# 1383 Maximum Performance of a Team

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.
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 `````` ``````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) 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``````