Maximum Performance of a Team

Here’s a solution to the problem using a greedy approach:

  1. Pair up the engineers by their efficiency and speed and then sort them by efficiency in descending order.
  2. Use a min-heap to keep track of the current team’s speed and manage the k engineers selected so far.
  3. Iterate through the sorted list, calculating the performance by adding each engineer’s speed, and taking into account the current minimum efficiency. If the team grows beyond k, we remove the engineer with the lowest speed.
  4. Track the maximum performance obtained.

Here’s the code:

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from heapq import heappush, heappop

class Solution:
    def maxPerformance(self, n: int, speed: List[int], efficiency: List[int], k: int) -> int:
        MOD = 10**9 + 7

        # Pair up the efficiency and speed and sort by efficiency in descending order
        engineers = sorted(zip(efficiency, speed), reverse=True)

        # Min-heap to keep track of the speeds of the k engineers in the team
        speed_heap = []
        total_speed = 0
        max_performance = 0

        for eff, spd in engineers:
            # Add the speed of the current engineer to the total_speed and push it to the min-heap
            total_speed += spd
            heappush(speed_heap, spd)

            # If the team size is greater than k, remove the engineer with the lowest speed
            if len(speed_heap) > k:
                total_speed -= heappop(speed_heap)

            # Calculate the current performance using the current efficiency and total speed
            performance = total_speed * eff

            # Track the maximum performance
            max_performance = max(max_performance, performance)

        return max_performance % MOD

This code returns the maximum performance of the team of at most k engineers, given their speeds and efficiencies.

Identifying Problem Isomorphism

“Maximum Performance of a Team” is isomorphic to “Minimum Cost to Hire K Workers”.

Both problems involve a form of optimization with constraints. In “Minimum Cost to Hire K Workers”, we aim to minimize the total wage bill for hiring K workers, where each worker has a certain quality and wage expectation. This situation is analogous to “Maximum Performance of a Team”, where we want to maximize the performance of a team under certain constraints.

In both problems, there’s a need to sort the workers (or team members) according to a certain metric (wage to quality ratio or speed to efficiency ratio) and then apply a heap-based strategy to maintain the best candidates.

“Minimum Cost to Hire K Workers” is simpler due to the one-dimensional optimization (minimize cost), compared to the two-dimensional optimization (maximize speed and efficiency) in “Maximum Performance of a Team”.

In both problems, the sorting and heap strategy enables us to determine the most optimal team or set of workers. Yet, understanding the nuance of how these strategies apply might be more challenging in “Maximum Performance of a Team” due to the two-dimensional nature of the problem.

10 Prerequisite LeetCode Problems

“1383. Maximum Performance of a Team” involves sorting, heap data structure, and greedy algorithm. Here are some simpler problems to prepare for this:

  1. LeetCode 215. Kth Largest Element in an Array

    • This problem introduces the concept of using a heap data structure to solve problems.

  2. LeetCode 378. Kth Smallest Element in a Sorted Matrix

    • This problem involves using a heap to solve a problem that involves both sorting and selecting elements.

  3. LeetCode 23. Merge k Sorted Lists

    • This problem can give you practice using a heap to merge sorted arrays, similar to the “Maximum Performance of a Team” problem.

  4. LeetCode 1046. Last Stone Weight

    • This problem involves using a heap to perform operations and find the maximum.

  5. LeetCode 253. Meeting Rooms II

    • This problem involves sorting and selecting elements based on certain conditions, similar to the “Maximum Performance of a Team” problem.

  6. LeetCode 406. Queue Reconstruction by Height

    • This problem also involves sorting and selecting elements based on certain conditions.

  7. LeetCode 452. Minimum Number of Arrows to Burst Balloons

    • This problem can help you understand the greedy algorithm where you make the locally optimal choice at each stage.

  8. LeetCode 621. Task Scheduler

    • This problem involves arranging tasks based on their frequencies, which is a scheduling problem that can be solved using a heap.

  9. LeetCode 767. Reorganize String

    • This problem also involves arranging characters based on their frequencies, a similar concept to arranging tasks in the “Maximum Performance of a Team” problem.

  10. LeetCode 295. Find Median from Data Stream

    • This problem involves maintaining a data structure to answer queries, which is a concept seen in the “Maximum Performance of a Team” problem.

These cover how to use heap data structures, sort and select elements, and apply the greedy algorithm, which are key to solving the “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