O(n) space complexity. The brute force approach involves generating all possible subsets of dishes and calculating the like-time coefficient for each subset. This helps us understand the problem by exploring all options, but it is inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Reducing Dishes?

Reducing Dishes uses the following concepts: Array, Dynamic Programming, Greedy, Sorting. The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Reducing Dishes?

Always clarify if you can discard dishes and if the order matters. Think about edge cases, such as all negative satisfaction levels. Explain your thought process clearly while coding; it shows your understanding.

What are common mistakes when solving Reducing Dishes?

Failing to sort the satisfaction array before processing. Not considering the impact of negative satisfaction levels on the total.

#1402

Reducing Dishes

Hard

ArrayDynamic ProgrammingGreedySortingGreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n² * 2^n)	O(n log n)
Space	O(n)	O(1)

💡

Intuition

Time O(n log n)Space O(1)

The optimal solution leverages sorting and a greedy approach. By sorting the satisfaction levels, we can maximize the like-time coefficient by considering only the most satisfying dishes in the right order.

⚙️

Algorithm

3 steps

1Step 1: Sort the satisfaction array in ascending order.
2Step 2: Initialize variables for the maximum sum and current sum.
3Step 3: Iterate from the end of the sorted array to the beginning, adding dishes to the current sum and updating the maximum sum if it increases.

solution.py10 lines

1def maxSatisfaction(satisfaction):
2    satisfaction.sort()
3    max_sum = 0
4    current_sum = 0
5    total = 0
6    for i in range(len(satisfaction) - 1, -1, -1):
7        total += satisfaction[i]
8        current_sum += total
9        max_sum = max(max_sum, current_sum)
10    return max_sum

ℹ

Complexity note: The time complexity is dominated by the sorting step, which is O(n log n). The space complexity is O(1) since we are using a constant amount of extra space.

1Sorting the satisfaction levels allows us to maximize the like-time coefficient effectively.
2Only the most satisfying dishes contribute positively to the total; negative dishes should be discarded.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.