2462. Total Cost to Hire K Workers

⭐️⭐️⭐️

# 題目敘述

You are given a 0-indexed integer array costs where costs[i] is the cost of hiring the ith worker.

You are also given two integers k and candidates . We want to hire exactly k workers according to the following rules:

• You will run k sessions and hire exactly one worker in each session.
• In each hiring session, choose the worker with the lowest cost from either the first candidates workers or the last candidates workers. Break the tie by the smallest index.
  • For example, if costs = [3,2,7,7,1,2] and candidates = 2 , then in the first hiring session, we will choose the 4th worker because they have the lowest cost [3,2,7,7,1,2] .
  • In the second hiring session, we will choose 1st worker because they have the same lowest cost as 4th worker but they have the smallest index [3,2,7,7,2] . Please note that the indexing may be changed in the process.
• If there are fewer than candidates workers remaining, choose the worker with the lowest cost among them. Break the tie by the smallest index.
• A worker can only be chosen once.

Return the total cost to hire exactly k workers.

# Example 1

Input: costs = [17,12,10,2,7,2,11,20,8], k = 3, candidates = 4
Output: 11
Explanation: We hire 3 workers in total. The total cost is initially 0.

• In the first hiring round we choose the worker from [17,12,10,2,7,2,11,20,8]. The lowest cost is 2, and we break the tie by the smallest index, which is 3. The total cost = 0 + 2 = 2.
• In the second hiring round we choose the worker from [17,12,10,7,2,11,20,8]. The lowest cost is 2 (index 4). The total cost = 2 + 2 = 4.
• In the third hiring round we choose the worker from [17,12,10,7,11,20,8]. The lowest cost is 7 (index 3). The total cost = 4 + 7 = 11. Notice that the worker with index 3 was common in the first and last four workers.
  The total hiring cost is 11.

# Example 2

Input: costs = [1,2,4,1], k = 3, candidates = 3
Output: 4
Explanation: We hire 3 workers in total. The total cost is initially 0.

• In the first hiring round we choose the worker from [1,2,4,1]. The lowest cost is 1, and we break the tie by the smallest index, which is 0. The total cost = 0 + 1 = 1. Notice that workers with index 1 and 2 are common in the first and last 3 workers.
• In the second hiring round we choose the worker from [2,4,1]. The lowest cost is 1 (index 2). The total cost = 1 + 1 = 2.
• In the third hiring round there are less than three candidates. We choose the worker from the remaining workers [2,4]. The lowest cost is 2 (index 0). The total cost = 2 + 2 = 4.
  The total hiring cost is 4.

# 解題思路

# Solution

import java.util.PriorityQueue;

class Solution {

    public long totalCost(int[] costs, int k, int candidates) {

        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> {

            if (a[0] == b[0]) {

                return a[1] - b[1];

            }

            return a[0] - b[0];

        });

        for (int i = 0; i < candidates; i++) {

            pq.offer(new int[] { costs[i], 0 });

        }

        for (int i = Math.max(candidates, costs.length - candidates); i < costs.length; i++) {

            pq.offer(new int[] { costs[i], 1 });

        }

        long answer = 0;

        int nextHead = candidates;

        int nextTail = costs.length - candidates - 1;

        for (int i = 0; i < k; i++) {

            int[] currWorker = pq.poll();

            int currCost = currWorker[0], currSectionId = currWorker[1];

            answer += currCost;

            if (nextHead <= nextTail) {

                if (currSectionId == 0) {

                    pq.offer(new int[] { costs[nextHead], 0 });

                    nextHead++;

                } else {

                    pq.offer(new int[] { costs[nextTail], 1 });

                    nextTail--;

                }

            }

        }

        return answer;

    }

}