Intervals - Meeting Rooms II (Minimum Conference Rooms)
What is the time complexity of the optimal min-heap approach for the Meeting Rooms II problem, given n intervals? Beware of common misconceptions about sorting and heap operations.
What is the time complexity of the optimal min-heap approach for the Meeting Rooms II problem, given n intervals? Beware of common misconceptions about sorting and heap operations.
medium🪤 Complexity Trap Q13 of Q15
Step-by-Step Solution
Step 1: Identify sorting cost.
Sorting intervals by start time costs O(n log n).Step 2: Analyze heap operations.
Each interval is pushed and possibly popped from a min-heap of size k (number of rooms). Each heap operation costs O(log k).Step 3: Combine complexities.
Total heap operations are O(n log k). Since k ≤ n, worst case is O(n log n).Final Answer:
Option A -> Option AQuick Check:
Sorting + heap operations dominate; heap size bounded by n [OK]
Quick Trick: Heap ops depend on number of rooms k, bounded by n [OK]
Common Mistakes:
MISTAKES- Assuming heap ops are O(1)
- Confusing n and k in complexity
- Thinking sorting is O(n)
Trap Explanation:
PITFALL- Option B looks correct but ignores that heap size k can be up to n, making O(n log k) effectively O(n log n).
Interviewer Note:
CONTEXT- Tests understanding of complexity nuances and heap operation costs.
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