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The number of elements that can be sorted in Q(log n) time using heap sort is
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- Θ(1)
- Θ( √log n)
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Θ log n log log n
- Θ (log n)
- Θ(1)
Correct Option: C
The number of elements that can be sorted in Θ (log n) time using heap
sort is Θ | ![]() | ![]() | |
log logn |
Consider the number of elements is "k", which can be sorted in θ (k log k) time. Analyzing the options in decreasing order of complexity since we need a tight bound i.e., θ.
i.e., θ (log n) , Θ | ![]() | ![]() | , Θ(√log n) , Θ(1) | |
log logn |
So if k ∈ Θ (log n) time required for heap sort is O (k log k) i.e.,
θ (log n × log log n), But this is not in Θ (log n)
If k ∈ Θ | ![]() | ![]() | time required for Heap Sort | |
log logn |
Θ | ![]() | × log | ![]() | ![]() | ![]() | ||
log logn | log logn |
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So, this is in Θ (log n)
Hence , answer is (c) Θ | ![]() | ![]() | ||
log logn |