Introduction

In this blog post, we will explore how to solve the LeetCode problem "1383" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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 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 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

Explanation

Here's the solution to the problem:

• Sort by Efficiency: Sort the engineers in descending order based on their efficiency. This allows us to iterate through potential minimum efficiency values in decreasing order.
• Maintain a Priority Queue (Min-Heap): Use a min-heap to keep track of the k engineers with the highest speeds seen so far, as we iterate through the sorted engineers. The heap will allow to maintain the k largest speeds, efficiently.

• Calculate Performance and Update Max: For each engineer (in sorted order of efficiency), add their speed to the total speed. If the heap size exceeds k, remove the engineer with the smallest speed from the heap and subtract his speed from the total speed. Compute performance = (total speed) * (current engineer's efficiency). Keep track of the maximum performance seen so far.

• Time Complexity: O(n log n), due to sorting and heap operations.

• Space Complexity: O(n), to store the sorted engineer indices, and the heap in the worst case.

Code

    import heapq

def maxPerformance(n: int, speed: list[int], efficiency: list[int], k: int) -> int:
    """
    Calculates the maximum performance of a team of at most k engineers.

    Args:
        n: The number of engineers.
        speed: A list of integers representing the speed of each engineer.
        efficiency: A list of integers representing the efficiency of each engineer.
        k: The maximum number of engineers allowed in the team.

    Returns:
        The maximum performance of the team modulo 10^9 + 7.
    """

    engineers = sorted(range(n), key=lambda i: efficiency[i], reverse=True)
    heap = []  # min-heap to store speeds
    total_speed = 0
    max_performance = 0
    modulo = 10**9 + 7

    for i in engineers:
        current_speed = speed[i]
        current_efficiency = efficiency[i]

        total_speed += current_speed
        heapq.heappush(heap, current_speed)

        if len(heap) > k:
            total_speed -= heapq.heappop(heap)

        max_performance = max(max_performance, total_speed * current_efficiency)

    return max_performance % modulo