Minimum Cost to Hire K Workers

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Steps

setup

Calculate ratios and sort workers

Calculate each worker's wage-to-quality ratio and pair it with their quality. Then sort all workers by this ratio in ascending order.

💡 Sorting by ratio ensures we consider workers in order of increasing wage demands per quality unit, which is crucial for minimizing cost.

Line:

workers = sorted([(w/q, q) for q, w in zip(quality, wage)])

💡 Workers are now ordered by their cost efficiency, setting the stage for the greedy selection process.

setup

Initialize variables before iteration

Initialize an empty max-heap for qualities, sum_quality as 0, and min_cost as infinity before processing workers.

💡 These variables track the current group of workers, their total quality, and the best cost found so far.

Line:

qualities = []
sum_quality = 0
min_cost = float('inf')

💡 Setting up these variables is essential for the iterative process of selecting workers and calculating costs.

insert

Insert first worker's quality into heap

Insert the quality 20 of the first worker (ratio 2.5) into the max-heap and update sum_quality to 20.

💡 Adding the first worker starts building the candidate group for hiring.

Line:

bisect.insort(qualities, q)
sum_quality += q

💡 The heap now contains one quality, representing the current group of workers considered.

insert

Insert second worker's quality into heap

Insert the quality 5 of the second worker (ratio 6.0) into the max-heap and update sum_quality to 25.

💡 Adding the second worker expands the candidate group to size 2, which matches K.

Line:

bisect.insort(qualities, q)
sum_quality += q

💡 Heap now contains qualities [20, 5], with 20 as the largest quality at the root.

compare

Calculate cost with first two workers

Calculate cost as ratio 6.0 * sum_quality 25 = 150. Update min_cost from infinity to 150.

💡 With exactly K workers, we can compute a valid total cost to compare against the minimum.

Line:

if len(qualities) == K:
    cost = ratio * sum_quality
    if cost < min_cost:
        min_cost = cost

💡 This step shows how cost depends on the current ratio and sum of qualities.

insert

Insert third worker's quality into heap

Insert the quality 10 of the third worker (ratio 7.0) into the max-heap and update sum_quality to 35.

💡 Adding the third worker temporarily exceeds K, so we must remove the largest quality next.

Line:

bisect.insort(qualities, q)
sum_quality += q

💡 Heap now contains qualities [20, 10, 5], with 20 as the largest quality at the root.

extract_max

Remove largest quality to maintain size K

Remove the largest quality 20 from the heap to keep only K=2 workers. Update sum_quality to 15.

💡 Removing the largest quality reduces total quality sum, which helps minimize cost.

Line:

if len(qualities) > K:
    removed = qualities.pop()
    sum_quality -= removed

💡 This step shows how pruning the largest quality keeps the group size fixed and cost low.

compare

Calculate cost with workers 0 and 2

Calculate cost as ratio 7.0 * sum_quality 15 = 105. Since 105 < current min_cost 150, update min_cost to 105.

💡 Recalculate cost with the new ratio and sum after pruning to check for a better minimum.

Line:

if len(qualities) == K:
    cost = ratio * sum_quality
    if cost < min_cost:
        min_cost = cost

💡 Shows how a lower ratio can reduce total cost even with the same sum of qualities.

insert

Insert first worker's quality again (ratio 2.5)

Insert quality 20 of the first worker (ratio 2.5) again to consider next iteration, sum_quality becomes 35.

💡 Reconsidering workers as we iterate through sorted list to find minimal cost group.

Line:

bisect.insort(qualities, q)
sum_quality += q

💡 Heap grows again, preparing for pruning if size exceeds K.

extract_max

Remove largest quality to maintain size K again

Remove largest quality 20 to keep heap size at K=2, sum_quality back to 15.

💡 Maintaining heap size ensures cost calculations are valid for exactly K workers.

Line:

if len(qualities) > K:
    removed = qualities.pop()
    sum_quality -= removed

💡 Heap pruning is repeated to maintain minimal sum of qualities.

compare

Calculate cost again with workers 0 and 2

Calculate cost as ratio 6.0 * sum_quality 15 = 90. Since 90 < current min_cost 105, update min_cost to 90.

💡 Check if this cost improves the minimum; it does in this case.

Line:

if len(qualities) == K:
    cost = ratio * sum_quality
    if cost < min_cost:
        min_cost = cost

💡 Not every iteration improves the minimum cost; some are skipped, but this one improves it.

complete

End of iteration, return minimum cost

All workers processed. Return the minimum cost found, which is 90.

💡 The algorithm finishes by returning the best cost computed during iteration.

Line:

return min_cost

💡 The final answer is the minimal total cost to hire exactly K workers.

import bisect

def mincostToHireWorkers(quality, wage, K):
    # STEP 1
    workers = sorted([(w/q, q) for q, w in zip(quality, wage)])
    qualities = []
    sum_quality = 0
    min_cost = float('inf')
    for ratio, q in workers:
        # STEP 3,4,6,9 Insert quality
        bisect.insort(qualities, q)
        sum_quality += q
        # STEP 7,10 Remove largest if size > K
        if len(qualities) > K:
            removed = qualities.pop()  # largest quality
            sum_quality -= removed
        # STEP 5,8,11 Calculate cost if size == K
        if len(qualities) == K:
            cost = ratio * sum_quality
            if cost < min_cost:
                min_cost = cost
    # STEP 12 Return answer
    return min_cost

if __name__ == '__main__':
    quality = [10, 20, 5]
    wage = [70, 50, 30]
    K = 2
    print(mincostToHireWorkers(quality, wage, K))  # Expected: 90.0

📊

Minimum Cost to Hire K Workers - Watch the Algorithm Execute, Step by Step

Watching each insertion and removal in the heap clarifies how the algorithm balances quality sums and wage ratios to find the minimum cost, which is hard to grasp from code alone.

setupmax-heap

empty heap

setupmax-heap

empty heap

insertmax-heap

20

0

insertmax-heap

20

0

5

1

peekmax-heap

20

0

5

1

Added: 150

insertmax-heap

20

0

10

1

5

2

extract_maxmax-heap

10

0

5

1

peekmax-heap

10

0

5

1

Added: 105

insertmax-heap

20

0

10

1

5

2

extract_maxmax-heap

10

0

5

1

peekmax-heap

10

0

5

1

Added: 90

peekmax-heap

10

0

5

1

Added: 90

Key Takeaways

✓ Sorting workers by their wage-to-quality ratio is the foundation for the greedy approach.

This ordering ensures we consider cheaper workers first, which is not obvious without seeing the sorted list.

✓ Using a max-heap to keep track of qualities allows efficient removal of the largest quality to minimize total quality sum.

The heap structure is critical to maintaining the minimal sum of qualities for exactly K workers, which is hard to visualize from code alone.

✓ The minimum cost is updated only when the heap size reaches K, showing the importance of group size in cost calculation.

Understanding when and why the cost updates happen clarifies the algorithm's correctness and efficiency.

Practice

1. Consider the following Python code implementing a frequency stack using a max-heap with timestamp. After executing the sequence of operations below, what is the output of the last pop() call?

fs = FreqStack()
fs.push(1)
fs.push(2)
fs.push(1)
print(fs.pop())
print(fs.pop())

easy

A. 1 and then 1

B. 2 and then 1

C. 1 and then 2

D. 2 and then 2

Solution

1. Step 1: Trace push operations

   Push 1: freq[1]=1, time=1, heap push (-1, -1, 1) Push 2: freq[2]=1, time=2, heap push (-1, -2, 2) Push 1: freq[1]=2, time=3, heap push (-2, -3, 1)

2. Step 2: Trace pop operations

   First pop: heap pops (-2, -3, 1) -> returns 1, freq[1]=1 Second pop: heap now has (-1, -2, 2) and (-1, -1, 1). Max freq is 1, tie broken by time: -2 > -1, so pop (-1, -2, 2) returns 2.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Most frequent element 1 popped first, then next most recent with freq 1 is 2 [OK]

Hint: Most frequent and most recent element pops first [OK]

Common Mistakes:

• Ignoring timestamp order in heap
• Mixing up pop order when frequencies tie

2. The following code attempts to find the kth largest XOR coordinate value but contains a subtle bug. Identify the line with the bug.

medium

A. Line replacing min_heap root when val is greater than min_heap[0]

B. Line initializing prefix matrix with zeros

C. Line computing prefix[i][j] with XOR of prefix[i - 1][j], prefix[i][j - 1], and matrix[i - 1][j - 1]

D. Line pushing val into min_heap when heap size is less than k

Solution

1. Step 1: Examine prefix XOR formula

   The prefix XOR formula must XOR four terms: prefix[i-1][j], prefix[i][j-1], prefix[i-1][j-1], and matrix[i-1][j-1]. The code misses prefix[i-1][j-1].

2. Step 2: Understand impact of missing term

   Omitting prefix[i-1][j-1] causes incorrect prefix XOR values, leading to wrong kth largest results.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Correct prefix XOR requires all four terms XORed [OK]

Hint: Prefix XOR requires XOR of four terms, not three [OK]

Common Mistakes:

• Forgetting prefix[i-1][j-1] in XOR
• Off-by-one errors in indexing
• Misusing heap operations

3. What is the time complexity of the optimal bucket sort approach for finding the top k frequent elements in an array of size n, and why?

medium

A. O(n²) because nested loops iterate over frequencies and elements.

B. O(n log n) because sorting is involved in frequency ordering.

C. O(k log n) because a heap of size k is maintained during iteration.

D. O(n) because frequency counting and bucket placement are linear, and iterating buckets is at most n.

Solution

1. Step 1: Analyze frequency counting

   Counting frequencies requires one pass over n elements -> O(n).

2. Step 2: Analyze bucket creation and iteration

   Creating buckets and placing elements is O(n) since total elements equal n. Iterating buckets from max frequency down is also O(n) in total.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   All steps combined remain linear in n [OK]

Hint: Bucket sort avoids sorting unique keys explicitly [OK]

Common Mistakes:

• Confusing with sorting complexity
• Assuming heap operations dominate

4. Suppose you want to extend the Kth Largest in a Stream problem to support queries for the kth largest element where elements can be negative, zero, or positive, and the stream can contain duplicates. Which modification to the optimal min-heap approach is necessary to correctly handle this scenario?

hard

A. No modification needed; the min-heap approach already handles negative and duplicate values correctly.

B. Maintain a balanced binary search tree instead of a heap to handle duplicates efficiently.

C. Use a max-heap instead of a min-heap to handle negative numbers correctly.

D. Store elements as absolute values in the heap to avoid issues with negative numbers.

Solution

1. Step 1: Understand data range and duplicates

   Negative numbers and duplicates do not affect the heap property or kth largest logic since heap compares values normally.

2. Step 2: Evaluate heap suitability

   Min-heap of size k maintains the kth largest element regardless of sign or duplicates. No special handling is required.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Min-heap approach works correctly with negative and duplicate values [OK]

Hint: Heap comparisons handle negatives and duplicates naturally [OK]

Common Mistakes:

• Thinking max-heap needed for negatives
• Trying to store absolute values
• Replacing heap with BST unnecessarily

5. Suppose the problem changes: the matrix is no longer sorted by columns, only rows are sorted, and you still need to find the kth smallest element. Which approach is the most suitable to solve this variant efficiently?

hard

A. Use binary search on the value range with the same counting method as before.

B. Use a greedy approach picking the smallest element from each row repeatedly.

C. Flatten the matrix and sort all elements to find the kth smallest.

D. Use a min heap to merge the k sorted rows, extracting the kth smallest element.

Solution

1. Step 1: Understand changed constraints

   Only rows are sorted, columns are not sorted, so binary search counting on columns is invalid.

2. Step 2: Identify suitable approach

   Min heap merging k sorted lists (rows) efficiently finds kth smallest in O(k log n) time.

3. Step 3: Reject other approaches

   Binary search counting fails without column sorting; flattening is O(n² log n²); greedy is incorrect.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Min heap merge exploits row sorting without column sorting [OK]

Hint: Without column sorting, binary search counting fails [OK]

Common Mistakes:

• Trying binary search counting without column sorting
• Assuming greedy works