1383. Maximum Performance of a Team

Hard

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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:
        arr = []
        result = 0
        speed_sum = 0
        for e, s in sorted(zip(efficiency, speed), reverse=1):
            heapq.heappush(arr, s)
            speed_sum += s
            if len(arr) > k:
                speed_sum -= heapq.heappop(arr)
            result = max(result, speed_sum*e)
        return result % (10**9+7)