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What is the time complexity of calculating waiting times for n processes scheduled using FCFS, assuming burst times are known?

medium🪤 Complexity Trap Q5 of Q15
Operating Systems - FCFS Scheduling - Convoy Effect & Waiting Time
What is the time complexity of calculating waiting times for n processes scheduled using FCFS, assuming burst times are known?
AO(log n), due to efficient prefix sum computations.
BO(n^2), because each process's waiting time requires summing all previous bursts.
CO(n), since waiting time for each process depends on the sum of previous bursts.
DO(1), as waiting times can be computed instantly with a formula.
Step-by-Step Solution
Solution:
  1. Step 1: Understand waiting time calculation

    Waiting time for process i is sum of bursts of all previous processes.
  2. Step 2: Analyze complexity

    Using prefix sums, waiting times for all n processes can be computed in O(n) total time.
    Naive summation per process is O(n^2), but prefix sums optimize to O(n).
  3. Final Answer:

    Option C -> Option C
  4. Quick Check:

    Prefix sums reduce complexity to linear [OK]
Quick Trick: Prefix sums enable O(n) waiting time calculation [OK]
Common Mistakes:
MISTAKES
  • Assuming naive O(n^2) without prefix sums
  • Thinking waiting time calculation is constant time
  • Confusing logarithmic complexity with prefix sums
Trap Explanation:
PITFALL
  • Candidates often forget prefix sums optimize waiting time calculation from O(n^2) to O(n).
Interviewer Note:
CONTEXT
  • Tests understanding of algorithmic optimization in waiting time computation.
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