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What is the time complexity of the line sweep algorithm for counting intervals containing each point, given n intervals and m points?

medium🪤 Complexity Trap Q5 of Q15
Intervals - Count of Intervals Containing Each Point
What is the time complexity of the line sweep algorithm for counting intervals containing each point, given n intervals and m points?
AO(n log m)
BO(n * m)
CO((n + m) log (n + m))
DO(n + m)
Step-by-Step Solution
Solution:

  1. Step 1: Count events

    There are 2n interval events (start and end) and m point events, total O(n + m) events.

  2. Step 2: Sorting events

    Sorting all events takes O((n + m) log (n + m)) time, which dominates the algorithm.

  3. Final Answer:

    Option C -> Option C

  4. Quick Check:

    Sorting all events dominates time complexity [OK]
Quick Trick: Sorting all events dominates time complexity [OK]
Common Mistakes:
MISTAKES
  • Assuming O(n*m) from nested loops
  • Ignoring sorting cost of combined events
  • Confusing binary search complexity with line sweep
Trap Explanation:
PITFALL
  • Candidates often mistake the complexity as O(n*m) or forget sorting cost of combined events.
Interviewer Note:
CONTEXT
  • Tests understanding of event-based algorithm complexity and sorting cost.
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