Minimum number of comparisons required in optimal algorithm

Question

Comments

Here my doubt is should we consider number of comparisons as m+n-1 as worst case is asked or min(m,n) as minimum number of comparisons is asked.

Answer 1

1.From the give array(say size =n) , using min heap algorithm -> create a minheap

2.Take two minimum elements and merge them and place the result back in the array again apply minheap on the new array.

    Repeat step 2 for "n-1 " times .

Minimum number of comparisions required =n-1(Here its 5)