Maximum Performance Of A Team Leetcode Problem 1383 [Python Solution]

In this blog post tutorial, we’ll dive into solving the Maximum Performance Of A Team problem, which is a LeetCode problem (number 1383).

This problem falls under the “Heap / Priority Queue” category and is rated as “Hard.” We will provide a Python solution to this problem, along with a detailed explanation of the approach and reasoning.

Problem Overview

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.

Your task is to 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 calculated as the sum of its engineers’ speeds multiplied by the minimum efficiency among its engineers.

Return the maximum performance of this team.

Since the answer can be a large number, return it modulo 10^9 + 7. Let’s break down the problem and understand it better.

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

Now that we have a clear understanding of the problem, let’s move on to the efficient Python solution.

Efficient Python Code Solution

View on GitHub ➹

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

        # Combine efficiency and speed into pairs and sort by efficiency in descending order
        for eff, spd in zip(efficiency, speed):
            eng.append([eff, spd])
        eng.sort(reverse=True)

        res, speed = 0, 0
        minHeap = []

        for eff, spd in eng:
            if len(minHeap) == k:
                speed -= heapq.heappop(minHeap)
            speed += spd
            heapq.heappush(minHeap, spd)
            res = max(res, eff * speed)

        return res % mod

Let’s break down this code step by step.

Combining and Sorting

We start by combining the efficiency and speed arrays into pairs and sorting them in descending order of efficiency.

This is crucial because we want to prioritize engineers with higher efficiency.

Initializing Variables

mod is set to 10^9 + 7 as required.
We create an empty list eng to store the engineer pairs and initialize res and speed to 0. res will keep track of the maximum performance, and speed will represent the total speed of the engineers we are currently considering.
We create an empty minHeap to keep track of the K engineers with the lowest speeds.

Main Loop

We iterate through the engineers in descending order of efficiency.

For each engineer:

If the size of minHeap is equal to k, we need to make room for a new engineer by removing the engineer with the lowest speed.

We subtract their speed from the total speed.

We add the speed of the current engineer to the total speed.
We push the speed of the current engineer into the minHeap.
We calculate the performance of the team with the current engineer as part of it and update res if this performance is greater than the current maximum performance.

Returning the Result

Finally, we return the maximum performance, res, modulo mod, as specified in the problem statement.

This efficient solution allows us to find the maximum performance with time complexity of O(n * log(n)) due to the sorting and space complexity of O(n) to store the engineer pairs.

Time and Space Complexity

Time Complexity: O(n * log(n))

Sorting the engineer pairs based on efficiency takes O(n * log(n)) time.
Iterating through the sorted pairs takes O(n) time.
Operations within the loop are O(log(k)) due to heap operations.

Space Complexity: O(n)

We store the engineer pairs in the eng list, which has a space complexity of O(n).
The minHeap can have at most k elements, and k is part of the input.

Reasoning Behind Our Approach

The key insight in this problem is to sort engineers by efficiency in descending order.

This allows us to consider engineers with higher efficiency first, as including them can potentially lead to a higher team performance.

We maintain a minHeap to keep track of the k engineers with the lowest speeds, ensuring we consider the fastest engineers.

By iterating through the sorted engineers and calculating the team performance for each combination, we find the maximum performance efficiently.

Conclusion

In this blog post, we tackled the Maximum Performance Of A Team problem on LeetCode.

We provided an efficient Python solution, explained the approach, and discussed the reasoning behind it.

This problem challenges your understanding of sorting, heap data structures, and the optimization of algorithmic performance.

We hope this guide helps you grasp the core concepts and strategies required to solve such challenging coding problems.

If you found this guide helpful, please feel free to comment, ask questions, make suggestions, and share the content.

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