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

medium🪤 Complexity Trap Q13 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 once to find the minimum number of arrows?
AO(n)
BO(n log n)
CO(n^2)
DO(log n)
Step-by-Step Solution
  1. Step 1: Identify sorting cost

    Sorting n intervals by end coordinate takes O(n log n) time.
  2. Step 2: Identify iteration cost

    Single pass through sorted intervals is O(n).
  3. Final Answer:

    Option B -> Option B
  4. Quick Check:

    Dominant cost is sorting, so total is O(n log n) [OK]
Quick Trick: Sorting dominates; iteration is linear [OK]
Common Mistakes:
MISTAKES
  • Assuming linear time because of single pass
  • Confusing with brute force O(n^2)
  • Ignoring sorting cost
Trap Explanation:
PITFALL
  • Candidates often overlook sorting cost and pick O(n) because iteration is linear.
Interviewer Note:
CONTEXT
  • Tests understanding of algorithmic complexity beyond just iteration loops.
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