Minimum Cost to Hire K Workers

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Problem

Imagine you are managing a project and need to hire exactly K workers from a pool, each demanding a minimum wage and having a quality rating. You want to pay them fairly based on their quality but minimize the total cost.

You have n workers. Each worker has a quality and a minimum wage expectation. You want to hire exactly K workers to form a group. To pay them fairly, every worker in the group must be paid at least their minimum wage, and the pay must be proportional to their quality (i.e., if a worker has twice the quality, they get twice the pay). Find the minimum total wage to hire exactly K workers under these conditions.

1 ≤ n ≤ 10^51 ≤ K ≤ n1 ≤ quality[i], wage[i] ≤ 10^4

Edge cases: All workers have the same quality and wage → output is K times the minimum wageK equals n → must hire all workers, sum all wages proportionallyOne worker has extremely high wage but low quality → avoid if possible

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IDE

def mincostToHireWorkers(quality: list[int], wage: list[int], K: int) -> float:public double mincostToHireWorkers(int[] quality, int[] wage, int K)double mincostToHireWorkers(vector<int>& quality, vector<int>& wage, int K)function mincostToHireWorkers(quality, wage, K)

def mincostToHireWorkers(quality, wage, K):
    # Write your solution here
    pass

class Solution {
    public double mincostToHireWorkers(int[] quality, int[] wage, int K) {
        // Write your solution here
        return 0.0;
    }
}

#include <vector>
using namespace std;

double mincostToHireWorkers(vector<int>& quality, vector<int>& wage, int K) {
    // Write your solution here
    return 0.0;
}

function mincostToHireWorkers(quality, wage, K) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: NaN or crashNo handling of empty input or K=0, causing division by zero or index errors.✅ Add base case: if K == 0 or input arrays empty, return 0.0 immediately.

Wrong: Sum of wages without ratio adjustmentIgnoring wage-to-quality ratio and paying only minimum wages directly.✅ Calculate ratio = wage[i]/quality[i] and pay all workers at max ratio in chosen group.

Wrong: Including high ratio workers leading to higher costGreedy approach picking workers with lowest wages ignoring ratio constraints.✅ Sort by ratio and maintain heap of K smallest qualities to minimize total cost.

Wrong: Incorrect cost when all workers have identical ratioNot handling equal ratio case properly, leading to wrong subset or cost.✅ When ratios are equal, pick K workers with smallest quality sum and multiply by ratio.

Wrong: Timeout or TLEUsing brute force subset enumeration instead of sorting and heap approach.✅ Implement O(n log n) solution using sorting by ratio and max-heap for qualities.

✓

Test Cases

Focus on handling edge cases like empty input and single worker scenarios.

Watch out for common algorithmic traps such as greedy selection ignoring ratio constraints.

Optimize your solution to use sorting and heap to handle large inputs efficiently.

t1_01basic

Input{"quality":[10,20,5],"wage":[70,50,30],"K":2}

Expected105

⏱ Performance - must finish in 2000ms

Hire workers 0 and 2. Pay ratio is max(wage[i]/quality[i]) = max(70/10, 30/5) = 7. Total cost = 7 * (10 + 5) = 105.

💡 Sort workers by their wage-to-quality ratio to find candidates.

💡 Use a min-heap to keep track of the K workers with the smallest total quality.

💡 Calculate cost as ratio * sum of qualities for the chosen K workers.

Why it failed: Incorrect output means ratio calculation or heap maintenance is wrong. Fix by ensuring ratio is max(wage[i]/quality[i]) among chosen workers and sum quality is correct.

✓ Correctly computes minimum cost using ratio sorting and heap.

t1_02basic

Input{"quality":[3,1,10,10,1],"wage":[4,8,2,2,7],"K":3}

Expected30.666666666666668

⏱ Performance - must finish in 2000ms

Hire workers with qualities [3,1,1] and wages [4,8,7]. Ratio max is 7/1=7. Total quality sum=5. Cost=7*5=35 is too high. Better is workers with qualities [3,10,10] and wages [4,2,2]. Ratio max=4/3=1.3333, sum quality=23, cost=30.6667 minimum.

💡 Check all workers' wage-to-quality ratios and sort ascending.

💡 Maintain a heap of K smallest qualities while iterating ratios.

💡 Update minimum cost as ratio * sum of qualities when heap size is K.

Why it failed: Failing this test means heap or ratio update logic is incorrect. Fix by correctly updating heap and cost at each ratio step.

✓ Handles multiple workers and finds minimal cost correctly.

t2_01edge

Input{"quality":[],"wage":[],"K":0}

Expected0

⏱ Performance - must finish in 2000ms

No workers to hire, cost is zero.

💡 Check for empty input arrays and K=0.

💡 Return 0 immediately if no workers are needed.

💡 Avoid processing when input arrays are empty.

Why it failed: Code crashes or returns non-zero for empty input. Fix by adding base case for empty input or K=0.

✓ Correctly handles empty input and zero K.

t2_02edge

Input{"quality":[5],"wage":[10],"K":1}

Expected10

⏱ Performance - must finish in 2000ms

Only one worker, must pay at least their wage, cost is 10.

💡 Single element input means cost equals that worker's wage.

💡 Ratio is wage/quality, but with one worker cost is wage.

💡 Return wage[0] as minimum cost when K=1 and single worker.

Why it failed: Fails single element test if ratio or cost calculation is incorrect. Fix by handling single worker case correctly.

✓ Correctly returns wage for single worker.

t2_03edge

Input{"quality":[10,10,10],"wage":[20,20,20],"K":2}

Expected40

⏱ Performance - must finish in 2000ms

All workers have same quality and wage, cost is K times minimum wage = 2 * 20 = 40.

💡 When all ratios are equal, cost is ratio * sum of qualities.

💡 Sum qualities of any K workers since all equal.

💡 Cost = ratio * K * quality of one worker.

Why it failed: Fails when identical ratios cause wrong cost calculation. Fix by handling equal ratio case correctly.

✓ Correctly computes cost when all workers have identical ratio.

t3_01corner

Input{"quality":[1,2,3,4,5],"wage":[10,10,10,10,100],"K":3}

Expected50

⏱ Performance - must finish in 2000ms

Avoid hiring worker with high wage to quality ratio (worker 4). Best group is workers 0,1,3 with ratio 10/1=10 and sum quality 6, cost=50.

💡 Beware of greedy approach picking lowest wages only.

💡 Sort by ratio and maintain heap of K smallest qualities.

💡 Exclude workers with high ratio to minimize cost.

Why it failed: Fails by including high ratio worker leading to higher cost. Fix by sorting by ratio and using heap to track minimal quality sum.

✓ Correctly avoids high ratio workers to minimize cost.

t3_02corner

Input{"quality":[4,4,4,4],"wage":[8,8,8,8],"K":4}

Expected32

⏱ Performance - must finish in 2000ms

All workers have same ratio 2.0, must hire all, cost = ratio * sum quality = 2 * 16 = 32.

💡 Check boundary case where K equals n.

💡 Sum all qualities and multiply by common ratio.

💡 No subset choice, must pay all workers.

Why it failed: Fails when K equals n by not summing all qualities or miscalculating ratio. Fix by handling K=n case explicitly.

✓ Correctly handles K equals n boundary case.

t3_03corner

Input{"quality":[1,2,3],"wage":[3,6,9],"K":2}

Expected6

⏱ Performance - must finish in 2000ms

Workers have identical wage/quality ratio 3. Hiring any two workers costs 3 * sum of their qualities. Minimum sum quality for 2 workers is 2 (1+1), but qualities are 1 and 2, sum=3, cost=3*2=6.

💡 Check for identical ratios and pick lowest sum quality subset.

💡 Sort by ratio and use heap to maintain smallest qualities.

💡 Calculate cost as ratio * sum of chosen qualities.

Why it failed: Fails when identical ratios cause incorrect subset selection. Fix by always choosing smallest sum quality subset.

✓ Correctly handles identical ratio workers and minimal quality sum.

t4_01performance

Input{"quality":[10000,9999,9998,9997,9996,9995,9994,9993,9992,9991,9990,9989,9988,9987,9986,9985,9984,9983,9982,9981,9980,9979,9978,9977,9976,9975,9974,9973,9972,9971,9970,9969,9968,9967,9966,9965,9964,9963,9962,9961,9960,9959,9958,9957,9956,9955,9954,9953,9952,9951,9950,9949,9948,9947,9946,9945,9944,9943,9942,9941,9940,9939,9938,9937,9936,9935,9934,9933,9932,9931,9930,9929,9928,9927,9926,9925,9924,9923,9922,9921,9920,9919,9918,9917,9916,9915,9914,9913,9912,9911,9910,9909,9908,9907,9906,9905,9904,9903,9902,9901,9900],"wage":[10000,9999,9998,9997,9996,9995,9994,9993,9992,9991,9990,9989,9988,9987,9986,9985,9984,9983,9982,9981,9980,9979,9978,9977,9976,9975,9974,9973,9972,9971,9970,9969,9968,9967,9966,9965,9964,9963,9962,9961,9960,9959,9958,9957,9956,9955,9954,9953,9952,9951,9950,9949,9948,9947,9946,9945,9944,9943,9942,9941,9940,9939,9938,9937,9936,9935,9934,9933,9932,9931,9930,9929,9928,9927,9926,9925,9924,9923,9922,9921,9920,9919,9918,9917,9916,9915,9914,9913,9912,9911,9910,9909,9908,9907,9906,9905,9904,9903,9902,9901,9900],"K":50}

Expectednull

⏱ Performance - must finish in 2000ms

Large input with n=100, K=50. Algorithm must run in O(n log n) time to avoid TLE.

💡 Sorting and heap operations are O(n log n).

💡 Avoid brute force subsets which are exponential.

💡 Use a max-heap to maintain K smallest qualities efficiently.

Why it failed: TLE occurs if brute force or inefficient approach used. Fix by implementing sorting by ratio and heap-based solution.

✓ Algorithm runs within time limits using O(n log n) approach.

Practice

1. Given the following Python code implementing Quickselect, what is the output of findKthLargest([4,1,3,2], 3)?

easy

A. 2

B. 3

C. 1

D. 4

Solution

1. Step 1: Understand k and indexing

   k=3 means the 3rd largest element, so index k-1=2 in zero-based indexing.

2. Step 2: Trace partition and quickselect

   Partition rearranges so elements > pivot are left; after recursive calls, the element at index 2 is 2.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   3rd largest in [4,1,3,2] is 2 [OK]

Hint: Remember zero-based index k-1 for kth largest [OK]

Common Mistakes:

• Off-by-one error in k indexing
• Confusing kth largest with kth smallest
• Misunderstanding partition direction

2. Consider the following Python code that finds the maximum CPU load from a list of jobs. What is the value of max_load after processing all events for the input jobs = [[1,4,3], [2,5,4], [7,9,6]]?

def find_max_cpu_load(jobs):
    events = []
    for start, end, load in jobs:
        events.append((start, load))
        events.append((end, -load))
    events.sort(key=lambda x: x[0])

    current_load = 0
    max_load = 0
    for _, load_change in events:
        current_load += load_change
        max_load = max(max_load, current_load)
    return max_load

jobs = [[1,4,3], [2,5,4], [7,9,6]]
print(find_max_cpu_load(jobs))

easy

A. 7

B. 6

C. 9

D. 4

Solution

1. Step 1: List all events sorted by time:

   Events: (1,+3), (2,+4), (4,-3), (5,-4), (7,+6), (9,-6)

2. Step 2: Trace current_load and max_load through events:

   At 1: current_load=3, max_load=3; at 2: current_load=7, max_load=7; at 4: current_load=4, max_load=7; at 5: current_load=0, max_load=7; at 7: current_load=6, max_load=7; at 9: current_load=0, max_load=7.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Maximum overlap load is 7 at time 2 [OK]

Hint: Sum loads at event times, track max load [OK]

Common Mistakes:

• Off-by-one in event processing
• Ignoring negative load at job end

3. Consider the following Python code implementing the optimal approach for top K frequent words. Given the input words = ["apple", "banana", "apple", "orange", "banana", "banana"] and k = 2, what is the returned list?

easy

A. ["banana", "apple"]

B. ["banana", "orange"]

C. ["apple", "banana"]

D. ["apple", "orange"]

Solution

1. Step 1: Count frequencies

   apple: 2, banana: 3, orange: 1; max_freq = 3

2. Step 2: Fill buckets and sort within buckets

   buckets[3] = ["banana"], buckets[2] = ["apple"], buckets[1] = ["orange"]; iterate from freq=3 down

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Top 2 frequent words are banana (3), apple (2) [OK]

Hint: Trace frequency buckets from highest to lowest [OK]

Common Mistakes:

• Confusing lex order with frequency order
• Off-by-one in bucket indices
• Returning words in ascending frequency order

4. What is the time complexity of the heap-based approach to find the k-th smallest pair distance in an array of size n?

medium

A. O(n log n + k log n)

B. O(n^2 log n)

C. O(n log n + k log k)

D. O(n log n + k n)

Solution

1. Step 1: Analyze sorting and heap initialization

   Sorting takes O(n log n). Initial heap push for n-1 pairs is O(n log n) but dominated by sorting.

2. Step 2: Analyze k iterations of heap operations

   Each pop and push operation costs O(log n), repeated k times -> O(k log n).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Heap size bounded by n, so log n factor dominates [OK]

Hint: Heap operations cost log n, not log k or n^2 [OK]

Common Mistakes:

• Confusing k log k with k log n
• Assuming O(n^2 log n) due to pairs
• Ignoring heap size limits

5. The following code attempts to implement the optimal bucket sort approach for top k frequent elements. Identify the line that contains a subtle bug that can cause incorrect results or runtime errors.

medium

A. Line updating frequency dictionary with freq.get(num, 0) + 1

B. Line iterating over freq.items() to append elements to buckets

C. Line creating buckets with size max_freq instead of max_freq + 1

D. Line iterating buckets from max_freq - 1 down to 0

Solution

1. Step 1: Check bucket list size

   Buckets are created with size max_freq, but frequencies range from 1 to max_freq inclusive, so index max_freq is needed.

2. Step 2: Consequence of incorrect bucket size

   Accessing buckets[count] where count == max_freq causes IndexError or missing elements if bucket list is too small.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Bucket list must have max_freq + 1 slots to index frequencies correctly [OK]

Hint: Bucket size must be max_freq + 1 to include max frequency [OK]

Common Mistakes:

• Off-by-one bucket size
• Misindexing buckets causing runtime errors