GATE | GATE CS 2012 | Question 37

A list of n strings, each of length n, is sorted into lexicographic order using the merge-sort algorithm. The worst case running time of this computation is
(A) O(n log n)
(B) O(n² log n)
(C) O(n² + log n)
(D) O(n²)


Answer: (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, 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.

See question 5 of https://www.geeksforgeeks.org/data-structures-and-algorithms-set-28/

This solution is contributed by Pranjul Ahuja

Quiz of this Question

My Personal Notes arrow_drop_up

Save