{ "data": { "question": { "questionId": "2553", "questionFrontendId": "2462", "boundTopicId": null, "title": "Total Cost to Hire K Workers", "titleSlug": "total-cost-to-hire-k-workers", "content": "
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:
k
sessions and hire exactly one worker in each session.candidates
workers or the last candidates
workers. Break the tie by the smallest index.\n\tcosts = [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]
.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.Return the total cost to hire exactly k
workers.
\n
Example 1:
\n\n\nInput: costs = [17,12,10,2,7,2,11,20,8], k = 3, candidates = 4\nOutput: 11\nExplanation: We hire 3 workers in total. The total cost is initially 0.\n- 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.\n- 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.\n- 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.\nThe total hiring cost is 11.\n\n\n
Example 2:
\n\n\nInput: costs = [1,2,4,1], k = 3, candidates = 3\nOutput: 4\nExplanation: We hire 3 workers in total. The total cost is initially 0.\n- 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.\n- 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.\n- 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.\nThe total hiring cost is 4.\n\n\n
\n
Constraints:
\n\n1 <= costs.length <= 105
1 <= costs[i] <= 105
1 <= k, candidates <= costs.length
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