Java: Interleaving multiple arrays into a single array

Question

I found similar question about interleaving two arraylists into one, but its in PHP. I was asked this question in interview as well but could'nt solve it, came back to SO to look if it was addressed already, but i could only find this paper

So any pointers to pseudo code or method definition ?

Big(O) restrictions : O(n) - time cost and O(1) - space cost

Example:
a[]= a1, a2, ..., an
b[]= b1, b2, ..., bn
Rearrange the arraylist to a1, b1, a2, b2, ..., an, bn

Editv1.0 : Arraylists a[] and b[] are of same size

Editv2.0 : What if the question is extended to rearrange in one of given two arrays, but not create a new array ?

Answer 1

For simplicity, assume that the arrays are the same length, and are int arrays.

int[] merge(int[] a, int[] b)
{
    assert (a.length == b.length);

    int[] result = new int[a.length + b.length];

    for (int i=0; i<a.length; i++)
    {
        result[i*2] = a[i];
        result[i*2+1] = b[i];
    }

    return result;
}

Answer 2

I think this is not doable with your given constraints (O(n) time and O(1) space, i.e. no additional space) for an array or array-based list. (Assuming of course, that we can't simply create a new List object delegating to the original ones.)

If you have two linked lists, this is doable - if we assume the garbage collector is fast enough, i.e. deleting an element from one list and adding it to another list does not violate the space limitation.

public <X> void interleaveLists(List<X> first, List<X> second)
{
    ListIterator<X> firstIt = first.listIterator();
    ListIterator<X> secondIt = second.listIterator();
    while(secondIt.hasNext()) {
        fistIt.next();
        firstIt.add(secondIt.next());
        secondIt.remove();
    }
}

This method works for any pair of lists, but is only O(n) for linked lists.

For a custom linked list where we can modify the pointers, we don't have to rely on the garbage collector, we would simply change the nodes. Here for a singly-linked list:

public void interleaveLinkedLists(Node<X> firstList, Node<X> secondList) {
    while(secondList != null) {
        Node<X> nextFirst = firstList.next;
        Node<X> nextSecond = secondList.next;
        firstList.next = secondList;
        secondList.next = nextFirst;
        firstList = nextFirst;
        secondList = nextSecond;
    }
}

For a doubly-linked list, we would also have to adapt the prev-pointers.

Here the wrapping variant mentioned in the first paragraph:

public List<X> interleaveLists(final List<X> first, final List<X> second)
{
   if (first.size() != second.size())
      throw new IllegalArgumentException();
   return new AbstractList<X>() {
      public int size() {
         return 2 * first.size();
      }
      public X get(int index) {
         return index % 2 == 0 ? first.get(index / 2) : second.get(index / 2);
      }
      // if necessary, add a similar set() method.  add/remove are not sensible here.
   };
}

This is actually O(1) in time, too.