Let's run through each part of the code. First, createSharedListViaStream:

public static List<SchoolObj> createSharedListViaStream(List<SchoolObj> listOne, List<SchoolObj> listTwo)
{
    // We create a stream of elements from the first list.
    List<SchoolObj> listOneList = listOne.stream()
    // We select any elements such that in the stream of elements from the second list
    .filter(two -> listTwo.stream()
    // there is an element that has the same name and school as this element,
        .anyMatch(one -> one.getName().equals(two.getName()) 
            && two.getSchool().equals(one.getSchool())))
    // and collect all matching elements from the first list into a new list.
    .collect(Collectors.toList());
    // We return the collected list.
    return listOneList;
}

After running through the code, it does exactly what you want it to do. Now, let's run through createSharedListViaLoop:

public static List<SchoolObj> createSharedListViaLoop(List<SchoolObj> listOne, List<SchoolObj> listTwo)
{
    // We build up a result by...
    List<SchoolObj> result = new ArrayList<SchoolObj>();
    // going through each element in the first list,
    for (SchoolObj one : listOne)
    {
    // going through each element in the second list,
        for (SchoolObj two : listTwo)
        {
    // and collecting the first list's element if it matches the second list's element.
            if (one.getName().equals(two.getName()) && one.getSchool().equals(two.getSchool()))
            {
                result.add(one);
            }
        }
    }
    // We return the collected list
    return result;
}

So far, so good... right? In fact, your code in createSharedListViaStream is fundamentally correct; instead, it is your createSharedListViaLoop that may be causing discrepancies in output.

Think about the following set of inputs:
List1 = [SchoolObj("nameA","SchoolX"), SchoolObj("nameC","SchoolZ")]
List2 = [SchoolObj("nameA","SchoolX"), SchoolObj("nameA","SchoolX"), SchoolObj("nameB","SchoolY")]

Here, createSharedListViaStream will return the only element of the first list that appears in both lists: SchoolObj("nameA","SchoolX"). However, createSharedListViaLoop will return the following list: [SchoolObj("nameA","SchoolX"),SchoolObj("nameA","SchoolX")]. More precisely, createSharedListViaLoop will collect the correct object, but it will do so twice. I suspect this to be the reason for the output of createSharedListViaStream to be "incorrect" based on comparison to the output of createSharedListViaLoop.

The reason that createSharedListViaLoop does this duplication is based on the lack of termination of its inner for loop. Although we iterate over all elements of the first list to check if they are present in the second, finding a single match will suffice to add the element to the result. We can avoid redundant element addition by changing the inner loop to the following:

for (SchoolObj one : listOne)
    {
    for (SchoolObj two : listTwo)
    {
        if (one.getName().equals(two.getName()) && one.getSchool().equals(two.getSchool()))
        {
            result.add(one);
            break;
        }
    }
}

Additionally, if you don't want duplicate Objects in your list (by location in memory), you can use distinct like so:

List<SchoolObj> result = ...;
result = result.stream().distinct().collect(Collectors.toList());

As a final caution, the above will keep the results distinct in the following scenario:

List<SchoolObj> list = new ArrayList<>();
SchoolObj duplicate = new SchoolObj("nameC", "schoolD");
listOne.add(duplicate);
listOne.add(duplicate);
list.stream().distinct().forEach(System.out::println); 
// prints:
// nameC schoolD

However, it will not work in the following scenario, unless you override the equals method for SchoolObj:

List<SchoolObj> list = new ArrayList<>();
listOne.add(new SchoolObj("nameC", "schoolD"));
listOne.add(new SchoolObj("nameC", "schoolD"));
list.stream().distinct().forEach(System.out::println); 
// prints (unless Object::equals overridden)
// nameC schoolD
// nameC schoolD

Let listOne.size() is N and listTwo.size() is M. Then 2-for-loops solution has complexity of O(M*N).

We can reduce it to O(M+N) by indexing listTwo by ids.

Case 1 - assuming listTwo has no objects with the same id

// pair each id with its marks
Map<String, String> marksIndex = listTwo.stream().collect(Collectors.toMap(ObjectTwo::getId, ObjectTwo::getMarks));
// go through list of `ObjectOne`s and lookup marks in the index
listOne.forEach(o1 -> o1.setScore(marksIndex.get(o1.getId())));

Case 2 - assuming listTwo has objects with the same id

    final Map<String, List<ObjectTwo>> marksIndex = listTwo.stream()
            .collect(Collectors.groupingBy(ObjectTwo::getId, Collectors.toList()));

    final List<ObjectOne> result = listOne.stream()
            .flatMap(o1 -> marksIndex.get(o1.getId()).stream().map(o2 -> {
                // make a copy of ObjectOne instance to avoid overwriting scores
                ObjectOne copy = copy(o1);
                copy.setScore(o2.getMarks());
                return copy;
            }))
            .collect(Collectors.toList());

To implement copy method you either need to create a new object and copy fields one by one, but in such cases I prefer to follow the Builder pattern. It also results in more "functional" code.

Your question’s code does not reflect what you describe in the comments. In the comments you say that all names should be present and the size should match, in other words, only the order may be different.

Your code is

List<Person> people = getPeopleFromDatabasePseudoMethod();
List<String> expectedValues = Arrays.asList("john", "joe", "bill");

assertTrue(people.stream().map(person -> person.getName())
                 .collect(Collectors.toList()).containsAll(expectedValues));

which lacks a test for the size of people, in other words allows duplicates. Further, using containsAll combining two Lists in very inefficient. It’s much better if you use a collection type which reflects you intention, i.e. has no duplicates, does not care about an order and has an efficient lookup:

Set<String> expectedNames=new HashSet<>(expectedValues);
assertTrue(people.stream().map(Person::getName)
                 .collect(Collectors.toSet()).equals(expectedNames));

with this solution you don’t need to test for the size manually, it is already implied that the sets have the same size if they match, only the order may be different.

There is a solution which does not require collecting the names of persons:

Set<String> expectedNames=new HashSet<>(expectedValues);
assertTrue(people.stream().allMatch(p->expectedNames.remove(p.getName()))
           && expectedNames.isEmpty());

but it only works if expectedNames is a temporary set created out of the static collection of expected names. As soon as you decide to replace your static collection by a Set, the first solution doesn’t require a temporary set and the latter has no advantage over it.

If I understand correctly, this is the example scenario:

• listOne [datab] items: [A, B, C, D]
• listTwo [front] items: [B, C, D, E, F]

and what you need to get as an effect is:

• added: [E, F]
• deleted: [A]

First thing first, I would use some type adapter or extend the different types from one common class and override the equals method so you can match them by id and name

Secondly, this is very easy operations on sets (you could use set's but list are fine too). I recommend using a library: https://commons.apache.org/proper/commons-collections/apidocs/org/apache/commons/collections4/CollectionUtils.html

And now basically:

• added is listTwo - listOne
• deleted is listOne - listTwo

and using java code:

• added: CollectionUtils.removeAll(listTwo, listOne)
• deleted: CollectionUtils.removeAll(listOne, listTwo)

Otherwise, all collections implementing Collection (Java Docs) also has removeAll method, which you can use.

EDIT

Here are two versions. One using ArrayList and other using HashSet

Compare them and create your own version from this, until you get what you need.

This should be enough to cover the:

P.S: It is not a school assignment :) So if you just guide me it will be enough

part of your question.

continuing with the original answer:

You may use a java.util.Collection and/or java.util.ArrayList for that.

The retainAll method does the following:

Retains only the elements in this collection that are contained in the specified collection

see this sample:

import java.util.Collection;
import java.util.ArrayList;
import java.util.Arrays;

public class Repeated {
    public static void main( String  [] args ) {
        Collection listOne = new ArrayList(Arrays.asList("milan","dingo", "elpha", "hafil", "meat", "iga", "neeta.peeta"));
        Collection listTwo = new ArrayList(Arrays.asList("hafil", "iga", "binga", "mike", "dingo"));

        listOne.retainAll( listTwo );
        System.out.println( listOne );
    }
}

EDIT

For the second part ( similar values ) you may use the removeAll method:

Removes all of this collection's elements that are also contained in the specified collection.

This second version gives you also the similar values and handles repeated ( by discarding them).

This time the Collection could be a Set instead of a List ( the difference is, the Set doesn't allow repeated values )

import java.util.Collection;
import java.util.HashSet;
import java.util.Arrays;

class Repeated {
      public static void main( String  [] args ) {

          Collection<String> listOne = Arrays.asList("milan","iga",
                                                    "dingo","iga",
                                                    "elpha","iga",
                                                    "hafil","iga",
                                                    "meat","iga", 
                                                    "neeta.peeta","iga");

          Collection<String> listTwo = Arrays.asList("hafil",
                                                     "iga",
                                                     "binga", 
                                                     "mike", 
                                                     "dingo","dingo","dingo");

          Collection<String> similar = new HashSet<String>( listOne );
          Collection<String> different = new HashSet<String>();
          different.addAll( listOne );
          different.addAll( listTwo );

          similar.retainAll( listTwo );
          different.removeAll( similar );

          System.out.printf("One:%s%nTwo:%s%nSimilar:%s%nDifferent:%s%n", listOne, listTwo, similar, different);
      }
}

Output:

$ java Repeated
One:[milan, iga, dingo, iga, elpha, iga, hafil, iga, meat, iga, neeta.peeta, iga]

Two:[hafil, iga, binga, mike, dingo, dingo, dingo]

Similar:[dingo, iga, hafil]

Different:[mike, binga, milan, meat, elpha, neeta.peeta]

If it doesn't do exactly what you need, it gives you a good start so you can handle from here.

Question for the reader: How would you include all the repeated values?

You can make the code shorter with filter and count, but using loops is cleaner here. Streams are not always the solution.

return Arrays.asList(
    IntStream.range(0, Math.min(a.size(), b.size())).filter(i -> a.get(i) > b.get(i)).count(),
    IntStream.range(0, Math.min(a.size(), b.size())).filter(i -> a.get(i) < b.get(i)).count()
);

It's not the most efficient solution but the most terse code would be:

boolean equalLists = listA.size() == listB.size() && listA.containsAll(listB);

Update:

@WesleyPorter is right. The solution above will not work if duplicate objects are in the collection.
For a complete solution you need to iterate over a collection so duplicate objects are handled correctly.

private static boolean cmp( List<?> l1, List<?> l2 ) {
    // make a copy of the list so the original list is not changed, and remove() is supported
    ArrayList<?> cp = new ArrayList<>( l1 );
    for ( Object o : l2 ) {
        if ( !cp.remove( o ) ) {
            return false;
        }
    }
    return cp.isEmpty();
}

Update 28-Oct-2014:

@RoeeGavriel is right. The return statement needs to be conditional. The code above is updated.