THIS IS FOR JAVA

Given an oversize array of size words and a word to remove,write a method that returns the array with each occurrence of thegiven word removed.

Shift the remaining words in the nonempty part of the array tothe left so that each occurrence of the given word is overwritten.(Leave the words in the empty part of the array unchanged.)

Hint: To understand the test cases, note that the size (but notthe capacity) of the array shrinks if removeWord appears in thearray. In the first test case, for example, "up" is removed and"down" and "left" shift to the left by one element. The size of thearray shrinks from 3 to 2, although the capacity is still 3. Thethird element is now in the empty part of the array. This elementis equal to "left", but it is treated as empty space.

here is the code:

import java.util.Arrays;
	import java.util.Scanner;
	public class RemoveOversize
	{
	public static void main(String[] args)
	{
	// Do not edit this code
	Scanner keyboard = new Scanner(System.in);
	final int SIZE = 20; // for the oversized array
	String[] array = new String[SIZE];
	// Read in data from keyboard--all array elements are presumedto be on one line
	String contents = keyboard.nextLine();
	String[] contentsInArray = contents.split(" ");
	// Make sure we're in the range of both arrays
	int index;
	for(index = 0; index
	{
	array[index] = contentsInArray[index];
	}
	int arraySize = index;
	String value = keyboard.nextLine(); // Get the value toremove
	int resultSize = removeOversize(array, arraySize, value);
	System.out.println("The resulting array contains " +Arrays.toString(array) + " and is of size " + resultSize);
	keyboard.close();
	}
	public static int removeOversize(String[] source, intsourceSize, String removeMe)
	{
	// Write the code below.
	return null;
	}
	}