How do you solve Mice and Cheese on LeetCode?

Mice and Cheese (LeetCode #2611) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Choosing the types of cheese based on the difference in rewards maximizes the total points.

What is the time complexity of Mice and Cheese?

The optimal solution for Mice and Cheese runs in O(n log n) time and uses O(n) space. The complexity is O(n log n) due to the sorting step, while we use O(n) space to store the differences.

What is the brute force solution for Mice and Cheese?

The brute force approach for Mice and Cheese is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying all possible combinations of cheese types that the first mouse can eat. We calculate the total points for each combination and keep track of the maximum points found. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Mice and Cheese?

Mice and Cheese uses the following concepts: Array, Greedy, Sorting, Heap (Priority Queue). The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Mice and Cheese?

Always think about how to optimize your brute force solution. Explain your thought process clearly when transitioning from brute force to optimal solutions. Practice similar problems to recognize patterns in greedy algorithms.

What are common mistakes when solving Mice and Cheese?

Not considering the greedy nature of the problem and trying to brute force instead. Overlooking the need to sort differences before selection.

#2611

Mice and Cheese

Medium

ArrayGreedySortingHeap (Priority Queue)GreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal solution uses a greedy approach by calculating the difference in rewards for each cheese type and sorting them. This allows us to efficiently select the k types of cheese that maximize the total points.

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Algorithm

3 steps

1Step 1: Create an array of tuples containing the difference between reward1 and reward2 for each cheese type.
2Step 2: Sort this array based on the differences in descending order.
3Step 3: Sum the rewards for the first mouse for the top k differences and the rewards for the second mouse for the remaining types.

solution.py11 lines

1# Full working Python code
2def maxPointsOptimal(reward1, reward2, k):
3    n = len(reward1)
4    differences = [(reward1[i] - reward2[i], reward1[i], reward2[i]) for i in range(n)]
5    differences.sort(reverse=True)
6    total_points = 0
7    for i in range(k):
8        total_points += differences[i][1]  # reward1 for first mouse
9    for i in range(k, n):
10        total_points += differences[i][2]  # reward2 for second mouse
11    return total_points

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Complexity note: The complexity is O(n log n) due to the sorting step, while we use O(n) space to store the differences.

1Choosing the types of cheese based on the difference in rewards maximizes the total points.
2Sorting helps us efficiently select the best options for the first mouse.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.