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What is the time complexity of the optimal greedy algorithm that sorts balloons by their end coordinate and then iterates through them to find the minimum number of arrows?

medium🪤 Complexity Trap Q5 of Q15
Intervals - Minimum Number of Arrows to Burst Balloons
What is the time complexity of the optimal greedy algorithm that sorts balloons by their end coordinate and then iterates through them to find the minimum number of arrows?
AO(n log n)
BO(n^2)
CO(n)
DO(log n)
Step-by-Step Solution
Solution:

  1. Step 1: Analyze sorting step

    Sorting n intervals by end coordinate takes O(n log n).

  2. Step 2: Analyze iteration step

    Single pass through sorted intervals is O(n).

  3. Final Answer:

    Option A -> Option A

  4. Quick Check:

    Dominant cost is sorting, total O(n log n) [OK]
Quick Trick: Sorting dominates time complexity [OK]
Common Mistakes:
MISTAKES
  • Assuming iteration is O(n log n) or quadratic
Trap Explanation:
PITFALL
  • Candidates often confuse iteration cost or think nested loops exist.
Interviewer Note:
CONTEXT
  • Tests understanding of sorting plus linear scan complexity.
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