Suppose intervals can be reused multiple times (i.e., you can select the same interval multiple times if it does not overlap with itself). Which modification to the greedy algorithm is correct to find the maximum number of non-overlapping intervals under this new condition?
ASort intervals by end time and greedily select intervals, resetting last_end after each selection to allow reuse.
BUse dynamic programming to consider multiple selections of intervals, since greedy no longer works.
CSort intervals by start time and greedily select intervals without updating last_end to allow reuse.
DSort intervals by end time and greedily select intervals, but allow intervals to be selected multiple times by not updating last_end.
Step-by-Step Solution
Step 1: Understand reuse changes problem
Allowing reuse means intervals can be selected multiple times, breaking the greedy assumption of one-time selection.
Step 2: Identify correct approach
Dynamic programming is needed to explore multiple selections and ensure maximum count without overlap.
Final Answer:
Option B -> Option B
Quick Check:
Greedy fails with reuse; DP handles multiple selections [OK]
Quick Trick:Reuse breaks greedy; DP needed for multiple selections [OK]
Common Mistakes:
MISTAKES
Trying to reuse intervals by resetting last_end incorrectly
Ignoring overlap constraints when reusing intervals
Assuming greedy still works unchanged
Trap Explanation:
PITFALL
Options that try to tweak greedy look plausible but fail on overlapping reuse cases.
Interviewer Note:
CONTEXT
Tests candidate's ability to adapt approach when problem constraints change.
Master "Non-overlapping Intervals (Max Non-Overlap)" in Intervals
3 interactive learning modes - each teaches the same concept differently
