notes each of size and merge the first list with the second list to form single sorted list of size similarly we merge the third and the fourth lists to form second single sorted list of size and so on this completes the one pass we then consider the first sorted list of size and second sorted list of size and merge them to form single sorted list of size similarly we merge the third and the fourth sorted lists each of size to form the second single sorted list of size and so onthis completes the second pass in the third pass we merge these adjacent sorted lists each of size to form sorted lists of size we continue this process till finally we end up with single sorted list of size as shown in figure [ , [ , [ , [ , , , [ , [ , , , [ , , , , , , , to carry out the above taskwe require procedure to merge the two sorted lists of size and respectively to form single sorted list of size ( + )we also require procedure to carry out one pass of the list merging the adjacent sorted lists of the specified size this is because we have to carry out the repeated passes of the given list in the first passwe merge the adjacent lists of size in the second passwe merge the adjacent lists of size and so on thereforewe will call this procedure by varying the size of the lists to be merged here is implementation of the same void merge(int []int []int lint mint nint , ,ki = + while (( < &( < )if( [ < [ ] lovely professional university