Question

Devise a heap-sorting-based algorithm for finding the k smallest EVEN elements of an unsorted set of n-element array. The corresponding average analytical time-complexity should also be provided. (Show your work; the time complexity for heap-building must be included; it is assumed that 50% of elements are even )

Answer 1

Basic idea:-

First of all we need to make a minheap as we are going to need smallest elements.

After that, we will perform deletion on minheap followed by necessary adjustments to minheap. and check if deleted item is even or not?. If it is then we will increment number of Even number deleted till by 1. we will continue this process until number of Even number deleted till becomes k.

Let algorithm be called kSmallestEven(A,p,q,k) where A is input unsorted array with starting index as p and ending index as q.and k as its numy of even numbers to be deleted.

Here n = q-p+1

pseudoCode in C-style:-

kSmallestEven (A,p,q,k)

{

buildHeap(A,p,q)// this will take O(n) time

NumOfEvensTillNow = 0

While(NumOfEvensTillNow<k)

{

x = delete(A,p,q)//delete from above heap.

if(x is even)

NumOfEvensTillNow++

}

}

Time complexity analysis:-

Best case:-

In best case , the first k elements to be deleted are even.so, only k times deletion operation on minheap is performed.

So, T(n) in best case = O(n) for buildHeap + k*O(logn) for deletion in total

= O(n + klogn)

Worst case:-

Worst case will occur when k^{​​​​​​th}​ even number is happened to be last element of the heap represented in array.so, here in order to find k even number almost all the elements in heap are deleted.

So,T(n) = O(n) for buildHeap + n*O(logn) for deletion in total

= O(n+nlogn)

= O(nlogn)