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What is the time complexity of the sweep line algorithm for finding the minimum number of platforms required given n trains, and why?

medium🪤 Complexity Trap Q13 of Q15
Intervals - Minimum Number of Platforms Required
What is the time complexity of the sweep line algorithm for finding the minimum number of platforms required given n trains, and why?
AO(n log n) due to sorting arrivals and departures separately and merging them.
BO(n^2) because each event is compared with all others to count overlaps.
CO(n) because we only iterate through arrivals and departures once.
DO(n log n) because events are sorted and then processed in a single pass.
Step-by-Step Solution
Solution:
  1. Step 1: Identify sorting as the main cost

    We create 2n events and sort them by time, which takes O(n log n).
  2. Step 2: Single pass processing after sorting

    After sorting, we iterate through events once, O(n), updating counts.
  3. Final Answer:

    Option A -> Option A
  4. Quick Check:

    Sorting dominates, linear sweep after [OK]
Quick Trick: Sorting events dominates time complexity [OK]
Common Mistakes:
MISTAKES
  • Assuming O(n) because of single pass ignoring sorting
  • Confusing nested loops brute force with sweep line
  • Thinking sorting arrivals and departures separately is enough
Trap Explanation:
PITFALL
  • Sorting all events together is required; sorting separately then merging is less efficient or incorrect.
Interviewer Note:
CONTEXT
  • Checks understanding of sorting cost and linear sweep in event-based algorithms.
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