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Mice and Cheese

Number: 2725

Difficulty: Medium

Paid? No

Companies: Meta, DoorDash

Problem Description

There are two mice and n types of cheese. Each cheese type can only be eaten by one mouse. If the first mouse eats cheese i, it scores reward1[i] points; if the second mouse eats it, it scores reward2[i] points. The first mouse must eat exactly k types of cheese, while the second mouse eats the remaining ones. The task is to maximize the total points obtained.

Key Insights

  • Start with a baseline where every cheese is eaten by the second mouse, earning a sum of reward2.
  • The benefit (or loss) of giving a cheese to the first mouse is reward1[i] - reward2[i].
  • To maximize the total, choose exactly k cheeses with the highest (reward1[i] - reward2[i]) differences.
  • Sorting the differences in descending order will allow selection of the top k cheeses.
  • Time complexity is dominated by sorting, O(n log n), and space complexity is O(n).

Space and Time Complexity

Time Complexity: O(n log n) due to sorting the difference values. Space Complexity: O(n) for storing the array of differences.

Solution

We begin by calculating the total points if the second mouse were to eat all cheeses. Then, we compute the benefit of switching each cheese to the first mouse which is calculated as (reward1[i] - reward2[i]). To maximize the total score, we need to pick exactly k cheeses that offer the highest benefits. Sort the benefits in descending order and add the top k benefits to the initial total (which was the sum of reward2). This strategy ensures we gain the largest possible extra points from switching cheeses to the first mouse. The necessary data structures include arrays for values and differences, and the algorithmic approach primarily employs sorting.

Code Solutions

def miceAndCheese(reward1, reward2, k):
    # Calculate the total points assuming the second mouse eats all the cheeses.
    total_points = sum(reward2)
    
    # Calculate the benefit of switching each cheese from the second mouse to the first mouse.
    benefits = [r1 - r2 for r1, r2 in zip(reward1, reward2)]
    
    # Sort the benefits in descending order to choose the k cheeses with maximum benefit.
    benefits.sort(reverse=True)
    
    # Add up the benefits of the top k cheeses.
    total_points += sum(benefits[:k])
    
    return total_points

# Example usage:
print(miceAndCheese([1, 1, 3, 4], [4, 4, 1, 1], 2))
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