Class Solution
- java.lang.Object
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- g2401_2500.s2462_total_cost_to_hire_k_workers.Solution
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public class Solution extends Object
2462 - Total Cost to Hire K Workers.Medium
You are given a 0-indexed integer array
costswherecosts[i]is the cost of hiring theithworker.You are also given two integers
kandcandidates. We want to hire exactlykworkers according to the following rules:- You will run
ksessions and hire exactly one worker in each session. - In each hiring session, choose the worker with the lowest cost from either the first
candidatesworkers or the lastcandidatesworkers. Break the tie by the smallest index.- For example, if
costs = [3,2,7,7,1,2]andcandidates = 2, then in the first hiring session, we will choose the4thworker because they have the lowest cost[3,2,7,7, 1 ,2]. - In the second hiring session, we will choose
1stworker because they have the same lowest cost as4thworker but they have the smallest index[3, 2 ,7,7,2]. Please note that the indexing may be changed in the process.
- For example, if
- 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
kworkers.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.
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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.
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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.
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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.
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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.
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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.
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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.
Constraints:
1 <= costs.length <= 1051 <= costs[i] <= 1051 <= k, candidates <= costs.length
- You will run
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Constructor Summary
Constructors Constructor Description Solution()
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