Solution.py

"""

You are given two integers n and k and two integer arrays speed and efficiency both of length n. There are n engineers numbered from 1 to n. speed[i] and efficiency[i] represent the speed and efficiency of the ith engineer respectively.
Choose at most k different engineers out of the n engineers to form a team with the maximum performance.
The performance of a team is the sum of their engineers' speeds multiplied by the minimum efficiency among their engineers.
Return the maximum performance of this team. Since the answer can be a huge number, return it modulo 109 + 7.
 
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 <= k <= n <= 105
speed.length == n
efficiency.length == n
1 <= speed[i] <= 105
1 <= efficiency[i] <= 108


"""


class Solution:
    def maxPerformance(self, n: int, speed: List[int], efficiency: List[int], k: int) -> int:
        hp = []
        ss = 0
        mm = 0
        for e, s in sorted(zip(efficiency, speed))[::-1]:
            ss += s
            if len(hp) == k:
                removed_s = heapq.heappop(hp)
                ss -= removed_s
            heapq.heappush(hp, s)
            mm = max(mm, ss * e)
        return mm % (10 ** 9 + 7)