1383 Maximum Performance of a Team

🔀 Pick One

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
  1. Keep track of the engineers by their efficiency in decreasing order.
  2. Starting from one engineer, to build a team, it suffices to bring K-1 more engineers who have higher efficiencies as well as high speeds.
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class Solution:
    def maxPerformance(
        self,
        n: int,
        speed: List[int],
        efficiency: List[int],
        k: int
    ) -> int:
        MOD = 10 ** 9 + 7
        engineers = list(zip(speed, efficiency))
        engineers.sort(key=lambda x: -x[1])
        speed_heap = []
        speed_sum = 0
        max_perf = -1
        for s, e in engineers:
            if len(speed_heap) > k - 1:
                speed_sum -= heapq.heappop(speed_heap)
            heapq.heappush(speed_heap, s)
            speed_sum += s
            max_perf = max(max_perf, e * speed_sum)
        return max_perf % MOD