Remove from Destination array all occurrences of all characters that occur in Source Array [closed]

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Given two character arrays a[] and b[], remove from b[] all occurrences of all characters that occur in array a[]. You need to do this in-place i.e. without using an extra array of characters. E.g.:

Input: a[] = [‘G’, ‘O’]          
Input  b[] = [‘G’, ’O’, ’O’, ’G’, ’L’, ’E’] 

Output: b[] = [‘L’, ‘E’]

Code :

public class ReplaceCharacterArray{         
       public static void main(String args[]){        
          char a[] = [‘G’, ‘O’]         
          char b[] = [‘G’, ’O’, ’O’, ’G’, ’L’, ’E’]     

         //to replace all the occurences of all the characters of     
         //a[] array in b[] array we have below logic.     

          for(int i=0;i<a.length;i++){     
             for(int j=0;j<b.length;j++){     
                  if(b[j] == a[i]){     
                    //im stuck here how to proceed      
             }     
          }    

---

You can't remove elements "in place" in Java arrays. They have fixed length. That is, in your example you'll have to return a new array, since you can't change the length of the b array.

Here are some pointers for you:

• Maintain a write-index for the b array (left of this index, you have only characters not present in a).
• Iterate through the b array
• While the current character is contained in a, step forward
• Swap current character (not contained in a) with the character at the write-index
• Increment the write-index, and continue from there.

Use for instance Arrays.copyOfRange to return the part of the array to the left of the write-index.

---

Regarding your update:

• Arrays are not written using [ and ] and characters are not written using ‘, change them to {, } and '.
• Having a helper method with the signature boolean arrayContains(char[] arr, char c) will make it easier to write the algorithm
• If you follow my approach, you will also benefit from having a helper method void swap(char[] arr, int index1, int index2).

---

You can go about like this

Iterate the loop a, that is 2 times and In each iteration, perform the copare that value exists in the b[] array, if yes delete that. Continue that until you iterate the array b[] completely.

So that in the end you will have b[] array only with the other elements which is not in the array a[].

I hope this will help you to complete your work.

---

I guess since this is a homework 3rd party library's may not be allowed.Any way i'll just post a code using Guava:

 char[] removeChars(char a[],char b[])
 {
    return  CharMatcher.anyOf(new String(b)).removeFrom(new String(a)).toCharArray();    
 }

UPDATE:(check this sample,it's one way to solve)

import java.util.*;
import java.lang.*;

class CharReplace{
public static void main(String[] args)
{
char a[] = "GOOGLE".toCharArray();
char b[] = "GO".toCharArray();
BitSet charToRemove=new BitSet();
for(char c:b)
 charToRemove.set(c);
StringBuilder str=new StringBuilder();
for(char c:a)
 if(!charToRemove.get(c))
  str.append(c);
b=str.toString().toCharArray();
System.out.println(Arrays.toString(b));

}

}

---

You have mistakes in your code,

• Arrays don't start and end with [ and ] respectively. They start and end with { and } respectively.
• Characters are enclosed with ' only.
• You will have to keep the state of your iterations (in my case I kept the indexing state and the state of to know whether the character in b exists in a).
• Have a remove(char[] array, char characterToRemove) and pass the b[] and it's respective character to remove to complete the flow.

---

Update without using String, StringBuffer, or any libraries, this is what works....

Warning the array b never gets reduced (it gets appended with NULL character). That I will leave it for you to crack.

/**
 * @author The Elite Gentleman
 * 
 */
public class ReplaceCharacterArray {
    public static boolean containsCharacter(char character, char[] array) {
        for (int i = 0; i < array.length; i++) {
            if (array[i] == character) {
                return true;
            }
        }

        return false;
    }

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        char a[] = {'G', 'O'};
        char b[] = {'G', 'O', 'O', 'G', 'L', 'E'};
        //to replace all the occurences of all the characters of a[] array in b[] array we have below logic.

        int index = 0;
        int count = 0;
        while (index < b.length) {
            while (ReplaceCharacterArray.containsCharacter(b[index], a)) {
                System.arraycopy(b, index + 1, b, 0, b.length - index - 1);
                count++;
            }

            index++;
        }

        for (int i = (b.length - count); i < b.length; i++) {
            b[i]= '\0';
        }

        for (int i = 0; i < b.length; i++) {
            System.out.println(b[i]);
        }
    }
}