3.Given two sorted lists of size m and n respectively.The number of comparisons needed in the worst case by the merge sort algorithm will be?A mnB max(m,n)C min(m,n)D m+n-1

The answer is D. m+n-1.

Here's why:

The merge process of merge sort algorithm compares each element of one list with the other. In the worst case scenario, it will compare each element from the first list with each element from the second list until it exhausts one of the lists.

So, if you have m elements in the first list and n elements in the second list, the maximum number of comparisons in the worst case will be when you have compared all elements in both lists.

This means you will have m comparisons for the elements in the first list and n comparisons for the elements in the second list. But, since the last comparison includes an element from both lists, you subtract 1 from the total.

Hence, the worst case number of comparisons will be m+n-1.