Answer:
Explanation:
We start from the bottom-most and rightmost internal node of min Heap and then heapify all internal modes in the bottom-up way to build the Max heap.
To build a heap, the following algorithm is implemented for any input array.
BUILD-HEAP(A)
heapsize := size(A)
for i := floor(heapsize/2) downto 1
do HEAPIFY(A, i)
end for
Convert the given array of elements into an almost complete binary tree.
Ensure that the tree is a max heap.
Check that every non-leaf node contains a greater or equal value element than its child nodes.
If there exists any node that does not satisfy the ordering property of max heap, swap the elements.
Start checking from a non-leaf node with the highest index (bottom to top and right to left).
Answer:
O(n²)
Explanation:
The worse case time complexity of insertion sort using binary search for positioning of data would be O(n²).
This is due to the fact that there are quite a number of series of swapping operations that are needed to handle each insertion.
Answer:
true
Explanation:
because it needs a interpreter so that it could be more developed
