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How does this affect the algorithm to find the minimum number of arrows needed?

hard🎤 Interviewer Follow-up Q15 of Q15
Intervals - Minimum Number of Arrows to Burst Balloons
Suppose balloons can be reused multiple times after bursting (i.e., an arrow can burst a balloon multiple times if shot at different points). How does this affect the algorithm to find the minimum number of arrows needed?
AThe original greedy algorithm still applies without changes.
BWe must use dynamic programming to track multiple bursts per balloon.
CThe problem reduces to counting unique balloons since reuse removes overlap constraints.
DWe can shoot arrows only at balloon start points to minimize arrows.
Step-by-Step Solution

  1. Step 1: Understand reuse effect

    If balloons can be burst multiple times independently, overlapping intervals no longer reduce arrow count.

  2. Step 2: Simplify problem

    Each balloon requires at least one arrow, so minimum arrows equal number of unique balloons.

  3. Step 3: Algorithm implication

    Overlap and interval sorting become irrelevant; counting balloons suffices.

  4. Final Answer:

    Option C -> Option C

  5. Quick Check:

    Reuse breaks overlap optimization, so count balloons directly [OK]
Quick Trick: Reuse removes overlap benefit; count balloons directly [OK]
Common Mistakes:
MISTAKES
  • Assuming greedy still works unchanged
  • Trying to track multiple bursts with DP unnecessarily
  • Ignoring that reuse breaks interval overlap logic
Trap Explanation:
PITFALL
  • Candidates often assume reuse doesn't change problem structure, leading to wrong algorithm choice.
Interviewer Note:
CONTEXT
  • Tests candidate's ability to adapt algorithm to problem variants and understand core assumptions.
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