# Solving Leetcode Interviews in Seconds with AI: Mice and Cheese


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2611" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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 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 Explanation: 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  

	# Explanation
	Here's the breakdown of the solution:

*   **Key Idea:** The core idea is to initially assign all cheese to the second mouse, then iteratively reassign the 'k' cheeses that give the largest point difference (reward1[i] - reward2[i]) to the first mouse. This ensures we maximize the total points.
*   **Difference Array:** Calculate the difference between `reward1[i]` and `reward2[i]` for each cheese type.
*   **Sorting:** Sort the differences in descending order to identify the cheeses that provide the most significant point gain when assigned to the first mouse.

*   **Complexity:**
    *   Runtime: O(n log n) due to sorting.
    *   Storage: O(n) to store the differences.

	
	# Code
	```python
	def max_points(reward1, reward2, k):
    """
    Calculates the maximum points the mice can achieve.

    Args:
        reward1: A list of points if the first mouse eats the cheese.
        reward2: A list of points if the second mouse eats the cheese.
        k: The number of cheese types the first mouse eats.

    Returns:
        The maximum total points the mice can achieve.
    """

    n = len(reward1)
    diff = []
    total_points = 0

    for i in range(n):
        diff.append((reward1[i] - reward2[i], i))  # Store the difference and original index
        total_points += reward2[i]  # Initially, all cheese is eaten by mouse 2

    diff.sort(key=lambda x: x[0], reverse=True)  # Sort by the difference in descending order

    for i in range(k):
        total_points += diff[i][0]  # Add the difference to the total points

    return total_points
	```