Answer:
Following are the solution to the given question:
Explanation:
Since a procedure has the Read10 parameter, the 10 characters from the input file are stored in the BYTE array as myString. The LOOP instruction, which includes indirect addressing and also the call to the ReadChar method, please find the attached file of the procedure:
Answer:
The correct answer to the following question will be "Θ(n2)
". The further explanation is given below.
Explanation:
If we're to show all the objects that exist from either the first as well as the second vector, though not all of them, so we'll have to cycle around the first vector, so we'll have to match all the objects with the second one.
So,
This one takes:
= 
And then the same manner compared again first with the second one, this takes.
= 
Therefore the total complexity,
= Θ(n2)