Question 10112 – GATE 1996
April 26, 2024UGC NET CS 2013 Dec-paper-2
April 26, 2024GATE 2012
| Question 5 |
The worst case running time to search for an element in a balanced binary search tree with n2n elements is
| Θ (n log n) | |
| Θ (n2n) | |
| Θ (n) | |
| Θ (log n) |
Question 5 Explanation:
→ Worst case running time to search for an element in a balanced binary search tree of ‘n’ elements is (log n).
→ No of elements = n.2n then search time = (log n.2n)
= (log n + log 2n)
= (log n + n log 2)
= O(n)
→ No of elements = n.2n then search time = (log n.2n)
= (log n + log 2n)
= (log n + n log 2)
= O(n)
Correct Answer: C
Question 5 Explanation:
→ Worst case running time to search for an element in a balanced binary search tree of ‘n’ elements is (log n).
→ No of elements = n.2n then search time = (log n.2n)
= (log n + log 2n)
= (log n + n log 2)
= O(n)
→ No of elements = n.2n then search time = (log n.2n)
= (log n + log 2n)
= (log n + n log 2)
= O(n)