Question

Suppose you have two arrayLists: A and B.

How do I create a new array which is the same size as B and stores the index values of A as integers in sequential order.

So say for example the size of B is 5 and A has 3 values.

A[0] = Ra
A[1] = Be
A[2] = Ce


B[0] = F
B[1] = M
B[2] = K
B[3] = P
B[4] = L

I want to then create in java different possible versions (say 5 versions) of an arrayC of size 5 (the same size as listB) with a sequential ordering of the indexes of listA.

So like:

arrayC[0] = 0
arrayC[1] = 1
arrayC[2] = 1
arrayC[3] = 2
arrayC[4] = 2

or

arrayC[0] = 0
arrayC[1] = 0
arrayC[2] = 1
arrayC[3] = 2
arrayC[4] = 2

are both valid combinations in arrayC. However

arrayC[0] = 0
arrayC[1] = 2
arrayC[2] = 1
arrayC[3] = 2
arrayC[4] = 2

is not.

Answer 1

This method returns all orderings. Limit it if you just want a few.

public static Set<List<Integer>> orderings(int i, int len, int max) {

    Set<List<Integer>> seqs = new HashSet<List<Integer>>();

    if (len <= 0 || i > max)
        return seqs;

    if (max - i == len) {
        List<Integer> l = new ArrayList<Integer>();
        while (i < max)
            l.add(i++);
        seqs.add(l);
        return seqs;
    }

    seqs.addAll(orderings(i    , len - 1, max));
    seqs.addAll(orderings(i + 1, len - 1, max));

    for (List<Integer> l : seqs)
        l.add(0, i);

    return seqs;
}


public static Set<List<Integer>> orderings(int[] arr1, int[] arr2) {
    return orderings(0, arr2.length, arr1.length);
}

Test code:

public static void main(String[] args) {
    int[] listA = { 0, 1, 2 };
    int[] listB = { 0, 1, 2, 3, 4 };
    for (List<Integer> seq : orderings(listA, listB))
        System.out.println(seq);
}

Output:

[0, 0, 1, 2, 2]  <-- your second example
[0, 1, 1, 1, 2]
[0, 1, 1, 2, 2]  <-- your first example
[0, 1, 2, 2, 2]
[0, 0, 0, 1, 2]
[0, 0, 1, 1, 2]

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