Meeting Rooms III (K Machines, Earliest Available)

▶

Steps

setup

Sort jobs by start time

Jobs are sorted by their start time with their original indices preserved. This ensures jobs are processed in chronological order.

💡 Sorting jobs by start time is crucial to assign machines in the correct order and handle overlapping jobs properly.

Line:

indexed_jobs = sorted(enumerate(jobs), key=lambda x: x[1][0])

💡 Sorting prepares the job list for sequential processing by start time.

setup

Initialize available machines heap

Initialize a min-heap with all machine indices [0, 1], representing machines currently available.

💡 The available heap lets us quickly find the machine with the smallest index that is free.

Line:

available = list(range(K))
heapq.heapify(available)

💡 Heapifying the available machines allows efficient extraction of the smallest machine index.

setup

Initialize running machines heap as empty

Create an empty min-heap to track running machines by their finish times.

💡 The running heap tracks when machines become free, enabling reuse.

Line:

running = []

💡 Starting with an empty running heap means no machines are busy yet.

traverse

Process first job (index 0, start=0, duration=5)

Check running machines for any that finished before job start (none). Assign machine 0 from available heap to job 0.

💡 No machines are busy yet, so the first job gets the smallest available machine immediately.

Line:

while running and running[0][0] <= start:
    _, machine = heapq.heappop(running)
    heapq.heappush(available, machine)
machine = heapq.heappop(available)

💡 The first job uses machine 0, showing how available machines are assigned in order.

insert

Add machine 0 to running heap with finish time 5

Push (5, 0) into running heap to mark machine 0 busy until time 5.

💡 Tracking finish times in running heap allows machines to be freed later.

Line:

finish_time = start + duration
heapq.heappush(running, (finish_time, machine))

💡 Running heap now contains the earliest finishing machine with its finish time.

traverse

Process second job (index 1, start=1, duration=2)

No machines finished before start=1 (running heap root finish=5). Assign machine 1 from available heap to job 1.

💡 Machine 0 is busy, so the next smallest available machine 1 is assigned.

Line:

while running and running[0][0] <= start:
    _, machine = heapq.heappop(running)
    heapq.heappush(available, machine)
machine = heapq.heappop(available)

💡 Shows how the algorithm waits for machines to become free before reusing them.

insert

Add machine 1 to running heap with finish time 3

Push (3, 1) into running heap to mark machine 1 busy until time 3.

💡 Running heap now tracks two machines with their finish times.

Line:

finish_time = start + duration
heapq.heappush(running, (finish_time, machine))

💡 Running heap maintains order by finish time: (3,1) is earlier than (5,0).

traverse

Process third job (index 2, start=5, duration=3)

Pop running machines that finished by start=5: (3,1) and (5,0). Push machines 1 and 0 back to available heap.

💡 Machines finishing before or at job start become available again.

Line:

while running and running[0][0] <= start:
    _, machine = heapq.heappop(running)
    heapq.heappush(available, machine)

💡 Shows how machines become free and ready for reassignment.

traverse

Assign machine 0 to third job

Pop machine 0 from available heap (smallest index) and assign it to job 2.

💡 The earliest available machine with smallest index is assigned to the job.

Line:

machine = heapq.heappop(available)

💡 Machine reuse is efficient and respects machine index priority.

insert

Add machine 0 to running heap with finish time 8

Push (8, 0) into running heap to mark machine 0 busy until time 8.

💡 Running heap tracks the new finish time for machine 0.

Line:

finish_time = start + duration
heapq.heappush(running, (finish_time, machine))

💡 Running heap now contains machine 0 finishing at 8 and machine 1 available.

reconstruct

All jobs assigned, final result array ready

The result array records machine assignments for each job in original order: [0, 1, 0].

💡 The final assignments show how jobs were scheduled on machines.

Line:

return result

💡 The algorithm successfully assigned jobs to earliest available machines respecting start times.

import heapq
from typing import List

def assign_jobs_optimal(K: int, jobs: List[List[int]]) -> List[int]:
    n = len(jobs)
    # STEP 1: Sort jobs by start time
    indexed_jobs = sorted(enumerate(jobs), key=lambda x: x[1][0])

    # STEP 2: Initialize available machines heap
    available = list(range(K))
    heapq.heapify(available)

    # STEP 3: Initialize running machines heap
    running = []
    result = [0] * n

    for idx, (start, duration) in indexed_jobs:
        # STEP 4: Free machines that finished before current job start
        while running and running[0][0] <= start:
            _, machine = heapq.heappop(running)
            heapq.heappush(available, machine)

        # STEP 5: Assign machine with smallest index
        machine = heapq.heappop(available)

        # STEP 6: Add machine to running heap with updated finish time
        finish_time = start + duration
        heapq.heappush(running, (finish_time, machine))

        # STEP 7: Record assignment
        result[idx] = machine

    # STEP 11: Return final assignments
    return result

if __name__ == '__main__':
    K = 2
    jobs = [[0,5],[1,2],[5,3]]
    print(assign_jobs_optimal(K, jobs))  # Output: [0, 1, 0]

📊

Meeting Rooms III (K Machines, Earliest Available) - Watch the Algorithm Execute, Step by Step

Watching the heaps update step-by-step reveals how the algorithm efficiently assigns jobs to the earliest available machine, a key insight that is hard to grasp from code alone.

peekmin-heap

0

0

1

1

heapifymin-heap

0

0

1

1

peekmin-heap

empty heap

extract_minmin-heap

1

0

Added: 0

insertmin-heap

5,0

0

extract_minmin-heap

empty heap

Added: 1

insertmin-heapswap [1↔0]

3,1

0

5,0

1

heapifymin-heap

0

0

1

1

extract_minmin-heap

1

0

Added: 0

insertmin-heap

8,0

0

peekmin-heap

8,0

0

Added: [0,1,0]

Key Takeaways

✓ Using two min-heaps allows efficient tracking of available machines and running jobs by finish time.

This insight is hard to see from code alone because the interplay of two heaps is subtle and dynamic.

✓ Jobs are assigned to the machine with the smallest index that is free at the job's start time.

Visualizing heap operations clarifies how machine indices prioritize assignments.

✓ Machines become available exactly when their running job finishes, enabling immediate reuse for subsequent jobs.

The trace shows how popping from the running heap frees machines, a key step often overlooked in static code.

Practice

1. You are given K sorted arrays and need to find the median of all elements combined without fully merging them. Which algorithmic approach guarantees an optimal solution in terms of time complexity for large inputs?

easy

A. Merge all arrays fully into one sorted array and then pick the median.

B. Use binary search on the value range combined with counting elements ≤ mid across arrays.

C. Apply a greedy approach by repeatedly picking the smallest current element from arrays until median is found.

D. Use dynamic programming to build a table of medians for subarrays and combine results.

Solution

1. Step 1: Understand problem constraints

   The problem requires finding the median without merging all arrays fully, which is costly for large inputs.

2. Step 2: Identify optimal approach

   Binary search on the value range combined with counting elements ≤ mid leverages sorted arrays and avoids full merge, achieving better time complexity.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Binary search on value range is known optimal for this problem [OK]

Hint: Binary search on value range beats full merge [OK]

Common Mistakes:

• Assuming full merge is always optimal
• Confusing greedy with binary search
• Misapplying DP to median finding

2. You need to generate the nth number whose prime factors come only from a given list of k primes. Which approach guarantees an optimal solution with time complexity better than brute force?

easy

A. Use a dynamic programming approach with a 2D table tracking multiples of each prime and the current ugly number index.

B. Use brute force by generating all multiples of primes and checking if they are super ugly numbers.

C. Use a greedy approach that picks the smallest prime factor at each step without tracking duplicates.

D. Use a min-heap to merge sorted sequences generated by multiplying each prime with previously found ugly numbers (k-way merge).

Solution

1. Step 1: Understand problem constraints

   The problem requires generating numbers whose prime factors are limited to a given set of primes, efficiently finding the nth such number.

2. Step 2: Identify the optimal pattern

   Using a min-heap to merge k sorted sequences (each sequence generated by multiplying a prime with previously found ugly numbers) efficiently produces the sequence in ascending order without duplicates.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Min-heap merges k sorted lists in O(n log k) time [OK]

Hint: Min-heap merges multiple sorted sequences efficiently [OK]

Common Mistakes:

• Greedy approach misses duplicates and ordering
• Brute force is too slow for large n
• DP without heap is less efficient

3. Consider the following buggy code for finding k pairs with smallest sums. Which line contains the subtle bug that can cause incorrect results or runtime errors?

medium

A. Line 16: Condition 'if j + 1 <= n2' before pushing new pair

B. Line 11: Looping up to min(k, n1) to initialize heap

C. Line 14: Popping from the min-heap

D. Line 7: Checking if nums1 or nums2 is empty

Solution

1. Step 1: Analyze boundary condition in line 16

   The condition 'if j + 1 <= n2' allows j+1 to be equal to n2, which is out of bounds for nums2 indexing.

2. Step 2: Correct condition

   It should be 'if j + 1 < n2' to prevent index out of range errors and missing pairs.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Off-by-one error in boundary check causes runtime or logic errors [OK]

Hint: Check array index boundaries carefully [OK]

Common Mistakes:

• Off-by-one in array indexing
• Incorrect heap initialization
• Ignoring empty input checks

4. Consider this modified code snippet for nth super ugly number. Which line contains a subtle bug that can cause incorrect results?

medium

A. Line where val is appended to ugly without checking duplicates

B. Line where initial primes are pushed into the heap

C. Line where heap is initialized as an empty list

D. Line where new heap element is pushed with updated index

Solution

1. Step 1: Identify difference from correct code

   The original code checks if val != ugly[-1] before appending to avoid duplicates.

2. Step 2: Understand impact of missing duplicate check

   Appending duplicates causes the sequence to have repeated numbers, leading to incorrect nth number.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Duplicate values break the strictly increasing sequence [OK]

Hint: Check for duplicate filtering when appending new ugly numbers [OK]

Common Mistakes:

• Forgetting to skip duplicates when popping from heap
• Incorrect heap push indices
• Misinitializing heap elements

5. Suppose the array elements can be reused multiple times (duplicates allowed and repeated). Which modification to the k-sorted array sorting algorithm is necessary to correctly handle this scenario?

hard

A. No modification needed; the existing min-heap approach works correctly with duplicates.

B. Use a max-heap instead of a min-heap to handle duplicates correctly.

C. Add a frequency map to track counts and adjust heap operations accordingly.

D. Sort the array fully using a standard sort to handle duplicates reliably.

Solution

1. Step 1: Understand duplicates in k-sorted array

   Duplicates do not affect the heap property; min-heap handles equal elements correctly.

2. Step 2: Confirm no algorithmic change needed

   Heap operations remain valid; duplicates are naturally ordered by heapq.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Min-heap supports duplicates without extra data structures [OK]

Hint: Min-heap naturally supports duplicates without changes [OK]

Common Mistakes:

• Thinking max-heap is needed for duplicates
• Adding unnecessary frequency maps
• Defaulting to full sort unnecessarily