Introduction

In this blog post, we will explore how to solve the LeetCode problem "857" 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

There are n workers. You are given two integer arrays quality and wage where quality[i] is the quality of the ith worker and wage[i] is the minimum wage expectation for the ith worker. We want to hire exactly k workers to form a paid group. To hire a group of k workers, we must pay them according to the following rules: Every worker in the paid group must be paid at least their minimum wage expectation. In the group, each worker's pay must be directly proportional to their quality. This means if a worker’s quality is double that of another worker in the group, then they must be paid twice as much as the other worker. Given the integer k, return the least amount of money needed to form a paid group satisfying the above conditions. Answers within 10-5 of the actual answer will be accepted. Example 1: Input: quality = [10,20,5], wage = [70,50,30], k = 2 Output: 105.00000 Explanation: We pay 70 to 0th worker and 35 to 2nd worker. Example 2: Input: quality = [3,1,10,10,1], wage = [4,8,2,2,7], k = 3 Output: 30.66667 Explanation: We pay 4 to 0th worker, 13.33333 to 2nd and 3rd workers separately. Constraints: n == quality.length == wage.length 1 <= k <= n <= 104 1 <= quality[i], wage[i] <= 104

Explanation

Here's a breakdown of the approach, followed by the Python code:

• Ratio Calculation and Sorting: Calculate the wage-to-quality ratio for each worker. Sort the workers based on this ratio in ascending order. This ensures we consider workers who offer the most "value" (lowest wage per unit quality) first.
• Priority Queue for Quality: Maintain a max-heap (priority queue) of the qualities of the k workers with the smallest ratios seen so far. This helps us efficiently track the k highest qualities and update the total quality sum.

• Iterative Minimum Cost Calculation: Iterate through the sorted workers. For each worker, add their quality to the priority queue and update the total quality sum. If the priority queue size exceeds k, remove the worker with the highest quality from the queue. The current wage ratio multiplied by the current total quality represents a potential minimum cost. Keep track of the overall minimum cost encountered.

• Runtime Complexity: O(n log n) due to sorting, and O(n log k) due to heap operations. Overall: O(n log n). Storage Complexity: O(n) to store ratios and worker indices and O(k) for the heap

Code

    import heapq

def mincostToHireWorkers(quality, wage, k):
    """
    Calculates the least amount of money needed to form a paid group of k workers.

    Args:
        quality: A list of integers representing the quality of each worker.
        wage: A list of integers representing the minimum wage expectation for each worker.
        k: The number of workers to hire.

    Returns:
        The least amount of money needed to form a paid group satisfying the conditions.
    """

    n = len(quality)
    ratios = [(wage[i] / quality[i], i) for i in range(n)]
    ratios.sort()

    heap = []  # Max heap to store qualities of the current k workers
    total_quality = 0
    min_cost = float('inf')

    for ratio, worker_index in ratios:
        qual = quality[worker_index]
        heapq.heappush(heap, -qual)  # Negate for max-heap
        total_quality += qual

        if len(heap) > k:
            total_quality += heapq.heappop(heap) #Remove the largest element (smallest when negated)

        if len(heap) == k:
            min_cost = min(min_cost, ratio * total_quality)

    return min_cost