Time Complexity of building a heap
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-Heapmakes
such calls.
This upper bound, though correct, is not asymptotically tight.
This upper bound, though correct, is not asymptotically tight.
We can derive a tighter bound by observing that the running time of Heapifydepends on the height of the tree ‘h’ (which is equal to lg(n), where n is number of nodes) and the heights of most sub-trees are small.
The height ’h’ increases as we move upwards along the tree. Line-3 of Build-Heapruns 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 different time for each node, which is
.
The height ’h’ increases as we move upwards along the tree. Line-3 of Build-Heapruns 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 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.
For this we use the fact that, A heap of size n has at most
Now 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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