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 nodes with height h.
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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