Question

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 the given word removed.

Shift the remaining words in the nonempty part of the array to the 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 not the capacity) of the array shrinks if removeWord appears in the array. In the first test case, for example, "up" is removed and "down" and "left" shift to the left by one element. The size of the array shrinks from 3 to 2, although the capacity is still 3. The third element is now in the empty part of the array. This element is 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 presumed to 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 <contentsInArray.length && index < array.length; ++index)
	{
	array[index] = contentsInArray[index];
	}
	int arraySize = index;
	String value = keyboard.nextLine(); // Get the value to remove
	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, int sourceSize, String removeMe)
	{
	// Write the code below.
	return null;
	}
	}

Answer 1

Here is the completed code for this problem. Comments are included, go through it, learn how things work and let me know if you have any doubts or if you need anything to change. If you are satisfied with the solution, please rate the answer. If not, PLEASE let me know before you rate, I’ll help you fix whatever issues. Thanks

Note: I only completed removeOversize method as requested. Did not touch main method. The output will have null values for empty spaces. This is normal (not any mistake by me), because inside the main method, using Arrays.toString(), you are printing the entire array and not just the filled values. If you need to return an array of exact size, you will need to change the signature of removeOversize method. Right now it is returning the new size of the array, not entirely the new array.

// RemoveOversize.java

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 presumed to 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 < contentsInArray.length && index < array.length; ++index) {

                  array[index] = contentsInArray[index];

            }

            int arraySize = index;

            String value = keyboard.nextLine(); // Get the value to remove

            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, int sourceSize,

                  String removeMe) {

            // looping from last to front (easier to remove)

            for (int i = sourceSize - 1; i >= 0; i--) {

                  // checking if value at i needs to be removed

                  if (source[i].equals(removeMe)) {

                        // looping and shifting every elements on the right to one place

                        // to the left

                        for (int j = i; j < sourceSize - 1; j++) {

                              source[j] = source[j + 1];

                        }

                        // making the last element null

                        source[sourceSize - 1] = null;

                        // updating size

                        sourceSize--;

                  }

            }

            // returning new size

            return sourceSize;

      }

}

/*OUTPUT (user input in green)*/

hello abc hi how abc hello abc this is fun

abc

The resulting array contains [hello, hi, how, hello, this, is, fun, null, null, null, null, null, null, null, null, null, null, null, null, null] and is of size 7