Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

UGC-NET | UGC-NET CS 2017 Nov – III | Question 32

  • Difficulty Level : Hard
  • Last Updated : 09 Jun, 2021

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 the sorting problem is
(A) Ω (n)
(B) Ω (n/k)
(C) Ω (nlogk )
(D) Ω (n/klogn/k)

Answer: (C)

Explanation: We have n/k subsequences each containing k elements.

If we use the quick sort technique to sort k elements in a subsequence then the complexity of sorting k elements of a subsequence is klogk.

And we have n/k such sequences. Then the time complexity of n/k subsequences having k elements will be:

= (n/k)* klogk

= nlogk

Quiz of this Question

My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!