857. Minimum Cost to Hire K Workers LeetCode Solution
In this guide, you will get 857. Minimum Cost to Hire K Workers LeetCode Solution with the best time and space complexity. The solution to Minimum Cost to Hire K Workers problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Minimum Cost to Hire K Workers
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.
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
Complexity Analysis
Time Complexity: O(\texttt{sort}) + n\log k)
Space Complexity: O(n)
857. Minimum Cost to Hire K Workers LeetCode Solution in C++
class Solution {
public:
double mincostToHireWorkers(vector<int>& quality, vector<int>& wage, int k) {
double ans = DBL_MAX;
int qualitySum = 0;
// (wagePerQuality, quality) sorted by wagePerQuality
vector<pair<double, int>> workers;
priority_queue<int> maxHeap;
for (int i = 0; i < quality.size(); ++i)
workers.emplace_back((double)wage[i] / quality[i], quality[i]);
ranges::sort(workers);
for (const auto& [wagePerQuality, q] : workers) {
maxHeap.push(q);
qualitySum += q;
if (maxHeap.size() > k)
qualitySum -= maxHeap.top(), maxHeap.pop();
if (maxHeap.size() == k)
ans = min(ans, qualitySum * wagePerQuality);
}
return ans;
}
}; /* code provided by PROGIEZ */
857. Minimum Cost to Hire K Workers LeetCode Solution in Java
class Solution {
public double mincostToHireWorkers(int[] quality, int[] wage, int k) {
double ans = Double.MAX_VALUE;
int qualitySum = 0;
// (wagePerQuality, quality) sorted by wagePerQuality
Pair<Double, Integer>[] workers = new Pair[quality.length];
Queue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
for (int i = 0; i < quality.length; ++i)
workers[i] = new Pair<>((double) wage[i] / quality[i], quality[i]);
Arrays.sort(workers, (a, b) -> Double.compare(a.getKey(), b.getKey()));
for (Pair<Double, Integer> worker : workers) {
final double wagePerQuality = worker.getKey();
final int q = worker.getValue();
maxHeap.offer(q);
qualitySum += q;
if (maxHeap.size() > k)
qualitySum -= maxHeap.poll();
if (maxHeap.size() == k)
ans = Math.min(ans, qualitySum * wagePerQuality);
}
return ans;
}
} // code provided by PROGIEZ
857. Minimum Cost to Hire K Workers LeetCode Solution in Python
class Solution:
def mincostToHireWorkers(
self,
quality: list[int],
wage: list[int],
k: int,
) -> float:
ans = math.inf
qualitySum = 0
# (wagePerQuality, quality) sorted by wagePerQuality
workers = sorted((w / q, q) for q, w in zip(quality, wage))
maxHeap = []
for wagePerQuality, q in workers:
heapq.heappush(maxHeap, -q)
qualitySum += q
if len(maxHeap) > k:
qualitySum += heapq.heappop(maxHeap)
if len(maxHeap) == k:
ans = min(ans, qualitySum * wagePerQuality)
return ans # code by PROGIEZ
