## Similar Problems

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# Maximum Performance Of A Team - Leetcode Solution

## Companies:

LeetCode: Maximum Performance Of A Team Leetcode Solution

Difficulty: Hard

Topics: greedy sorting heap-priority-queue array

Problem Statement: 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 defined as 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.

Solution:

Approach: The main objective of this problem is to find the best performance we can get in a team of engineers according to their speeds and efficiency. To do that, we need to keep track of all the engineers and their speeds/efficiency. We can keep their speed in a separate array and their efficiency in another. We need to apply two steps to find the solution:

- Sort the engineers by their efficiency so that we can have the engineer with the least efficiency at the start. Starting from the first engineer, select 'k' different engineers and calculate our team's performance.
- Repeat this process until we have considered all engineers.

Algorithm:

- Create a list of tuples to store engineer's speed and efficiency in decreasing order according to efficiency.
- Sort the list based on efficiency so that the engineer with the least efficiency will be present at the start of the list.
- Initialize our team's speed to 0, and create a max heap to keep track of the top k speeds.
- Traverse the sorted list; For each engineer, add the engineer's speed to the speed of the entire team so far.
- Calculate the performance metric and update the result if it's greater than the current value.
- Push the speed of the current engineer onto the heap and pop the smallest element if the size of the heap is greater than k.
- Return the result.

Time complexity: The above algorithm's time complexity is O(n log n), where n is the number of engineers. The sorting operation takes O(n log n) time, and the traversal operation takes O(n log k) time, which is the time taken to push and pop values from the heap.

Space complexity: The space complexity of the above algorithm is O(n), which is used to hold the engineer's speed and efficiency.

Code: (Python)

## Maximum Performance Of A Team Solution Code

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