Mice and Cheese

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ArraysGreedy AlgorithmsDynamic Programming

There are two mice and n different types of cheese, each type of cheese should be eaten by exactly one mouse.

A point of the cheese with index i (0-indexed) is:

reward1[i] if the first mouse eats it.
reward2[i] if the second mouse eats it.

You are given a positive integer array reward1, a positive integer array reward2, and a non-negative integer k.

Return the maximum points the mice can achieve if the first mouse eats exactly k types of cheese.

Example 1:

Input: reward1 = [1,1,3,4], reward2 = [4,4,1,1], k = 2
Output: 15
Explanation: In this example, the first mouse eats the 2^nd (0-indexed) and the 3^rd types of cheese, and the second mouse eats the 0^th and the 1^st types of cheese.
The total points are 4 + 4 + 3 + 4 = 15.
It can be proven that 15 is the maximum total points that the mice can achieve.

Example 2:

Input: reward1 = [1,1], reward2 = [1,1], k = 2
Output: 2
Explanation: In this example, the first mouse eats the 0^th (0-indexed) and 1^st types of cheese, and the second mouse does not eat any cheese.
The total points are 1 + 1 = 2.
It can be proven that 2 is the maximum total points that the mice can achieve.

Constraints:

1 <= n == reward1.length == reward2.length <= 10⁵
1 <= reward1[i], reward2[i] <= 1000
0 <= k <= n

Try coding it in LeetCode

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 are the constraints on the sizes of `reward1` and `reward2`, and the value of `k`? Specifically, what are the minimum and maximum possible values for each?
Can the values in `reward1` and `reward2` be negative, zero, or non-integer?
Is `k` always a valid number of positions that mouse1 can take, i.e., is it always between 0 and the length of the arrays, inclusive?
If `reward1` and `reward2` have different lengths, how should I handle this? Is this an invalid input?
Are there any specific time or space complexity constraints I should aim for beyond the standard goal of optimization?

Brute Force Solution

Approach

The brute force approach means trying every possible way to allocate resources. In this problem, we explore every single combination of choices to find the best one. This guarantees we will find the optimal solution, even if it takes a very long time.

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

Consider all possible ways to assign each item to one of the available options.
For each possible assignment, calculate the overall score or cost.
Compare the scores from each possible assignment.
Select the assignment that results in the highest score or the lowest cost, based on the specific requirements of the problem.

Code Implementation

def mice_and_cheese_brute_force(reward1, reward2, numberOfMice):    number_of_items = len(reward1)
    max_reward = 0

    # Iterate through all possible combinations of mice allocations
    for i in range(1 << number_of_items):
        mice_for_cheese_one = 0
        current_reward = 0

        # Calculate the rewards and mice used for each combination
        for j in range(number_of_items):
            if (i >> j) & 1:
                current_reward += reward1[j]
                mice_for_cheese_one += 1
            else:
                current_reward += reward2[j]

        # Check if the number of mice used is within the limit
        if mice_for_cheese_one <= numberOfMice:

            # Update the maximum reward if the current reward is higher
            max_reward = max(max_reward, current_reward)

    return max_reward

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores every possible way to assign each item. If we have n items and each item can be assigned to one of two options (mouse A or mouse B), then there are 2 possible choices for the first item, 2 for the second, and so on. Thus, the total number of combinations we need to evaluate is 2 * 2 * ... * 2 (n times), which is 2^n. Calculating the score for each combination takes O(n) time, but since we need to consider all 2^n combinations, the dominant factor is O(2^n).

Space Complexity

O(1) – The brute force approach outlined generates all possible assignments of items. This involves creating combinations, but the explanation does not explicitly specify storing all combinations simultaneously. It suggests calculating and comparing scores for each assignment individually. Without any explicit creation of data structures to store intermediate assignments or visited combinations, the space complexity remains constant, using only a few variables to track the current assignment being evaluated and the best score found so far.

Optimal Solution

Approach

The goal is to maximize the total cheese collected by the mice. We should prioritize assigning tasks that yield the greatest difference in cheese between the two mice, ensuring we get the most benefit from each assignment.

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

Figure out, for each task, how much more cheese Mouse 1 gets compared to Mouse 2.
Sort the tasks from the biggest difference (Mouse 1 getting way more cheese) to the smallest difference (or even Mouse 2 getting more).
Assign the first set of tasks (based on the number Mouse 1 can do) to Mouse 1. These are the tasks where Mouse 1 gets a significantly larger amount of cheese.
Assign the remaining tasks to Mouse 2. These are the tasks where Mouse 2 might even be slightly better, or where the difference isn't big enough to prioritize them for Mouse 1.
Calculate the total cheese collected by both mice by adding up the cheese values based on who performed each task. The tasks for the first group should be assigned to mouse 1 and the rest to mouse 2.

Code Implementation

def mice_and_cheese(reward1, reward2, number_of_mice1):
    number_of_tasks = len(reward1)
    task_differences = []

    for i in range(number_of_tasks):
        task_differences.append((reward1[i] - reward2[i], i))

    # Sort tasks based on the difference in rewards.
    task_differences.sort(reverse=True)

    total_reward = 0
    for i in range(number_of_mice1):
        # Assign the first 'number_of_mice1' tasks to mouse 1.
        task_index = task_differences[i][1]
        total_reward += reward1[task_index]

    # Assign the remaining tasks to mouse 2.
    for i in range(number_of_mice1, number_of_tasks):
        task_index = task_differences[i][1]
        total_reward += reward2[task_index]

    return total_reward

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm first calculates the difference in cheese for each task, which takes O(n) time. The dominant operation is sorting the tasks based on these differences. Sorting n elements using an efficient algorithm like merge sort or quicksort takes O(n log n) time. After sorting, assigning tasks and calculating the total cheese takes O(n) time. Therefore, the overall time complexity is dominated by the sorting step, resulting in O(n log n).

Space Complexity

O(N) – The algorithm creates a list of differences, where N is the number of tasks (length of the input arrays). This list stores the difference in cheese gained by Mouse 1 versus Mouse 2 for each task, requiring space proportional to the input size N. The sorting operation is performed in-place or creates another copy of the difference array which also takes O(N) space, dominating any constant space usages. Therefore, the overall auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
reward1 or reward2 is null or empty	Return 0 if either reward array is null or empty, indicating no rewards can be collected.
k is 0 or k is equal to the length of the arrays	If k is 0, return the sum of reward2; if k equals the array length, return the sum of reward1.
reward1 and reward2 have different lengths	Throw an IllegalArgumentException indicating that the input arrays must have the same length.
Arrays contain large positive/negative numbers that could cause integer overflow	Use long data type to store intermediate sums and the final result to prevent integer overflow.
k is negative or greater than the length of the arrays	Throw an IllegalArgumentException indicating that k must be within the valid range of [0, length of arrays].
Arrays contain all identical values, and k is neither 0 nor equal to the array length	The algorithm should correctly choose k elements from reward1 and the rest from reward2, regardless of the identical values.
Arrays contain a mix of positive, negative, and zero values	The algorithm should still function correctly, maximizing the total reward considering both positive and negative contributions.
Arrays are extremely large (performance considerations)	A sorting-based approach should be optimized (e.g., using quicksort or mergesort) or a linear time solution should be employed if possible (e.g., finding the k largest differences between reward1 and reward2).

reward1 or reward2 is null or empty

How to Handle:

Return 0 if either reward array is null or empty, indicating no rewards can be collected.

k is 0 or k is equal to the length of the arrays

How to Handle:

If k is 0, return the sum of reward2; if k equals the array length, return the sum of reward1.

reward1 and reward2 have different lengths

How to Handle:

Throw an IllegalArgumentException indicating that the input arrays must have the same length.

Arrays contain large positive/negative numbers that could cause integer overflow

How to Handle:

Use long data type to store intermediate sums and the final result to prevent integer overflow.

k is negative or greater than the length of the arrays

How to Handle:

Throw an IllegalArgumentException indicating that k must be within the valid range of [0, length of arrays].

Arrays contain all identical values, and k is neither 0 nor equal to the array length

How to Handle:

The algorithm should correctly choose k elements from reward1 and the rest from reward2, regardless of the identical values.

Arrays contain a mix of positive, negative, and zero values

How to Handle:

The algorithm should still function correctly, maximizing the total reward considering both positive and negative contributions.

Arrays are extremely large (performance considerations)

How to Handle:

A sorting-based approach should be optimized (e.g., using quicksort or mergesort) or a linear time solution should be employed if possible (e.g., finding the k largest differences between reward1 and reward2).