Question

Can any one explain why the time complexity for generating a binary heap from a unsorted array using bottom-up heap construction is O(n) ?

(Solution found so far: I found in Thomas and Goodrich book that the total sum of sizes of paths for internal nodes while constructing the heap is 2n-1, but still don’t understand their explanation)

Thanks.

Answer 1

Normal BUILD-HEAP Procedure for generating a binary heap from an unsorted array is implemented as below :

BUILD-HEAP(A)
 heap-size[A] ← length[A]
  for i ← length[A]/2 downto 1
   do HEAPIFY(A, i)

Here HEAPIFY Procedure takes O(h) time, where h is the height of the tree, and there
are O(n) such calls making the running time O(n h). Considering h=lg n, we can say that BUILD-HEAP Procedure takes O(n lg n) time.

For tighter analysis, we can observe that heights of most nodes are small.
Actually, at any height h, there can be at most CEIL(n/ (2^h +1)) nodes, which we can easily prove by induction.
So, the running time of BUILD-HEAP can be written as,

lg n                     lg n
∑ n/(2^h+1)*O(h)  = O(n* ∑ O(h/2^h)) 
h=0                      h=0  

Now,

∞   
∑ k*x^k = X/(1-x)^2
k=0              
               ∞ 
Putting x=1/2, ∑h/2^h = (1/2) / (1-1/2)^2 = 2
               h=0     

Hence, running time becomes,

lg n                     ∞   
O(n* ∑ O(h/2^h))  = O(n* ∑ O(h/2^h))  = O(n)
h=0                      h=0  

So, this gives a running time of O(n).

N.B. The analysis is taken from this.