How do you solve Total Cost to Hire K Workers on LeetCode?

Total Cost to Hire K Workers (LeetCode #2462) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using heaps allows us to efficiently manage and retrieve the lowest cost workers.

What is the brute force solution for Total Cost to Hire K Workers?

The brute force approach for Total Cost to Hire K Workers is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will repeatedly scan the first and last segments of the costs array to find the minimum cost worker for each hiring session. This method is straightforward but inefficient as it requires multiple passes through the array. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Total Cost to Hire K Workers?

Total Cost to Hire K Workers uses the following concepts: Array, Two Pointers, Heap (Priority Queue), Simulation. The recommended patterns to study are: Heap (Priority Queue), Two Pointers.

What are the interview tips for Total Cost to Hire K Workers?

Always clarify the constraints and edge cases with the interviewer. Explain your thought process while coding to demonstrate your understanding. Consider the efficiency of your approach and be ready to discuss improvements.

What are common mistakes when solving Total Cost to Hire K Workers?

Not handling edge cases where the number of candidates is less than the number of workers. Forgetting to update the heaps correctly after hiring a worker.

#2462

Total Cost to Hire K Workers

Medium

ArrayTwo PointersHeap (Priority Queue)SimulationHeap (Priority Queue)Two Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal solution uses two min-heaps to efficiently track the lowest cost workers from both ends of the array. This allows us to quickly find and hire the cheapest worker in each session without scanning the entire array repeatedly.

⚙️

Algorithm

3 steps

1Step 1: Initialize two min-heaps, one for the left candidates and one for the right candidates.
2Step 2: Populate the heaps with the first 'candidates' workers from the left and the last 'candidates' workers from the right.
3Step 3: For each hiring session, compare the top of both heaps, hire the worker with the lowest cost, and update the total cost. If a worker is hired from one side, push the next worker from that side into the corresponding heap.

solution.py31 lines

1import heapq
2
3def totalCost(costs, k, candidates):
4    total_cost = 0
5    n = len(costs)
6    left_heap, right_heap = [], []
7    hired = [False] * n
8
9    for i in range(min(candidates, n)):
10        heapq.heappush(left_heap, (costs[i], i))
11    for i in range(max(0, n - candidates), n):
12        heapq.heappush(right_heap, (costs[i], i))
13
14    for _ in range(k):
15        left_cost = left_heap[0] if left_heap else (float('inf'), -1)
16        right_cost = right_heap[0] if right_heap else (float('inf'), -1)
17
18        if left_cost[0] <= right_cost[0]:
19            total_cost += left_cost[0]
20            hired[left_cost[1]] = True
21            heapq.heappop(left_heap)
22            if left_cost[1] + candidates < n:
23                heapq.heappush(left_heap, (costs[left_cost[1] + candidates], left_cost[1] + candidates))
24        else:
25            total_cost += right_cost[0]
26            hired[right_cost[1]] = True
27            heapq.heappop(right_heap)
28            if right_cost[1] - candidates >= 0:
29                heapq.heappush(right_heap, (costs[right_cost[1] - candidates], right_cost[1] - candidates))
30
31    return total_cost

ℹ

Complexity note: The time complexity is O(n log n) due to the use of heaps for managing the workers, where each insertion and extraction operation takes logarithmic time. The space complexity is O(n) because we store the workers in heaps.

1Using heaps allows us to efficiently manage and retrieve the lowest cost workers.
2Maintaining two heaps for both ends of the array helps in managing candidates effectively.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.