Minimum Cost to Hire K Workers

Hard

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ArraysGreedy AlgorithmsHeap (Priority Queue)

There are n workers. You are given two integer arrays quality and wage where quality[i] is the quality of the i^th worker and wage[i] is the minimum wage expectation for the i^th 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 0^th worker and 35 to 2^nd 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 0^th worker, 13.33333 to 2^nd and 3^rd workers separately.

Constraints:

n == quality.length == wage.length
1 <= k <= n <= 10⁴
1 <= quality[i], wage[i] <= 10⁴

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the values of `wage`, `quality`, and `k`? Should I expect large numbers that could lead to integer overflow?
Can `quality` or `wage` be zero or negative? What should I return if `k` is greater than the number of workers?
If multiple sets of `k` workers result in the same minimum cost, is any one of them acceptable, or is there a specific tie-breaking condition?
Is there a specific data type required for the return value (e.g., integer, float, double)? Should the returned cost be rounded to the nearest integer/decimal place?
If no group of `k` workers can be hired while satisfying the wage/quality ratio for each worker, what should I return (e.g., -1, infinity, or an error)?

Brute Force Solution

Approach

The goal is to find the cheapest way to hire a certain number of workers. A brute force strategy means we explore every possible combination of workers and calculate the cost for each combination.

Here's how the algorithm would work step-by-step:

First, consider every possible group of workers that has the exact number of workers we need.
For each possible group, figure out the highest 'quality per pay' ratio within that group.
Using that ratio, calculate the total cost of hiring that specific group of workers.
Compare the total cost of hiring each possible group.
Finally, pick the group that has the lowest total cost to hire.

Code Implementation

def min_cost_to_hire_k_workers_brute_force(quality, wage, k):
    number_of_workers = len(quality)
    minimum_cost = float('inf')

    # Iterate through all possible combinations of workers
    for i in range(1 << number_of_workers):
        group = []
        for worker_index in range(number_of_workers):
            if (i >> worker_index) & 1:
                group.append(worker_index)

        if len(group) == k:

            #Find the maximum wage per quality ratio
            max_ratio = 0.0
            for worker_index in group:
                max_ratio = max(max_ratio, float(wage[worker_index]) / quality[worker_index])

            # Calculate the cost of this group
            current_cost = 0.0
            for worker_index in group:
                current_cost += quality[worker_index] * max_ratio

            # Update the minimum cost if necessary
            minimum_cost = min(minimum_cost, current_cost)

    return minimum_cost

Big(O) Analysis

Time Complexity

O(n choose k * k) – The dominant operation is generating all possible combinations of k workers from a pool of n workers, which has a complexity of 'n choose k', often written as nCk or (n k). For each combination, we iterate through its k elements to find the maximum quality/wage ratio. After we find the most costly combination of k workers, we need to iterate through each of the k workers in the combination. Thus, the algorithm's time complexity is dominated by generating these combinations, and then iterating through the k elements in each combination to perform operations like calculating the total wage. Therefore, the overall complexity is O(n choose k * k).

Space Complexity

O(1) – The provided plain English explanation outlines a brute force approach. While it describes iterating through combinations of workers and calculating costs, it doesn't explicitly mention any auxiliary data structures that grow with the input size N (where N is the total number of workers). The operations involve comparing costs and tracking the minimum, which can be done with a constant amount of extra space using variables to hold these values. Therefore, the auxiliary space complexity is constant, O(1).

Optimal Solution

Approach

The best way to solve this problem is to focus on finding the most cost-effective 'ratio' of wage to quality. We sort the workers by this ratio and then keep track of the *k* best workers in terms of quality to minimize the total cost.

Here's how the algorithm would work step-by-step:

Calculate the ratio of wage to quality for each worker.
Sort all workers based on their ratio, from smallest to largest.
Pick the first *k* workers from the sorted list. These are our initial 'best' workers.
Calculate the total cost of hiring these *k* workers. The cost for each worker is their ratio multiplied by the highest quality among the *k* workers.
Now, consider the next worker in the sorted list. This worker *could* be better than one of our current *k* workers.
If the current worker has a lower ratio than any of our existing *k* workers, or if the current worker's quality is higher than the lowest quality worker currently in the group, remove the lowest quality worker from the group and add the current worker to the group.
Recalculate the total cost with the new set of *k* workers. The cost is still determined by the highest quality in the group multiplied by the individual worker's ratio.
Repeat steps 5-7 for all remaining workers in the sorted list, each time potentially updating the group of *k* workers and the total cost.
Keep track of the minimum total cost found so far. The smallest total cost after considering all workers is the answer.

Code Implementation

import heapq

def minimum_cost_to_hire_k_workers(qualities, wages, k_workers):
    number_of_workers = len(qualities)
    ratios = [wages[i] / qualities[i] for i in range(number_of_workers)]
    workers = sorted(range(number_of_workers), key=lambda i: ratios[i])

    total_quality = 0
    quality_heap = []
    minimum_cost = float('inf')

    for worker_index in workers:
        worker_quality = qualities[worker_index]
        worker_ratio = ratios[worker_index]

        # Add quality to the heap and running total.
        heapq.heappush(quality_heap, -worker_quality)
        total_quality += worker_quality

        # If the heap size exceeds k, remove the lowest quality worker.
        if len(quality_heap) > k_workers:
            total_quality += heapq.heappop(quality_heap)

        # We have exactly k workers, so calculate cost.
        if len(quality_heap) == k_workers:
            minimum_cost = min(minimum_cost, worker_ratio * total_quality)

    return minimum_cost

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation is sorting the workers by their wage/quality ratio, which takes O(n log n) time. We also maintain a group of k workers, which involves operations like finding the lowest quality worker, potentially removing a worker and adding a new worker. Using a heap (priority queue) data structure of size k for this can take O(log k) for each worker processed. Since we process n workers after sorting, the heap operations contribute O(n log k) to the complexity. However, since sorting has a higher growth rate, the overall time complexity is dominated by the sorting step, making the total time complexity O(n log n).

Space Complexity

O(N) – The auxiliary space is dominated by the sorted list of workers. We create a list of size N, where N is the number of workers, to store the worker ratios. We also maintain a group of k workers; however, since k is at most N, this list of size k does not impact overall Big O. Therefore, the auxiliary space used scales linearly with the number of workers. No significant recursion occurs, so call stack space is negligible.

Edge Cases

Case	How to Handle
Empty quality or wage arrays	Return 0 if either array is empty as no workers can be hired.
K is greater than the number of workers (N)	Return -1 or throw an exception, as hiring more workers than available is impossible.
All workers have the same wage/quality ratio.	The result is the K smallest qualities multiplied by that ratio.
Quality or wage contains very large numbers causing potential overflow	Use long data types or check for potential overflows before multiplications and additions.
Quality or wage contains zeros	Handle division by zero when calculating wage/quality ratio, possibly returning a very large value to effectively ignore this worker.
K = 1	Return the minimum wage across all workers, since we only need to hire one.
Large N (number of workers) affecting heap performance	Ensure heap operations scale logarithmically to avoid time limit exceeded errors.
Multiple workers have same wage to quality ratio	The algorithm must select the K workers with lowest quality among the workers with smallest ratio.

Empty quality or wage arrays

How to Handle:

Return 0 if either array is empty as no workers can be hired.

K is greater than the number of workers (N)

How to Handle:

Return -1 or throw an exception, as hiring more workers than available is impossible.

All workers have the same wage/quality ratio.

How to Handle:

The result is the K smallest qualities multiplied by that ratio.

Quality or wage contains very large numbers causing potential overflow

How to Handle:

Use long data types or check for potential overflows before multiplications and additions.

Quality or wage contains zeros

How to Handle:

Handle division by zero when calculating wage/quality ratio, possibly returning a very large value to effectively ignore this worker.

K = 1

How to Handle:

Return the minimum wage across all workers, since we only need to hire one.

Large N (number of workers) affecting heap performance

How to Handle:

Ensure heap operations scale logarithmically to avoid time limit exceeded errors.

Multiple workers have same wage to quality ratio

How to Handle:

The algorithm must select the K workers with lowest quality among the workers with smallest ratio.