Problem Description

You have n workers, each with a quality value and a minimum wage expectation. The goal is to hire exactly k workers while minimizing the total cost.

The hiring rules are:

1. Each hired worker must receive at least their minimum wage expectation
2. Within the hired group, pay must be proportional to quality (if worker A has twice the quality of worker B, then worker A must be paid exactly twice as much)

The key insight is that once you form a group, the pay ratio is determined by the worker with the highest wage-to-quality ratio in that group. This "captain" sets the wage rate for everyone else.

The solution sorts workers by their wage/quality ratio. For each potential captain (worker with highest ratio), it considers forming a group with that captain and the k-1 workers with lowest quality among those with equal or lower wage/quality ratios. This ensures all workers meet their minimum wage requirement.

The algorithm uses a max heap to track the k workers with smallest quality. For each captain position, it calculates the total cost as (captain's wage/quality ratio) × (sum of k workers' qualities). The heap helps efficiently maintain the k smallest qualities as we iterate through potential captains.

The time complexity is O(n log n) for sorting and heap operations, and space complexity is O(n) for storing the sorted array and heap.

Intuition

Let's think about what happens when we hire a group of workers. Since everyone's pay must be proportional to their quality, we can express each worker's pay as pay[i] = quality[i] × rate, where rate is the same for all workers in the group.

For each worker to meet their minimum wage expectation, we need quality[i] × rate ≥ wage[i], which means rate ≥ wage[i] / quality[i].

Here's the crucial observation: the rate for the entire group must be at least the maximum wage[i] / quality[i] ratio among all workers in the group. This worker with the highest ratio becomes the "bottleneck" - they determine the minimum rate everyone must be paid at.

So if we fix a particular rate (determined by some worker as the bottleneck), the total cost becomes rate × (sum of qualities of k workers). To minimize this cost for a fixed rate, we want to choose the k workers with the smallest qualities.

But which workers can we actually choose? We can only include workers whose wage/quality ratio is less than or equal to our chosen rate. Otherwise, they wouldn't meet their minimum wage expectation.

This leads us to the strategy:

1. Sort workers by their wage/quality ratio
2. Consider each worker as a potential "captain" (the one setting the rate)
3. For each captain, form a group with them and the k-1 workers with smallest qualities among those who appear before them in the sorted order (these have lower or equal ratios)
4. Calculate the cost for each such group and keep track of the minimum

The sorting ensures that when we consider a worker as captain, all previous workers in the sorted list are eligible to be in their group. We use a heap to efficiently maintain the k smallest qualities as we iterate through potential captains.

Learn more about Greedy, Sorting and Heap (Priority Queue) patterns.

Solution Approach

The implementation follows our intuition step by step:

1. Sort workers by wage/quality ratio: We create tuples of (quality, wage) and sort them by wage/quality ratio using a lambda function lambda x: x[1] / x[0]. This ensures workers are ordered from lowest to highest ratio.

2. Initialize variables:

   • ans = inf: Tracks the minimum cost found so far
   • tot = 0: Maintains the sum of qualities in our current group
   • h = []: A max heap (using negative values in Python's min heap) to store qualities

3. Iterate through sorted workers: For each worker (q, w) in sorted order:

   • Add their quality to the running total: tot += q
   • Push their negative quality to the heap: heappush(h, -q) (negative because Python only has min heap)

4. When we have k workers (len(h) == k):

   • Calculate cost with current worker as captain: w / q * tot
     • w / q is the captain's wage/quality ratio (the rate)
     • tot is the sum of k smallest qualities
     • Their product gives the total cost
   • Update minimum cost: ans = min(ans, w / q * tot)
   • Remove the worker with highest quality to make room for next iteration: tot += heappop(h)
     • We pop the largest quality (most negative value in heap)
     • Adding it back to tot effectively removes it from the sum

5. Return the minimum cost found across all valid groups.

The key insight is that as we iterate through workers in increasing wage/quality ratio order, each worker becomes the new captain (highest ratio in group). The heap maintains the k workers with smallest qualities that can be paired with this captain, ensuring we always compute the minimum possible cost for each captain choice.

Example Walkthrough

Let's walk through a small example with n = 4 workers and we want to hire k = 2:

• Worker 0: quality = 10, wage = 70 (ratio = 7.0)
• Worker 1: quality = 20, wage = 50 (ratio = 2.5)
• Worker 2: quality = 5, wage = 30 (ratio = 6.0)
• Worker 3: quality = 4, wage = 28 (ratio = 7.0)

Step 1: Sort by wage/quality ratio After sorting:

• Worker 1: quality = 20, wage = 50 (ratio = 2.5)
• Worker 2: quality = 5, wage = 30 (ratio = 6.0)
• Worker 0: quality = 10, wage = 70 (ratio = 7.0)
• Worker 3: quality = 4, wage = 28 (ratio = 7.0)

Step 2: Iterate through sorted workers

Iteration 1 (Worker 1 as potential captain):

• Add quality 20 to heap: h = [-20], tot = 20
• Heap size < k, continue

Iteration 2 (Worker 2 as captain):

• Add quality 5 to heap: h = [-20, -5], tot = 25
• Heap size = k = 2, calculate cost:
  • Captain's ratio: 30/5 = 6.0
  • Total qualities: 25
  • Cost = 6.0 × 25 = 150
• Update ans = 150
• Remove largest quality (20): tot = 25 - 20 = 5

Iteration 3 (Worker 0 as captain):

• Add quality 10 to heap: h = [-10, -5], tot = 15
• Heap size = k = 2, calculate cost:
  • Captain's ratio: 70/10 = 7.0
  • Total qualities: 15
  • Cost = 7.0 × 15 = 105
• Update ans = min(150, 105) = 105
• Remove largest quality (10): tot = 15 - 10 = 5

Iteration 4 (Worker 3 as captain):

• Add quality 4 to heap: h = [-5, -4], tot = 9
• Heap size = k = 2, calculate cost:
  • Captain's ratio: 28/4 = 7.0
  • Total qualities: 9
  • Cost = 7.0 × 9 = 63
• Update ans = min(105, 63) = 63

Final answer: 63

The optimal group consists of Workers 2 and 3, with Worker 3 as the captain setting the rate at 7.0. Worker 2 gets paid 5 × 7.0 = 35 (above their minimum of 30), and Worker 3 gets paid 4 × 7.0 = 28 (exactly their minimum).

Solution Implementation

Time and Space Complexity

Time Complexity: O(n log n) where n is the length of the quality/wage arrays.

• Sorting the workers by their wage-to-quality ratio takes O(n log n)
• The main loop iterates through all n workers
• Inside the loop, each heappush operation takes O(log k) where k is the size of the heap
• When the heap size reaches k, heappop also takes O(log k)
• Since k ≤ n, the heap operations are bounded by O(log n)
• Total: O(n log n) for sorting + O(n log n) for heap operations = O(n log n)

Space Complexity: O(n)

• The sorted list t stores all n worker pairs: O(n)
• The max heap h stores at most k elements: O(k) where k ≤ n
• Other variables (ans, tot, q, w) use constant space: O(1)
• Total: O(n) + O(k) = O(n) since k ≤ n

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Division Instead of Float Division

A critical pitfall is using integer division when calculating the wage/quality ratio, which can lead to incorrect sorting and wrong results.

Incorrect:

workers = sorted(zip(quality, wage), key=lambda worker: worker[1] // worker[0])  # Wrong!

Correct:

workers = sorted(zip(quality, wage), key=lambda worker: worker[1] / worker[0])

2. Forgetting to Use Negative Values for Max Heap

Python's heapq module only provides a min heap. To simulate a max heap, you must push negative values. Forgetting this will keep the workers with highest quality instead of lowest.

Incorrect:

heapq.heappush(max_heap, worker_quality)  # Wrong - keeps smallest values
removed_quality = heapq.heappop(max_heap)  # Wrong - removes smallest

Correct:

heapq.heappush(max_heap, -worker_quality)  # Push negative
removed_quality = -heapq.heappop(max_heap)  # Pop and negate back

3. Updating Total Quality Incorrectly After Popping

When removing a worker from the heap, you need to subtract their quality from the total, not add the negative value returned by heappop.

Incorrect:

total_quality += heapq.heappop(max_heap)  # Wrong - adds negative value

Correct:

removed_quality = -heapq.heappop(max_heap)
total_quality -= removed_quality

4. Calculating Cost Before Having k Workers

Computing the cost when you have fewer than k workers will give invalid results. Always check that the heap size equals k.

Incorrect:

for worker_quality, worker_wage in workers:
    total_quality += worker_quality
    heapq.heappush(max_heap, -worker_quality)
    # Wrong - calculating cost regardless of group size
    current_cost = (worker_wage / worker_quality) * total_quality
    min_cost = min(min_cost, current_cost)

Correct:

if len(max_heap) == k:
    current_cost = (worker_wage / worker_quality) * total_quality
    min_cost = min(min_cost, current_cost)

5. Not Handling Edge Cases

Ensure your solution handles edge cases like when k equals n (all workers must be hired) or when k equals 1.

Solution: The current implementation handles these naturally, but always verify with test cases:

# Edge case: k = 1 (hire cheapest worker)
# Edge case: k = n (must hire all workers)