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Suppose intervals can have negative values and zero length (start == end). Which modification to the optimal solution is necessary to correctly remove covered intervals?

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Intervals - Remove Covered Intervals

Suppose intervals can have negative values and zero length (start == end). Which modification to the optimal solution is necessary to correctly remove covered intervals?

AChange sorting to (start ascending, end ascending) to handle zero-length intervals correctly.

BUse strict inequality 'if end > max_end' to avoid counting zero-length intervals as uncovered.

CSwitch to brute force nested loops to handle negative and zero-length intervals safely.

DKeep sorting by (start ascending, end descending) and use 'if end >= max_end' to include zero-length coverage cases.

Step-by-Step Solution

Step 1: Consider zero-length intervals
Zero-length intervals (start == end) require coverage detection to include intervals with equal ends.
Step 2: Adjust inequality and sorting
Sorting by start ascending and end descending remains correct; change condition to 'if end >= max_end' to count intervals not strictly greater but equal ends.
Final Answer:
Option D -> Option D
Quick Check:
Using >= includes zero-length coverage correctly without changing sorting [OK]

Quick Trick: Use >= in condition to handle zero-length intervals [OK]

Common Mistakes:

MISTAKES

Changing sorting breaks coverage detection
Using strict > misses zero-length coverage

Trap Explanation:

PITFALL

Naively changing sorting or using strict inequality breaks correctness for zero-length intervals.

Interviewer Note:

CONTEXT

Tests candidate's ability to adapt solution to edge cases and relaxed constraints.

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