Simple Encrypt Decrypt String with array in Android

Question

My problem seems easy, I am trying to encrypt a string in android.

The objective of the program is, put a word or a letter in textbox1, click on button encrypt and textbox2 should to show the symbol in the same position in second array of the letter that you put in.

For example I write letter A (array 1 [0]) must show symbol $ (array 2 [0]) if a click on ENCRYPT button, if I put a symbol and click on DECRYPT must show the letter which is equivalent in position of array

Need help. Sorry for grammar and edit question, first time and no English speaker, i tried to make it more simple.

Arreglo Class, just for declare 2 arrays.

package PaqueteArreglo;

public class Arreglo {
public String [] encrypt = new String[3];
public String [] decrypt = new String[3];
}

Main Activity

package com.example.encrypt;
import android.os.Bundle;
import android.support.v7.app.AppCompatActivity;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import PaqueteArreglo.Arreglo;

public class MainActivity extends AppCompatActivity implements View.OnClickListener{

//Instance the class with the arrays
PaqueteArreglo.Arreglo ins = new Arreglo();
public Button button1,button2;
public EditText text1, text2;

@Override
protected void onCreate(Bundle savedInstanceState) {
  super.onCreate(savedInstanceState);
  setContentView(R.layout.activity_main);
  initialize();

  ins.encrypt[0] = "a";
  ins.encrypt[1] = "b";
  ins.encrypt[2] = "c";

  ins.decrypt[0] = "$";
  ins.decrypt[1] = "%";
  ins.decrypt[2] = "&";

}

private void initialize(){
  text1 = (EditText)findViewById(R.id.txtWord);
  text2 = (EditText)findViewById(R.id.txtResult);

  button1 = (Button)findViewById(R.id.btnEncrypt);
  button1.setOnClickListener(this);

  button2 =(Button)findViewById(R.id.btnDecrypt);
  button2.setOnClickListener(this);

}

@Override
public void onClick(View view) {
  String word = String.valueOf(this.text1.getText());

  if (view.getId() == R.id.btnEncrypt)
  {
     String demo = "";
     int lon = 0;
     lon = word.length();
     for (int i = 0; i < lon; i++ )
     {
        if (ins.encrypt[i].equalsIgnoreCase(String.valueOf(word.charAt(i))))
        {
           demo = demo + ins.decrypt[i];
           text2.setText(demo);
        }
     }
  }
}
}

You should probably use some kind of map like a HashMap to map your encryption character to it's matching decryption character. Also it is unclear what the specific problem you are having is. Could you clarify the question down a bit? — Nic Robertson
– Nic Robertson, Commented Apr 12, 2016 at 3:41
Thank you for answer, i edit the question for make an explanation more simple, and I do not know use (and for this purpose do not need) HashMap, i need resolve the problem with this method, thanks. — Roger Jimenez
– Roger Jimenez, Commented Apr 12, 2016 at 4:27

Pooya · Accepted Answer · 2016-04-12 09:34:19Z

There are two solutions to your problem using two different approach:

1- Using your defined structure (which is not optimized by the way) you need to put two nested loop for checking:

for (int i = 0; i < lon; i++ )
 {
    for(int j = 0; j<ins.encrypt.length;j++)
    {
       if (ins.encrypt[j].equalsIgnoreCase(String.valueOf(word.charAt(i))))
       {
          demo = demo + ins.decrypt[j];
       }
    }
 }

text2.setText(demo);

2- Using much better data structure like HashMap:

Change your Arreglo to the following:

public class Arreglo {
   public HashMap<Character, Character> encrypt = new HashMap<>();
   public HashMap<Character, Character> decrypt = new HashMap<>();
}

Now change the way you add your data like:

ins.encrypt.put('a','@');
ins.decrypt.put('@','a');
...

and finally do the encryption like:

for (int i = 0; i < lon; i++ )
{
   demo = demo + ins.encrypt.get(word.charAt(i));
}
text2.setText(demo);

The second implementation is more efficient

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