#2462

Medium

auto_awesomeTwo-pointer scanning with invariant tracking

LeetCode Problem Workspace

Total Cost to Hire K Workers

Optimize the total cost of hiring exactly k workers using a two-pointer approach with invariant tracking and priority queues.

array two pointers heap priority queue simulation

Problem Statement

You are given an array costs where costs[i] represents the cost of hiring the i-th worker. You are also given two integers, k and candidates, where k is the exact number of workers you need to hire, and candidates define the range of workers to consider for each selection round. The objective is to hire exactly k workers with the minimum total cost.

In each hiring round, you need to choose workers from both the leftmost and rightmost parts of the array. The selection process continues until exactly k workers have been hired. The problem asks you to return the total cost of hiring the k workers under these conditions.

Examples

Example 1

Input: costs = [17,12,10,2,7,2,11,20,8], k = 3, candidates = 4

Output: 11

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

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

Solution Approach

Two Minheaps for Efficient Worker Selection

Use two minheaps, one for the leftmost segment and another for the rightmost segment of the workers. In each hiring round, select the worker with the lowest cost from either heap and update the total cost. This approach minimizes the number of comparisons required when selecting workers.

Tracking Workers with Two Pointers

Utilize a two-pointer approach to track the leftmost and rightmost candidates. As workers are hired, the pointers move inward, adjusting the range of available workers in each round. This ensures that we consider both ends of the array and always select the worker with the lowest cost.

Efficient Cost Calculation Using Heaps

At each step, insert the workers' costs from the valid candidate range into the corresponding heaps. After selecting a worker, remove them from the heap, and repeat until k workers are selected. This ensures that the hiring process is efficient and minimizes the total cost.

Complexity Analysis

Metric	Value
Time	O((k + m) \cdot\log m)
Space	O(m)

The time complexity is O((k + m) * log m), where m is the number of candidates being considered in the heaps, and k is the number of workers to be hired. The space complexity is O(m) due to the storage of the candidates in the heaps.

What Interviewers Usually Probe

Candidate should understand the two-pointer approach and the use of minheaps in optimization problems.
Look for candidates who are comfortable with tracking invariant conditions across array segments.
Watch for the ability to efficiently calculate costs using heaps without unnecessary comparisons.

Common Pitfalls or Variants

Common pitfalls

Failing to maintain the invariant that the two pointers represent valid candidate ranges for selection.
Using only one heap and not considering the left and right parts of the array efficiently.
Not optimizing the heap operations, leading to suboptimal performance.

Follow-up variants

Consider different candidate ranges (not just two sides of the array) for more complex scenarios.
Use a single heap in certain cases to test if a simpler solution might be more efficient for smaller input sizes.
Introduce variations where the workers are ordered differently to test the efficiency of heap-based approaches.

FAQ

What is the primary algorithmic pattern for solving the Total Cost to Hire K Workers problem?

The primary algorithmic pattern for this problem is two-pointer scanning combined with invariant tracking and heap-based selection to minimize the total hiring cost.

How can heaps improve the efficiency of selecting workers in this problem?

Heaps allow for efficient extraction of the lowest-cost workers by maintaining a priority queue that ensures the smallest element is always accessible in O(log m) time.

What is the time complexity of the solution to the Total Cost to Hire K Workers problem?

The time complexity is O((k + m) * log m), where m is the number of candidates being considered, and k is the number of workers to hire.

Why is a two-pointer approach essential in this problem?

A two-pointer approach ensures that the workers are selected from both ends of the array, allowing for the optimal selection of workers with the lowest costs while maintaining efficient tracking of candidates.

What is the significance of the candidates parameter in this problem?

The candidates parameter defines the valid range of workers available for selection in each round. This range shrinks as workers are hired, ensuring that only the most relevant workers are considered.

terminal

Solution

Solution 1: Priority Queue (Min Heap)

First, we check if $candidates \times 2$ is greater than or equal to $n$. If it is, we directly return the sum of the costs of the first $k$ smallest workers.

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class Solution:
    def totalCost(self, costs: List[int], k: int, candidates: int) -> int:
        n = len(costs)
        if candidates * 2 >= n:
            return sum(sorted(costs)[:k])
        pq = []
        for i, c in enumerate(costs[:candidates]):
            heappush(pq, (c, i))
        for i in range(n - candidates, n):
            heappush(pq, (costs[i], i))
        heapify(pq)
        l, r = candidates, n - candidates - 1
        ans = 0
        for _ in range(k):
            c, i = heappop(pq)
            ans += c
            if l > r:
                continue
            if i < l:
                heappush(pq, (costs[l], l))
                l += 1
            else:
                heappush(pq, (costs[r], r))
                r -= 1
        return ans

Continue Practicing

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