# UGC NET COMPUTER SCIENCE SOLVED PAPERS 2017-19 - UGC NET Computer Science November 2017 Paper 2

>>>>>>>>UGC NET Computer Science November 2017 Paper 2

• Option : B
• Explanation :
When we are sorting an array of n integers, Recurrence relation for Total number of comparisons involved will be,
T(n) = 2T(n/2) + (n) where (n) is the number of comparisons in order to merge 2 sorted subarrays of size n/2. = (nlog2n)
Instead of integers whose comparison take O(1) time, we are given n strings. We can compare 2 strings in O(n) worst case.
Therefore, the total number of comparisons now will be (n2log2n) where each comparison takes O(n) time now.
In general, merge sort makes (nlog2n) comparisons, and runs in (nlog2n) time if each comparison can be done in O(1) time.

• Option : D
• Explanation :
Since Inorder traversal of a BST always gives elements in increasing order. For this question sorted order of keys will always be the in-order tree traversal for post-order traversal. So, option (D) is correct.

• Option : C
• Explanation :
Switch takes data from one network and forward it to other network based on MAC address. So option (C) is correct.

• Option : C
• Explanation :
Non-adoptive algorithms don't take their decisions on measurements or estimates of the current traffic and topology. So, option (C) is correct.