#2611

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Mice and Cheese

In 'Mice and Cheese', you must maximize the total reward of two mice eating cheese while respecting their preferences and a given limit.

array greedy sorting heap priority queue

Problem Statement

In this problem, you are given two mice and n different types of cheese. Each cheese type should be eaten by exactly one mouse. You are provided with two arrays: reward1, where reward1[i] represents the reward for the first mouse eating the i-th type of cheese, and reward2, where reward2[i] represents the reward for the second mouse. You are also given a non-negative integer k, representing the maximum number of cheeses that the first mouse can eat.

The task is to assign cheeses to both mice such that each cheese is eaten by exactly one mouse. You need to maximize the total reward by choosing cheeses in an optimal way. The first mouse can eat up to k cheeses, and the second mouse can eat the remaining cheeses. Your goal is to calculate the maximum total reward the mice can achieve under these constraints.

Examples

Example 1

Input: reward1 = [1,1,3,4], reward2 = [4,4,1,1], k = 2

Output: 15

In this example, the first mouse eats the 2nd (0-indexed) and the 3rd types of cheese, and the second mouse eats the 0th and the 1st 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

In this example, the first mouse eats the 0th (0-indexed) and 1st 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

Solution Approach

Greedy Approach

The problem is a perfect fit for a greedy approach. We first calculate the difference between the rewards for each cheese type (reward1[i] - reward2[i]). Sorting these differences in descending order allows us to choose the top k cheeses for the first mouse, maximizing the reward. The remaining cheeses are assigned to the second mouse.

Invariant Validation

After sorting based on the differences, we ensure that the first mouse gets the best k cheeses. This invariant remains valid because we only change the assignment of cheeses based on their relative reward differences, maintaining the maximum reward for both mice.

Efficiency Considerations

Given the large constraints, sorting the cheeses based on the reward difference ensures that we can compute the result efficiently. The overall time complexity is O(n log n) due to the sorting step, which is optimal for this problem.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this solution is O(n log n) due to sorting the reward differences. The space complexity is O(n) as we store the reward differences and perform operations on the arrays.

What Interviewers Usually Probe

Test the candidate's understanding of greedy algorithms and their ability to handle sorting in optimization problems.
Evaluate if the candidate can recognize the pattern of greedy choice plus invariant validation and apply it correctly.
Look for understanding of the trade-off between the first mouse's reward and the second mouse's reward, ensuring the final solution is maximized.

Common Pitfalls or Variants

Common pitfalls

Incorrectly assigning cheese types to both mice, ignoring the greedy approach that maximizes rewards.
Forgetting to validate that the first mouse eats at most k cheeses, which could violate the problem's constraints.
Neglecting to account for sorting the cheese reward differences before assignment, leading to suboptimal solutions.

Follow-up variants

Change the constraint on k, allowing for more or fewer cheeses for the first mouse to eat.
Modify the problem by having more than two mice, testing the scalability of the approach.
Alter the reward arrays to introduce some randomness or additional constraints to test robustness.

FAQ

What is the greedy strategy for 'Mice and Cheese'?

In 'Mice and Cheese', the greedy strategy involves sorting the cheese types by the difference in rewards and then assigning the best k cheeses to the first mouse, maximizing the total reward.

How do we handle the constraint on k in the problem?

The first mouse can eat at most k cheeses, so we choose the top k cheeses with the highest reward for the first mouse and assign the rest to the second mouse.

What is the time complexity of the optimal solution for 'Mice and Cheese'?

The optimal solution has a time complexity of O(n log n) due to the sorting step of the cheese rewards, where n is the number of cheese types.

Can the greedy approach for 'Mice and Cheese' fail?

The greedy approach is optimal for this problem, as long as we correctly choose the top k cheeses for the first mouse and respect the problem constraints.

What are some possible variations of the 'Mice and Cheese' problem?

Possible variations include changing the number of mice, adjusting the value of k, or introducing more complex constraints on cheese assignments.

terminal

Solution

Solution 1: Greedy + Sort

We can first give all the cheese to the second mouse. Next, consider giving $k$ pieces of cheese to the first mouse. How should we choose these $k$ pieces of cheese? Obviously, if we give the $i$-th piece of cheese from the second mouse to the first mouse, the change in the score is $reward1[i] - reward2[i]$. We hope that this change is as large as possible, so that the total score is maximized.

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class Solution:
    def miceAndCheese(self, reward1: List[int], reward2: List[int], k: int) -> int:
        n = len(reward1)
        idx = sorted(range(n), key=lambda i: reward1[i] - reward2[i], reverse=True)
        return sum(reward1[i] for i in idx[:k]) + sum(reward2[i] for i in idx[k:])

Solution 2

#### Python3

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class Solution:
    def miceAndCheese(self, reward1: List[int], reward2: List[int], k: int) -> int:
        n = len(reward1)
        idx = sorted(range(n), key=lambda i: reward1[i] - reward2[i], reverse=True)
        return sum(reward1[i] for i in idx[:k]) + sum(reward2[i] for i in idx[k:])

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