Question

Consider two vectors that are NOT sorted, each containing n comparable items. How long would it take to display all items (in any order) which appear in either the first or second vector, but not in both, if you are only allowed LaTeX: \Theta\left(1\right)Θ ( 1 ) additional memory? Give the worst-case time complexity of the most efficient algorithm.

Answer 1

The most efficient algorithm to display all items that appear in either the first or second vector, but not in both, without using any additional memory, is to use a modified version of the merge sort algorithm.

Step-by-step explanation:

The most efficient algorithm to display all items that appear in either the first or second vector, but not in both, without using any additional memory, is to use a modified version of the merge sort algorithm. Here are the steps:

Sort both vectors using any efficient sorting algorithm, such as quicksort or mergesort. This will take ω(n log n) time complexity in the worst case.

Initialize two pointers, one for each vector, pointing to the beginning of the vectors.

Compare the elements pointed by the pointers:

If the two elements are equal, increment both pointers.

If the first element is smaller, it means it appears only in the first vector. Display the element and increment the pointer of the first vector.

If the second element is smaller, it means it appears only in the second vector. Display the element and increment the pointer of the second vector.

Repeat steps 3 and 4 until both pointers reach the end of their respective vectors.

This algorithm will take ω(n log n) time complexity in the worst case, as the sorting step dominates the time complexity. The ω(1) additional memory constraint is met as we are not using any additional memory, only a few pointers.

Answer 2

The correct answer to the following question will be "Θ(​n​2​)

". The further explanation is given below.

Step-by-step 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:

=
`O(n^2)`

And then the same manner compared again first with the second one, this takes.

=
`O(n^2)`

Therefore the total complexity,

= Θ(​n​2​)