How can building a heap be O(n) time complexity?

Question

Can someone help explain how can building a heap be O(n) complexity?

Inserting an item into a heap is O(log n), and the insert is repeated n/2 times (t

Answer 1

As we know the height of a heap is log(n), where n is the total number of elements.Lets represent it as h
   When we perform heapify operation, then the elements at last level(h) won't move even a single step.
   The number of elements at second last level(h-1) is 2^h-1 and they can move at max 1 level(during heapify).
   Similarly, for the i^th, level we have 2ⁱ elements which can move h-i positions.

Therefore total number of moves=S= 2^h*0+2^h-1*1+2^h-2*2+...2⁰*h

                                               S=2^h {1/2 + 2/2² + 3/2³+ ... h/2^h} -------------------------------------------------1
this is AGP series, to solve this divide both sides by 2
                                               S/2=2^h {1/2² + 2/2³+ ... h/2^h+1} -------------------------------------------------2
subtracting equation 2 from 1 gives
                                               S/2=2^h {1/2+1/2² + 1/2³+ ...+1/2^h+ h/2^h+1}
                                               S=2^h+1 {1/2+1/2² + 1/2³+ ...+1/2^h+ h/2^h+1}
now 1/2+1/2² + 1/2³+ ...+1/2^h is decreasing GP whose sum is less than 1 (when h tends to infinity, the sum tends to 1). In further analysis, let's take an upper bound on the sum which is 1.
This gives S=2^h+1{1+h/2^h+1}
                    =2^h+1+h
                    ~2^h+h
as h=log(n), 2^h=n

Therefore S=n+log(n)
T(C)=O(n)