Mice and Cheese

Maximum Points From Cheese Problem

Problem Description

You are given two mice and n different types of cheese. Each type of cheese should be eaten by exactly one mouse. For each cheese i, the first mouse gets reward1[i] points if it eats the cheese, and the second mouse gets reward2[i] points if it eats the cheese. You are also given an integer k, representing the number of cheese types the first mouse must eat. The objective is to maximize the total points earned by both mice.

Naive Solution

A brute-force solution would involve trying all possible combinations of k cheeses for the first mouse and assigning the remaining cheeses to the second mouse. Calculate the total points for each combination, and return the maximum. This involves generating all possible subsets of size k from a set of size n, which is highly inefficient.

Complexity Analysis

• Time Complexity: O(C(n, k) * n), where C(n, k) is the binomial coefficient "n choose k", representing the number of ways to choose k elements from a set of n elements. This grows factorially, making it very slow.
• Space Complexity: O(k), to store the current combination of cheeses.

Optimal Solution

A more efficient approach involves using a greedy algorithm combined with sorting. We prioritize the cheeses for which the difference between reward1[i] and reward2[i] is the greatest. Intuitively, we want the first mouse to eat the cheeses where it gets significantly more points than the second mouse. Here's the algorithm:

1. Calculate the difference diff[i] = reward1[i] - reward2[i] for each cheese.
2. Sort the cheeses based on diff[i] in descending order.
3. Assign the first k cheeses (after sorting) to the first mouse.
4. Assign the remaining n - k cheeses to the second mouse.
5. Calculate the total points.

Edge Cases

• k = 0: The first mouse eats no cheese, so the total score is the sum of reward2.
• k = n: The first mouse eats all cheeses, so the total score is the sum of reward1.
• reward1 and reward2 are identical, meaning the order of selection doesn't matter beyond satisfying the k cheeses eaten.

Code Example (Python)

def max_points(reward1, reward2, k):
    n = len(reward1)
    diff = [reward1[i] - reward2[i] for i in range(n)]
    indices = sorted(range(n), key=lambda i: diff[i], reverse=True)
    
    total_points = 0
    for i in range(k):
        total_points += reward1[indices[i]]
    for i in range(k, n):
        total_points += reward2[indices[i]]
        
    return total_points

Complexity Analysis

• Time Complexity: O(n log n), due to sorting the differences. The rest of the operations take O(n) time.
• Space Complexity: O(n), to store the diff array and the indices array.

Example

Let reward1 = [1, 1, 3, 4], reward2 = [4, 4, 1, 1], and k = 2. Here's how the algorithm works:

1. diff = [-3, -3, 2, 3]
2. indices = [3, 2, 0, 1] (sorted indices based on diff in descending order)
3. The first mouse eats cheeses 3 and 2 (indices 3 and 2 in the original arrays), earning 4 + 3 = 7 points.
4. The second mouse eats cheeses 0 and 1 (indices 0 and 1 in the original arrays), earning 4 + 4 = 8 points.
5. The total points are 7 + 8 = 15.