#1383

Maximum Performance of a Team

candidate master · 1480 · lc hard +32 · verified · 47.7% accepted · 3,205 likes · top 33%

Description

There are n engineers where engineer i has speed[i] and efficiency[i]. Select at most k engineers to maximize team performance, defined as the sum of the selected engineers' speeds multiplied by the minimum efficiency among them. Return the maximum performance 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

Code

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class Solution:
    def maxPerformance(self, n: int, speed: List[int], efficiency: List[int], k: int) -> int:

class Solution:
    def maxPerformance(self, n: int, speed: List[int], efficiency: List[int], k: int) -> int: