Intervals - Meeting Rooms I
What is the time complexity of the optimal algorithm that determines if a person can attend all meetings given n intervals, where the algorithm sorts intervals by start time and then checks for overlaps in a single pass?
What is the time complexity of the optimal algorithm that determines if a person can attend all meetings given n intervals, where the algorithm sorts intervals by start time and then checks for overlaps in a single pass?
medium🪤 Complexity Trap Q13 of Q15
Step-by-Step Solution
Step 1: Identify sorting cost
Sorting n intervals by start time takes O(n log n) time.Step 2: Identify overlap check cost
Single pass to check overlaps is O(n), which is dominated by sorting.Final Answer:
Option C -> Option CQuick Check:
Sorting dominates total time complexity [OK]
Quick Trick: Sorting dominates complexity -> O(n log n) [OK]
Common Mistakes:
MISTAKES- Confusing single pass check as O(n log n)
- Assuming nested loops cause O(n^2) here
Trap Explanation:
PITFALL- Candidates often mistake the overlap check as O(n^2) or ignore sorting cost, making O(n) or O(n^2) plausible but incorrect.
Interviewer Note:
CONTEXT- Tests understanding of sorting cost and single pass checks in interval problems.
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