Creating Java Linear Algorithm

I'm finding common elements in a collection of arrays. This is my code so far.

import java.util.ArrayList;
import java.util.List;
public class CommonElements //< T extends Comparable<T> >
{
Comparable[] comparable;
Comparable[] tempArr;
Comparable[] tempArr1;
Comparable[] newArray = new Comparable[tempArr.length];
int comparisonCount;
 
 [code]....

I'm using this method to sort through my array. The method accepts a collection of arrays (of varying length and of any type) as input, the algorithm input should be no greater than n(k-1). I know I can solve it using a quadratic algorithm, but it won't meet the requirement of the assignment. I did have an idea of storing all my algorithm in one giant array. After storing as one giant array I was going to sort it. After sorting I was going to compare each array side by side and see if they are the same. Here is note my teacher provided.

Note About Testing You will need to develop several sets of test collections for testing your algorithm. The grading rubric mentions covering the case where all the test collections have the same length, as well as covering the case where the test collections are of different lengths. You will also need to think about what constitutes the worst case scenario for this algorithm, since only that scenario will make your analysis of total comparisons performed a meaningful one. You can use the formulas in the grading rubric to tell you how many comparisons you should expect in the quadratic and linear cases. For example, if you have 5 total collections (1 query collection and 4 test collections), each of which contains 10 elements, the total number of comparisons performed in the worst case should be: (k - 1)N2, which for k = 10 and N = 10 is: (5 - 1)102, or 400 comparisons. For the linear algorithm, you should only have N*(k - 1), which is 10*(5 - 1), or 40 comparisons.

Here is Q/A that my teacher provided:

1. Are the elements in the collections already sorted?The input collections may or may not be sorted. Don’t assume that they are sorted.

2. Can I sort the elements in the collections? Am I supposed to sort the elements in thecollections?Yes, you can sort the elements in the collections. It is not required that you do so, however..

3. If I sort the collections, should I count the element comparisons that are performed by the sorting algorithm?No. The only element comparisons you should count are the ones that are directly used to find the common elements. Ignore comparisons performed by any sorting algorithms you use.

4. How do I extract an individual collection from the Object[] collections argument of the findCommonElements method?You will need to typecast each collection as type Comparable[]. For example:Comparable[] currentCollection = (Comparable[])collections[i];

5. Why are we using a one-dimensional Object array instead of a two-dimensional array of type Comparable (e.g., Comparable[][])?In Java, arrays are treated as objects, so only a variable of type Object[] can contain arrays. A variable of type Comparable[] can only contain Comparable elements, not arrays. A variable of type Comparable[][] also can only store Comparable elements. Although it is possible to organize the elements such that all the elements at indexes [1][i], for example, are considered to comprise a collection, this is not the same as having an array that contains other arrays. The additional bookkeeping is required to keep a 2D array organized is more complex than simply performing the typecast described above.

6. Can I use other data structures besides arrays; e.g., HashMaps?No. Although it is easier from a programming perspective to use higher order data structures like HashMaps to find common elements, using these structures simply hides the details. If you were to examine the implementations of those structures, you would probably find that they use relatively inefficient algorithms such as sequential searches to find elements. Additionally, if you use library components, you have no way of counting the number of comparisons performed, since you don’t have access to the source code of those components.

7. Is this a trick question? I can’t think of any way to get below NlogN comparisons.It is not a trick question. It is possible to solve this problem using a number of comparisons proportional to (k – 1) * N.