How to compare multiple items of an arraylist into another arraylist at specific position?

Question

Background: there is product listing page and i have to grab all the product name (including out of stock product) and then have to verify that all out of stock product are in the end.

Problem : i have navigated all the page and stored the product names in an ArrayList.

lets say list1 and contents are -

[instant bcaa, vegan bcaa, complete bcaa energy™, branched chain amino acid (bcaa) tablets 1000mg, endure™, branched chain amino acids (bcaa), instant leucine, leucine tablets 1000mg, complete intra-workout™, leucine, bcaa jelly mix, complete hydration drink™, informed bcaa™, instant bcaa cocktail bundle]

Now i have another list which have only Out Of Stock product

list2 and contents are -

[informed bcaa™, instant bcaa cocktail bundle]

I have to make sure whether list1 has all the list2 items in the end in same sequence

Answer 1

This problem is essentially attempting to validate that a given list A, ends with a second list B.

You could implement this by determining the length of list B, backtracking that many spaces from the end of list A, and then doing a pair-wise comparison of both lists:

public static boolean listEndsWith(List<?> A, List<?> B) {
    if (B.size() > A.size()) {
        return false;
    }

    for (int i = A.size() - B.size(), j = 0; i < A.size(); i++, j++) {
        if (!A.get(i).equals(B.get(j))) {
            return false;
        }
    }

    return true;
}

Answer 2

In Java 8+, you can use stream().skip() to skip intial list1.size() - list2.size() objects, then compare with list2.

    if (list1.size() > list2.size()) {
        AtomicInteger ordinal = new AtomicInteger(0);
        boolean matched = list1.stream().skip(list1.size() - list2.size())
                .allMatch(item -> item == list2.get(ordinal.getAndIncrement()));

        System.out.println(matched);
    }

Answer 3

Generate a new list, that is the last n items of list1, where n is the length of list 2. Then compare list 3 to list 1.

The 3rd list can be delivered like this:

ArrayList list3 = new ArrayList(list1.subList(list1.size() - list2.size(), list2.size())

Answer 4

To check, that list1 contain list2 from specific position, you can use function:

 public static boolean compareArrsFromPosition(List<?> list1, List<?> list2, int fromPosition) {
    if (list1.size()-fromPosition < list2.size()) return false;
    return list1.subList(fromPosition,fromPosition+list2.size()).equals(list2);
}

For check of end of list1 you can call like this

compareArrsFromPosition(list1, list2, list1.size()-list2.size());

Answer 5

Alternatively, you can reverse both of your Lists (linear time though might require space to store a copy) and match all the elements iterating based on the size of list B(assuming it would be of smaller size) as :

public static boolean listEndsWith(List<?> A, List<?> B) {
    Collections.reverse(B); // modifies B, so you can choose to clone and reverse
    Collections.reverse(A);
    return IntStream.range(0, B.size())
            .allMatch(i -> A.get(i).equals(B.get(i)));
}