It is not possible in on step or you need to keep the reference to the array. If you can change the reference this can help:

      String[] n = new String[]{"google","microsoft","apple"};
      final List<String> list =  new ArrayList<String>();
      Collections.addAll(list, n); 
      list.remove("apple");
      n = list.toArray(new String[list.size()]);

I not recommend the following but if you worry about performance:

      String[] n = new String[]{"google","microsoft","apple"};
      final String[] n2 = new String[2]; 
      System.arraycopy(n, 0, n2, 0, n2.length);
      for (int i = 0, j = 0; i < n.length; i++)
      {
        if (!n[i].equals("apple"))
        {
          n2[j] = n[i];
          j++;
        }      
      }

I not recommend it because the code is a lot more difficult to read and maintain.

If you must use arrays, System.arraycopy is the most efficient, scalable solution. However, if you must remove one element from an array several times, you should use an implementation of List rather than an array.

The following utilizes System.arraycopy in order to achieve the desired effect.

public static Object[] remove(Object[] array, Object element) {
    if (array.length > 0) {
        int index = -1;
        for (int i = 0; i < array.length; i++) {
            if (array[i].equals(element)) {
                index = i;
                break;
            }
        }
        if (index >= 0) {
            Object[] copy = (Object[]) Array.newInstance(array.getClass()
                    .getComponentType(), array.length - 1);
            if (copy.length > 0) {
                System.arraycopy(array, 0, copy, 0, index);
                System.arraycopy(array, index + 1, copy, index, copy.length - index);
            }
            return copy;
        }
    }
    return array;
}

Also, you can increase the method's efficiency if you know that your array consists of only Comparable objects. You can use Arrays.sort to sort them before passing them through the remove method, modified to use Arrays.binarySearch to find index rather than a for loop, raising that portion of the method's efficiency from O(n) to O(nlogn).

There's the error that @Eran pointed out in your code, which can solve your problem. But I'm going to discuss another approach.

For now, you're first iterating over the entire array to find the number of occurrences to remove, and then, you're iterating over the array to remove them. Why don't you just iterate over the array, just to remove them. (I know, your first loop is helping you to determine the size of the output array, but you don't need that if you use some List like ArrayList etc.)

List<String> resultList = new ArrayList<String>();
for(int i = 0; i < arr.length; i++)
{
    if(!arr[i].contains(toRemove))
    {
        resultList.add(arr[i]);
    }
}

And you can return the resultList, but if you really need to return an array, you can convert the resultList to an array like this:

String [] resultArray = resultList.toArray(new String[resultList.size()]);

And then return this array. See this approach live here on ideone.

Your question isn't very clear. From your own answer, I can tell better what you are trying to do:

public static String[] removeElements(String[] input, String deleteMe) {
    List result = new LinkedList();

    for(String item : input)
        if(!deleteMe.equals(item))
            result.add(item);

    return result.toArray(input);
}

NB: This is untested. Error checking is left as an exercise to the reader (I'd throw IllegalArgumentException if either input or deleteMe is null; an empty list on null list input doesn't make sense. Removing null Strings from the array might make sense, but I'll leave that as an exercise too; currently, it will throw an NPE when it tries to call equals on deleteMe if deleteMe is null.)

Choices I made here:

I used a LinkedList. Iteration should be just as fast, and you avoid any resizes, or allocating too big of a list if you end up deleting lots of elements. You could use an ArrayList, and set the initial size to the length of input. It likely wouldn't make much of a difference.

You don't.

Arrays can not be resized.

You would need to create a new (smaller) array, and copy the elements you wished to preserve into it.

A better Idea would be to use a List implementation that was dynamic. An ArrayList<Integer> for example.

example = new String[example.length];

If you need dynamic collection, you should consider using one of java.util.Collection implementations that fits your problem. E.g. java.util.List.