[LeetCode] Total Cost to Hire K Workers

Problem: 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:

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.

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.

Constraints:

• 1 <= costs.length <= 105 
• 1 <= costs[i] <= 105
• 1 <= k, candidates <= costs.length

Approach: We can use two heaps / prioriy queues here. One to read from front and one to read from back. If front heap has an element at the top which less than or equal to back heap then we will pop from front heap add the cost in the final cost. We will add the next cost from the front into the front heap. 

If back heap has smaller element than front heap then we will pop from back heap, add it to final cost and then we will add next element from the back to the back heap.

We need to repeat it for k times. We also need to take care that we won't add same elements to both heaps. You can see that in the impelementation. Basicall whenever the nextfront index is greater than nextback index, that means all the elements are already added and we don't need to add the elements again.

That's all!

Implementation in C#:

    public long TotalCost(int[] costs, int k, int candidates)

    {

        int length = costs?.Length ?? 0;

        if (length < k)

        {

            return -1;

        }

        var frontPQ = new PriorityQueue<int, int>();

        var backPQ = new PriorityQueue<int, int>();

        int frontIndex = Math.Min(length, candidates);

        for (int i = 0; i < frontIndex; ++i)

        {

            frontPQ.Enqueue(costs[i], costs[i]);

        }

        int backIndex = Math.Max(frontIndex, length - candidates);

        for (int i = length - 1; i >= backIndex; --i)

        {

            backPQ.Enqueue(costs[i], costs[i]);

        }

        --backIndex;

        long finalCost = 0;

        for (int i = 0; i < k; ++i)

        {

            if (backPQ.Count == 0 ||

                (frontPQ.Count > 0 && frontPQ.Peek() <= backPQ.Peek()))

            {

                int currCost = frontPQ.Dequeue();

                finalCost += currCost;

                if (frontIndex <= backIndex && frontIndex < length)

                {

                    frontPQ.Enqueue(costs[frontIndex], costs[frontIndex]);

                    ++frontIndex;

                }

            }

            else

            {

                int currCost = backPQ.Dequeue();

                finalCost += currCost;

                if (frontIndex <= backIndex && backIndex >= 0)

                {

                    backPQ.Enqueue(costs[backIndex], costs[backIndex]);

                    --backIndex;

                }

            }

        }

        return finalCost;

    }

Complexity: O(candidates + k * log(candidates))