What is the primary time complexity difference when implementing preemptive SJF scheduling using a priority queue compared to non-preemptive SJF?

medium🪤 Complexity Trap Q5 of Q15

Operating Systems - Shortest Job First (SJF) - Preemptive vs Non-Preemptive

ANon-preemptive SJF requires O(n^2) time due to scanning all processes repeatedly

BPreemptive SJF has higher overhead due to frequent insertions and removals, leading to O(n log n) complexity

CBoth have the same O(n) complexity since burst times are known upfront

DPreemptive SJF is faster because it always picks the shortest job immediately

Step-by-Step Solution

Solution:

Step 1: Analyze preemptive SJF operations with priority queue
Preemptive SJF requires updating the priority queue on every arrival or completion, causing O(log n) per operation.
Step 2: Compare with non-preemptive SJF
Non-preemptive SJF selects once per process, resulting in fewer queue operations.
Final Answer:
Option B -> Option B
Quick Check:
Preemptive SJF has higher overhead due to dynamic queue updates [OK]

Quick Trick: Preemptive SJF updates queue more often [OK]

Common Mistakes:

MISTAKES

Trap Explanation:

PITFALL

Candidates underestimate the cost of frequent priority queue operations in preemptive SJF.

Interviewer Note:

CONTEXT

Tests understanding of algorithmic overhead differences between scheduling types.

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