ISRO-2016
September 30, 2023Process-Scheduling
October 2, 2023Algorithms
|
Question 1
|
An array of integers of size n can be converted into a heap by adjusting the heaps rooted at each internal node of the complete binary tree starting at the node ⌊(n – 1) /2⌋, and doing this adjustment up to the root node (root node is at index 0) in the order ⌊(n – 1)/2⌋, ⌊(n – 3)/2⌋, ….., 0. The time required to construct a heap in this manner is
|
O(log n)
|
|
|
O(n)
|
|
|
O(n log log n)
|
|
|
O(n log n)
|
Question 1 Explanation:
The above statement is actually algorithm for building a heap of an input array A.
And we know that time complexity for building the heap is O(n).
And we know that time complexity for building the heap is O(n).
Correct Answer: B
Question 1 Explanation:
The above statement is actually algorithm for building a heap of an input array A.
And we know that time complexity for building the heap is O(n).
And we know that time complexity for building the heap is O(n).
Subscribe
Login
0 Comments