Meeting Rooms III (K Machines, Earliest Available)

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Meeting Rooms III (K Machines, Earliest Available)

Edge cases:

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IDE

def assign_jobs(K: int, jobs: list[list[int]]) -> list[int]:public int[] assignJobs(int K, int[][] jobs)vector<int> assignJobs(int K, vector<vector<int>>& jobs)function assignJobs(K, jobs)

def assign_jobs(K, jobs):
    # Write your solution here
    pass

class Solution {
    public int[] assignJobs(int K, int[][] jobs) {
        // Write your solution here
        return new int[0];
    }
}

#include <vector>
using namespace std;

vector<int> assignJobs(int K, vector<vector<int>>& jobs) {
    // Write your solution here
    return {};
}

function assignJobs(K, jobs) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: [0, 0, 0]Assigning all jobs to machine 0 without considering machine availability or other machines.✅ Use min-heaps to track machine availability and assign jobs to earliest available machine, not always machine 0.

Wrong: [0, 1, 0]Assigning job to machine with smallest index ignoring machine availability times (greedy trap).✅ Compare machine availability times and assign to machine that frees earliest, breaking ties by smallest index.

Wrong: [0, 1, 2, 3]Not reusing machines when they become free exactly at job start time (boundary condition error).✅ Allow assignment if machine free time <= job start time, not strictly less.

Wrong: [0, 0, 0]Failing to handle zero duration jobs properly, causing incorrect machine availability updates.✅ Update machine free time to start + duration even if duration is zero to avoid infinite wait or overlap.

Wrong: Timeout or no outputUsing brute force O(n*K) approach for large inputs causing TLE.✅ Implement min-heap based approach with O(n log K) complexity to handle large inputs efficiently.

✓

Test Cases

Focus on handling minimal and boundary inputs correctly before tackling edge cases.

Understand common algorithmic pitfalls like greedy traps and exact boundary conditions.

Optimize your solution using appropriate data structures like min-heaps to handle large inputs efficiently.

t1_01basic

Input{"K":2,"jobs":[[0,5],[1,2],[5,3]]}

Expected[0,1,0]

⏱ Performance - must finish in 2000ms

Job 0 starts at 0 on machine 0. Job 1 starts at 1 on machine 1. Job 2 starts at 5; machine 0 is free at 5, so assign job 2 to machine 0.

💡 Consider tracking when each machine becomes free to assign jobs efficiently.

💡 Use two min-heaps: one for available machines by index, one for running jobs by finish time.

💡 Assign jobs by earliest available machine; if multiple free, pick smallest index.

Why it failed: Output incorrect: likely not updating machine availability correctly or not choosing smallest index machine when multiple are free. Fix by maintaining two min-heaps and always assign job to earliest available machine with smallest index.

✓ Correct assignment of jobs to machines with earliest availability and smallest index.

t1_02basic

Input{"K":3,"jobs":[[0,3],[1,9],[2,6],[3,1],[10,2]]}

Expected[0,1,2,0,0]

⏱ Performance - must finish in 2000ms

Jobs assigned to machines 0,1,2 initially. Job 3 starts at 3; machine 0 is free at 3, so assigned to 0. Job 4 starts at 10; machine 0 free at 4, assigned to 0.

💡 Track machine availability times and assign jobs accordingly.

💡 When multiple machines free at job start, pick smallest index.

💡 Update machine free time after assigning each job.

Why it failed: Wrong output due to ignoring machine index priority when multiple machines free simultaneously. Fix by using min-heap keyed by machine index for available machines.

✓ Correctly handles multiple machines and job assignments respecting earliest availability and machine index.

t2_01edge

Input{"K":1,"jobs":[[0,5],[1,3],[4,2]]}

Expected[0,0,0]

⏱ Performance - must finish in 2000ms

Only one machine; all jobs assigned sequentially to machine 0 regardless of overlap.

💡 Test behavior when only one machine is available.

💡 Ensure jobs are queued and assigned sequentially on the single machine.

💡 Machine availability time must be updated after each job.

Why it failed: Incorrect output due to not queuing jobs properly on single machine or ignoring machine availability updates. Fix by updating machine free time after each job.

✓ Correctly assigns all jobs sequentially to single machine.

t2_02edge

Input{"K":3,"jobs":[[0,2],[0,3],[0,1],[0,4]]}

Expected[0,1,2,0]

⏱ Performance - must finish in 2000ms

All jobs start at same time 0; assign to machines 0,1,2 in order; last job waits for machine 0 to free at 2, then assigned.

💡 Check handling of jobs with identical start times.

💡 Assign jobs to machines with smallest index first when multiple free.

💡 Update machine availability correctly for jobs starting simultaneously.

Why it failed: Wrong output due to ignoring machine index priority for jobs with same start time. Fix by using min-heap keyed by machine index for available machines.

✓ Correctly assigns jobs with same start time respecting machine index priority.

t2_03edge

Input{"K":5,"jobs":[]}

Expected[]

⏱ Performance - must finish in 2000ms

No jobs to assign; output is empty list.

💡 Test behavior with no jobs.

💡 Function should handle empty input gracefully.

💡 Return empty list when no jobs.

Why it failed: Function crashes or returns non-empty output on empty jobs input. Fix by checking for empty jobs list and returning empty result.

✓ Correctly returns empty list when no jobs.

t2_04edge

Input{"K":4,"jobs":[[0,0],[0,0],[0,0],[0,0]]}

Expected[0,1,2,3]

⏱ Performance - must finish in 2000ms

Jobs with zero duration assigned to machines 0,1,2,3; machines become free immediately after assignment.

💡 Test handling of zero duration jobs.

💡 Ensure machine availability updates correctly even if duration is zero.

💡 Machines should be immediately available after zero duration jobs.

Why it failed: Fails zero duration test by not updating machine availability or causing infinite wait. Fix by updating machine free time to start + duration even if duration is zero.

✓ Correctly handles zero duration jobs and machine availability updates.

t3_01corner

Input{"K":2,"jobs":[[0,5],[1,2],[2,3]]}

Expected[0,1,1]

⏱ Performance - must finish in 2000ms

Job 0 assigned to machine 0. Job 1 assigned to machine 1. Job 2 starts at 2; machine 1 is free at 3 (after job 1), machine 0 free at 5; assign job 2 to machine 1.

💡 Beware of greedy approach that assigns job 2 to machine 0 just because it has smaller index.

💡 Check machine availability times carefully before assignment.

💡 Assign job to machine that becomes free earliest, not just smallest index.

Why it failed: Fails greedy trap: assigning job 2 to machine 0 ignoring that machine 1 frees earlier. Fix by comparing machine availability times before assignment.

✓ Correctly assigns jobs respecting earliest machine availability, avoiding greedy trap.

t3_02corner

Input{"K":3,"jobs":[[0,2],[2,2],[4,2],[6,2]]}

Expected[0,0,0,0]

⏱ Performance - must finish in 2000ms

Jobs assigned sequentially to machine 0 as it becomes free exactly at job start times.

💡 Check boundary condition when machine frees exactly at job start time.

💡 Ensure machine is considered free if free time <= job start time.

💡 Update machine availability correctly to allow immediate reuse.

Why it failed: Fails exact boundary test by requiring machine free time < job start time instead of <=. Fix by allowing assignment when machine free time equals job start time.

✓ Correctly handles exact boundary condition for machine availability.

t3_03corner

Input{"K":3,"jobs":[[0,0],[0,0],[0,0]]}

Expected[0,1,2]

⏱ Performance - must finish in 2000ms

Jobs with zero duration assigned to machines 0,1,2; machines become free immediately after assignment.

💡 Test handling of zero duration jobs.

💡 Ensure machine availability updates correctly even if duration is zero.

💡 Machines should be immediately available after zero duration jobs.

Why it failed: Fails zero duration test by not updating machine availability or causing infinite wait. Fix by updating machine free time to start + duration even if duration is zero.

✓ Correctly handles zero duration jobs and machine availability updates.

t4_01performance

Input{"K":100000,"jobs":[]}

Expectednull

⏱ Performance - must finish in 2000ms

Large input with n=100000 jobs and K=100000 machines; solution must run in O(n log K) time within 2 seconds.

💡 Brute force O(n*K) approach will time out.

💡 Use two min-heaps to track available and running machines efficiently.

💡 Assign jobs by earliest available machine using heaps to achieve O(n log K).

Why it failed: TLE due to O(n*K) brute force approach. Fix by implementing min-heap based solution with O(n log K) complexity.

✓ Efficient O(n log K) solution passes performance test.

Practice

1. You are given k sorted linked lists and need to merge them into one sorted list efficiently. Which approach guarantees the optimal time complexity for this problem?

easy

A. Concatenate all lists into one and then sort the combined list using a comparison-based sort.

B. Use a greedy approach by repeatedly picking the smallest head node among all lists without any auxiliary data structure.

C. Apply a divide and conquer strategy by recursively merging pairs of lists until one remains.

D. Use dynamic programming to store intermediate merged results and build the final list.

Solution

1. Step 1: Understand the problem constraints

   We have k sorted lists and want to merge them efficiently.

2. Step 2: Analyze each approach's time complexity

   Greedy without auxiliary structures leads to O(kN) per extraction, concatenation then sorting is O(N log N), dynamic programming is not applicable here. Divide and conquer merges pairs in O(N log k) time, which is optimal.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Divide and conquer merges pairs efficiently [OK]

Hint: Divide and conquer merges pairs in O(N log k) time [OK]

Common Mistakes:

• Assuming greedy without heap is efficient
• Thinking sorting concatenated list is optimal
• Confusing DP applicability here

2. Given the following code to find the 7th super ugly number with primes [2,3,5], what is the returned value?

easy

A. 8

B. 9

C. 10

D. 12

Solution

1. Step 1: Trace initial heap and ugly list

   Start with ugly=[1], heap=[(2,0,0),(3,1,0),(5,2,0)]. Pop smallest val=2, append 2 -> ugly=[1,2]. Push (2*ugly[1])=4.

2. Step 2: Continue popping and pushing

   Sequence generated: 1,2,3,4,5,6,8,10,... The 7th super ugly number is 10.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Manual heap trace confirms 7th number is 10 [OK]

Hint: Trace heap pops and pushes carefully to avoid off-by-one errors [OK]

Common Mistakes:

• Off-by-one in counting ugly numbers
• Confusing index for ugly array
• Skipping duplicates incorrectly

3. Given the following Python code to find the nth ugly number, what is the output when nthUglyNumber(4) is called?

easy

A. 6

B. 5

C. 8

D. 4

Solution

1. Step 1: Trace heap pops and pushes for n=4

   Initial heap: [1]. Iteration 1 pops 1, pushes 2,3,5. Iteration 2 pops 2, pushes 4,6,10. Iteration 3 pops 3, pushes 6,9,15 (6 already in seen, skip). Iteration 4 pops 4, pushes 8,12,20.

2. Step 2: Identify the 4th popped ugly number

   The popped numbers in order are 1,2,3,4. The 4th ugly number popped is 4, so the output is 4.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   4 is the 4th ugly number [OK]

Hint: Trace heap pops carefully to find nth ugly number [OK]

Common Mistakes:

• Off-by-one error in loop iterations
• Confusing popped value with pushed values

4. Consider the following buggy code for finding the nth ugly number. Which line contains the subtle bug that causes incorrect results or performance issues?

medium

A. Line 3: heap initialization with [1]

B. Line 4: Missing initialization of the 'seen' set

C. Line 7: Loop runs n times popping from heap

D. Line 10: Multiplying ugly by factors 2, 3, and 5

Solution

1. Step 1: Identify missing duplicate check

   The 'seen' set is not initialized and the check to avoid pushing duplicates is missing, causing duplicates to be pushed repeatedly.

2. Step 2: Understand impact of missing 'seen'

   Without 'seen', the heap grows unnecessarily large with duplicates, causing performance degradation and possibly incorrect nth number due to repeated values.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Missing 'seen' set causes duplicates and performance issues [OK]

Hint: Always track seen elements to avoid duplicates in heap [OK]

Common Mistakes:

• Forgetting to initialize or use 'seen'
• Assuming heap automatically avoids duplicates

5. Suppose the problem is modified so that elements from nums2 can be reused multiple times in pairs (i.e., pairs can reuse the same element from nums2 multiple times). Which modification to the min-heap approach correctly handles this variant?

hard

A. Push all pairs (nums1[i] + nums2[j]) for all i, j into the heap initially and pop k times.

B. After popping a pair (i, j), push (i, j+1) and also push (i+1, j) if within bounds, allowing reuse of nums2 elements.

C. After popping a pair (i, j), push (i, j) again to allow reuse of nums2[j].

D. Modify the heap to store sums with counts of usage and only push new pairs when usage count is less than k.

Solution

1. Step 1: Understand reuse requirement

   Allowing reuse of nums2 elements means after using nums2[j], we can also try nums2[j+1] and nums2[j] again with next nums1 element.

2. Step 2: Modify heap push logic

   After popping (i, j), pushing (i, j+1) continues along nums2, and pushing (i+1, j) allows pairing next nums1 element with the same nums2[j], enabling reuse.

3. Step 3: Reject other options

   Push all pairs (nums1[i] + nums2[j]) for all i, j into the heap initially and pop k times. is brute force and inefficient. After popping a pair (i, j), push (i, j) again to allow reuse of nums2[j]. causes infinite loops by pushing same pair repeatedly. Modify the heap to store sums with counts of usage and only push new pairs when usage count is less than k. is unnecessarily complex and not standard.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   After popping a pair (i, j), push (i, j+1) and also push (i+1, j) if within bounds, allowing reuse of nums2 elements. correctly expands pairs to allow reuse without infinite loops [OK]

Hint: Push next nums1 with same nums2 to allow reuse [OK]

Common Mistakes:

• Pushing same pair repeatedly causing infinite loops
• Brute forcing all pairs initially
• Overcomplicating usage counts