Minimum Cost to Hire K Workers

There are N workers. The i-th worker has a quality quality[i] and a minimum wage expectation wage[i]. Now you want to hire exactly K workers to form a paid group. When hiring a group of K workers, you must pay them according to the following rules:

1. Every worker in the paid group should be paid in the ratio of their quality compared to other workers in the paid group.
2. Every worker in the paid group must be paid at least their minimum wage expectation.

Return the least amount of money needed to form a paid group satisfying the above conditions.

Sample Input/Output:

Let’s consider a sample input:

quality = [10, 20, 5]
wage = [70, 50, 30]
K = 2

Expected Output:

105.0

Explanation:
For the given input, if we hire the first worker with quality 10 and wage expectation 70, then we must hire the third worker with quality 5 (since their wage/quality ratio is the smallest among all workers) to form a paid group of size K = 2. The total cost would be (70 + 30) * (5 / 10) = 105.0.

Time and Space Complexity:

The time complexity of the solution will be O(N log N), where N is the number of workers. This is because we sort the workers based on their wage/quality ratio.

The space complexity will be O(N) to store the sorted array and a priority queue.

C++ Solution:

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>

using namespace std;

double minCostToHireWorkers(vector<int>& quality, vector<int>& wage, int K) {
    int N = quality.size();
    vector<pair<double, int>> workers;

    // Calculate wage/quality ratio for each worker
    for (int i = 0; i < N; ++i) {
        workers.push_back({static_cast<double>(wage[i]) / quality[i], quality[i]});
    }

    // Sort workers by wage/quality ratio
    sort(workers.begin(), workers.end());

    priority_queue<int> maxHeap;
    int sumQuality = 0;
    double result = DBL_MAX;

    for (const auto& worker : workers) {
        sumQuality += worker.second;
        maxHeap.push(worker.second);

        if (maxHeap.size() > K) {
            sumQuality -= maxHeap.top();
            maxHeap.pop();
        }

        if (maxHeap.size() == K) {
            result = min(result, sumQuality * worker.first);
        }
    }

    return result;
}

int main() {
    vector<int> quality = {10, 20, 5};
    vector<int> wage = {70, 50, 30};
    int K = 2;

    double result = minCostToHireWorkers(quality, wage, K);

    cout << "Minimum Cost to Hire " << K << " Workers: " << result << endl;

    return 0;
}

In this implementation, we first calculate the wage/quality ratio for each worker and store it along with their quality in a pair. We then sort the workers based on their wage/quality ratio. We use a max-heap to keep track of the K workers with the highest quality. As we iterate through the sorted workers, we update the heap and calculate the total quality of the K workers. We calculate the cost based on the minimum wage/quality ratio among these K workers.

The main function demonstrates the usage of minCostToHireWorkers by calculating the minimum cost to hire K workers from the given input quality, wage, and K.

The output for the provided test case will be:

Minimum Cost to Hire 2 Workers: 105