human
Solve in Python:
There are n engineers numbered from 1 to n and two arrays: speed and efficiency, where speed[i] and efficiency[i] represent the speed and efficiency for the i-th engineer respectively. Return the maximum performance of a team composed of at most k engineers, since the answer can be a huge number, return this modulo 10^9 + 7.
The performance of a team is the sum of their engineers' speeds multiplied by the minimum efficiency among their engineers. 
 
Example 1:
Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2
Output: 60
Explanation: 
We have the maximum performance of the team by selecting engineer 2 (with speed=10 and efficiency=4) and engineer 5 (with speed=5 and efficiency=7). That is, performance = (10 + 5) * min(4, 7) = 60.

Example 2:
Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3
Output: 68
Explanation:
This is the same example as the first but k = 3. We can select engineer 1, engineer 2 and engineer 5 to get the maximum performance of the team. That is, performance = (2 + 10 + 5) * min(5, 4, 7) = 68.

Example 3:
Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4
Output: 72

 
Constraints:

1 <= n <= 10^5
speed.length == n
efficiency.length == n
1 <= speed[i] <= 10^5
1 <= efficiency[i] <= 10^8
1 <= k <= n
 ... What is a the Python code you requested?
assistant
#Here is the Python code you requested.

class Solution:
    def maxPerformance(self, n: int, speed: List[int], efficiency: List[int], k: int) -> int:
        items = sorted(zip(speed, efficiency), key=lambda item: (-item[1], -item[0]))
        heap = []

        s = 0
        res = 0

        for item in items:
            if len(heap) < k or item[0] > heap[0][0]:
                if len(heap) == k:
                    popped = heapq.heappop(heap)
                    s -= popped[0]
                heapq.heappush(heap, item)
                s += item[0]
                res = max(res, s * item[1])

        return res % (10 ** 9 + 7)