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Mice and Cheese

Difficulty: Medium

Problem Description

There are two mice and n different types of cheese, each type of cheese should be eaten by exactly one mouse. You are given two arrays, reward1 and reward2, representing the points each mouse would earn by eating each type of cheese, and a non-negative integer k representing the number of types of cheese the first mouse must eat. The task is to return the maximum points the mice can achieve if the first mouse eats exactly k types of cheese.

Key Insights

  • Each type of cheese can only be eaten by one mouse.
  • The first mouse must eat exactly k types of cheese.
  • The total points are maximized by strategically choosing which k types of cheese the first mouse should eat.
  • The remaining cheese will be eaten by the second mouse.

Space and Time Complexity

Time Complexity: O(n log n) - due to sorting the cheese types based on the difference in reward between the two mice. Space Complexity: O(n) - for storing the differences and rewards in a new list.

Solution

To solve the problem, we can follow these steps:

  1. Create a list of tuples where each tuple contains the index of the cheese and the difference in rewards for the two mice (reward1[i] - reward2[i]).
  2. Sort this list based on the absolute value of the differences in descending order. This helps us prioritize the cheeses where the choice of mouse has the most impact on the total score.
  3. Select the first k items from the sorted list for the first mouse and the rest for the second mouse.
  4. Calculate the total points by summing the rewards based on which mouse eats which cheese.

This approach uses sorting, which is efficient for determining the optimal allocation of cheese types.

Code Solutions

def mice_and_cheese(reward1, reward2, k):
    n = len(reward1)
    differences = [(i, reward1[i] - reward2[i]) for i in range(n)]
    
    # Sort based on the absolute differences
    differences.sort(key=lambda x: abs(x[1]), reverse=True)
    
    total_points = 0
    for idx, diff in differences:
        if k > 0:
            total_points += reward1[idx]  # First mouse eats this cheese
            k -= 1
        else:
            total_points += reward2[idx]  # Second mouse eats this cheese
            
    return total_points
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