2462. Total Cost to Hire K Workers

Array Two Pointers Heap (Priority Queue)Simulation

Explanation

To solve this problem, we can follow these steps:

Create a list of workers sorted by cost in ascending order.
Iterate through each worker, keeping track of the lowest cost for each session.
Use a priority queue to maintain the candidates for each session and update the total cost.
Return the total cost after hiring k workers.

Time complexity: O(n log n) where n is the number of workers
Space complexity: O(n) for the priority queue

import java.util.Arrays;
import java.util.PriorityQueue;

class Solution {
    public int minCostToHireWorkers(int[] costs, int k, int candidates) {
        int n = costs.length;
        int[][] workers = new int[n][2];
        for (int i = 0; i < n; i++) {
            workers[i] = new int[]{costs[i], i};
        }
        Arrays.sort(workers, (a, b) -> a[0] - b[0]);

        PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
        int totalCost = Integer.MAX_VALUE;
        int sum = 0;
        for (int[] worker : workers) {
            sum += worker[0];
            pq.offer(worker[0]);
            if (pq.size() > candidates) {
                sum -= pq.poll();
            }
            if (pq.size() == candidates) {
                totalCost = Math.min(totalCost, sum);
            }
        }
        return totalCost;
    }
}

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