Question

You are given a sequence of n elements to sort. The input sequence consists of n/k subsequences, each containing k elements. The elements in a given subsequence are all smaller than the elements in the succeeding subsequence and larger than the elements in the preceding subsequence. Thus, all that is needed to sort the whole sequence of length n is to sort the k elements in each of the n/k subsequences.
The lower bound on the number of comparisons needed to solve this variant of sorting problem is

• A. $\Omega \left(nlogk\right)$
• B. $\Omega \left(n/k\right)$
• C. $\Omega \left(nklogn/k\right)$
• D. $\Omega \left(n\right)$

Answer 1

The correct option is A $\Omega \left(nlogk\right)$

Since, each sub sequence of length is 'k', so to sort each subsequence it will take $\Omega \left(klogk\right)$, but we have $\frac{n}{k}$ subsequences, so total time required is $\Omega (\frac{n}{k}\times klogk)$

$=\Omega \left(nlogk\right)$