# Time Complexity of building a heap

Last Updated : 17 Jun, 2022

Consider the following algorithm for building a Heap of an input array A.

BUILD-HEAP(A)

heapsize := size(A);

for i := floor(heapsize/2) downto 1

do HEAPIFY(A, i);

end for

END

A quick look over the above algorithm suggests that the running time is  since each call to Heapify costs and Build-Heap makes such calls.

This upper bound, though correct, is not asymptotically tight.

We can derive a tighter bound by observing that the running time of Heapify depends on the height of the tree â€˜hâ€™ (which is equal to lg(n), where n is a number of nodes) and the heights of most sub-trees are small. The height â€™hâ€™ increases as we move upwards along the tree. Line-3 of Build-Heap runs a loop from the index of the last internal node (heapsize/2) with height=1, to the index of root(1) with height = lg(n). Hence, Heapify takes a different time for each node, which is:

For finding the Time Complexity of building a heap, we must know the number of nodes having height h. For this we use the fact that, A heap of size n has at most nodes with height h.

a  to derive the time complexity, we express the total cost of Build-Heap as-

Step 2 uses the properties of the Big-Oh notation to ignore the ceiling function and the constant 2(). Similarly in Step three, the upper limit of the summation can be increased to infinity since we are using Big-Oh notation. Sum of infinite G.P. (x < 1)

On differentiating both sides and multiplying by x, we get

Putting the result obtained in (3) back in our derivation (1), we get

Hence Proved that the Time complexity for Building a Binary Heap is

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