Well Explained Easy Python 3 solution
Intuition\nThe problem can be solved using a greedy approach. We first calculate the difference in rewards between the two mice for each type of cheese and sort
# Intuition\nThe problem can be solved using a greedy approach. We first calculate the difference in rewards between the two mice for each type of cheese and sort them in decreasing order. Then, we select the top k differences and assume that the first mouse will eat those types of cheese. For the remaining types of cheese, we need to decide which mouse will eat them. We can do this by checking the remaining differences in decreasing order and assigning them to the mouse that has a higher reward for that type of cheese. However, we need to be careful to avoid assigning a type of cheese to the first mouse if it was not one of the top k differences.\n# Approach\n1. Calculate the difference in rewards between the two mice for each type of cheese.\n2. Sort the differences in decreasing order and select the top k.\n3. For the remaining types of cheese, iterate through them in decreasing order of the difference.\n4. If the type of cheese was in the top k differences, assign it to the first mouse.\n5. If the difference is positive, assign it to the mouse with the higher reward for that type of cheese.\n6. Otherwise, assign it to the second mouse.\n7. Calculate the total reward for the mice.\n# Complexity\n- Time complexity:\nO(nlogn) for sorting the differences, where n is the number of types of cheese.\n- Space complexity:\n O(n) for storing the differences.\n# Code\n```\nclass Solution:\n    def miceAndCheese(self, reward1: List[int], reward2: List[int], k: int) -> int:\n        if k == 0:\n            return sum(reward2)\n        difference = [reward1[i] - reward2[i] for i in range(len(reward1))]\n        sorted_diff = sorted(difference, reverse=True)\n        max_diff = sorted_diff[:k][-1]\n        skip = sorted_diff[k:].count(max_diff)\n        i, j, res = 0, 0, 0\n        while i < len(reward1):\n            if j == k:\n                res += reward2[i]\n            elif difference[i] == max_diff and skip > 0:\n                res += reward2[i]\n                skip -= 1\n            elif difference[i] >= max_diff:\n                res += difference[i] + reward2[i]\n                j += 1\n            else:\n                res += reward2[i]\n            i += 1\n        return res\n```
['Python3']