recategorized by
1,172 views

3 Answers

1 votes
1 votes

In binary max heaps, smallest element will be present at leaves which resides from $\lceil\frac{n}{2}\rceil$ to $n$ location so we need to traverse at max $\frac{n}{2}$ elements which takes O(n)$

GATE 2006

Hence Option A) is correct

0 votes
0 votes

In a max heap, the smallest element is always present at a leaf node. So we need to check for all leaf nodes for the minimum value. Worst case complexity will be O(n)


 
Answer:

Related questions

1 votes
1 votes
3 answers
1
1 votes
1 votes
2 answers
3
go_editor asked Nov 20, 2020
798 views
Match $\text{List I}$ with $\text{List II}$Let $R_1=\{(1,1), (2,2), (3,3)\}$ and $R_2=\{(1,1), (1,2), (1,3), (1,4)\}$$\begin{array}{llll} & \text{List I} & & \text{List I...
3 votes
3 votes
3 answers
4
go_editor asked Nov 20, 2020
3,440 views
The number of positive integers not exceeding $100$ that are either odd or the square of an integer is _______$63$$59$$55$$50$