leetcode 1383. Maximum Performance of a Team

There are n engineers numbered from 1 to n and two arrays: speed and efficiency, where speed[i] and efficiency[i] represent the speed and efficiency for the i-th engineer respectively. Return the maximum performance of a team composed of at most k engineers, since the answer can be a huge number, return this modulo 10^9 + 7.

The performance of a team is the sum of their engineers' speeds multiplied by the minimum efficiency among their engineers. 

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 <= n <= 10^5
• speed.length == n
• efficiency.length == n
• 1 <= speed[i] <= 10^5
• 1 <= efficiency[i] <= 10^8
• 1 <= k <= n

题目 最多选k个工程师，求最大工作绩效sum(efficiencys)*min(speed)。

n选k的可能性有n!/k!种。需要进行一定的转换。对efficiency作为参数排序后不会起到任何作用。对speed排序后，只需要得到大于speed[i]的max(sum(efficiencys))即可。可以使用一个堆实现。代码如下:

class Solution(object):
    def maxPerformance(self, n, speed, efficiency, k):
        """
        :type n: int
        :type speed: List[int]
        :type efficiency: List[int]
        :type k: int
        :rtype: int
        """
        if k == 1:
            return max([speed[i]*efficiency[i] for i in range(n)])%(10**9 + 7)
        sortedWorker = sorted([[efficiency[i],speed[i]] for i in range(n)])
        heap = []
        totalSpeed = 0
        ret = 0
        for i in range(n-1,n-k,-1):
            heapq.heappush(heap,sortedWorker[i][1]) # 获取最大的k-1的总值
            totalSpeed += sortedWorker[i][1]
            ret = max(ret,totalSpeed*sortedWorker[i][0]) # 选择低于k的最大值
        
        for i in range(n-k,-1,-1):
            ret = max(ret,sortedWorker[i][0]*(totalSpeed+sortedWorker[i][1]))
            if sortedWorker[i][1] > heap[0]: #有更优的选择
                totalSpeed += sortedWorker[i][1] - heap[0]
                heapq.heappop(heap)
                heapq.heappush(heap,sortedWorker[i][1])
        return ret%(10**9 + 7)