How find common number from two textview

Question

I have two Textview and two buttons.

TextView TV1,TV2;
Button YES,NO;

Now

   TV1.setText("1 2 3 4 5");
   TV2.setText("3 4 5 6 7");

And

   YES.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View view) {

           //here i want to show common number that is (3,4,5)

            }
        });

    NO.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View view) {

    //here i want to show number  no matched in both Textview

           }
        });

Now my question is that when i click on yes it show for (3,4,5 because common) FOR TV1 and when i click on no it show (1,2,6,7) TV1.

I can do it by set text but what if i have more numbers.

the question is a bit unclear. When you click yes, how come it will show 3,4,5 ? — John Joe, Apr 07 '17 at 03:08
@PradeepSheoran You should try to write this in pure Java before you do it in Android — OneCricketeer, Apr 07 '17 at 03:13
Your title should describe the problem you're having or a specific question you're asking, and should contain content that will be useful to a future reader here searching for an answer to a problem. Your current title is useless both as a problem description and a search result. Please [edit] it to something that has some relevant content. See [ask]. While you're in a reading mood, see [mcve] as well. — Ken White, Apr 07 '17 at 03:13

score 0 · Answer 1 · edited May 23 '17 at 12:25

You may put number into Array or List so that easy to find common number.

public List<Integer> findCommonElement(int[] a, int[] b) {

    List<Integer> commonElements = new ArrayList<Integer>();

    for(int i = 0; i < a.length ;i++) {
        for(int j = 0; j< b.length ; j++) {
            if(a[i] == b[j]) {
                //Check if the list already contains the common element
                if(!commonElements.contains(a[i])) {
                    //add the common element into the list
                    commonElements.add(a[i]);
                }
            }
        }
    }
    return commonElements;
}

And you may check it using:

int myArray[] = { 2, 2, 7, 7, 2, 1, 5, 4, 5, 1, 1 };
int myArray2[] = { 2, 3, 4, 7, 10 };

Log.d("Common", String.valueOf(findCommonElement(myArray, myArray2)));

Referral: here

Okay, that's the common elements, but the question also asked for difference too — OneCricketeer, Apr 07 '17 at 03:27

jace · Answer 2 · 2017-04-07T03:35:20.137

First, you must have a way for the user to input numbers if what you mean in "what if i have more numbers" is flexibility.

List<int> set1 = new ArrayList<int>();   //your set 1
List<int> set2 = new ArrayList<int>();   //your set 2
List<int> answer = new ArrayList<int>(); // container of common or difference

//Assuming you already have a value in your set 1 and set 2

 YES.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            answer.clear() //always reset the List
            for(int i=0; i<set1.size(); i++)
            {
                for(int j=0; j<set2.size(); j++)
                {
                    if(set1.get(i)==set2.get(j))
                    {
                        answer.add(set2.get(j));
                        break;
                    }
                }
            }
            //you now have the value of all common @ this point
            //it's up to you what you will do now
        }
    });

 NO.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            answer.clear() //always reset the List
            int exist = 0;
            for(int i=0; i<set1.size(); i++)
            {
                for(int j=0; j<set2.size(); j++)
                {
                    if(set1.get(i)==set2.get(j))
                    {
                        exist += 1;
                    }
                }
                if(exist==0)
                {
                    answer.add(set1.get(i));
                }
                exist = 0;//reset
            }
            //you now have the value of all uncommon @ this point
            //it's up to you what you will do now
       }
    });

if i have a mistake here, feel free to comment :) I may include the C# coding here in java :D — jace, Apr 07 '17 at 03:33

score 0 · Answer 3 · answered Apr 07 '17 at 03:33

Try this-

    TV1.setText("1 2 3 4 5");
    TV2.setText("3 4 5 6 7"); 
    String array1[]= TV1.getText().toString().split(" ");
    String array2[]= TV2.getText().toString().split(" ");
    List<Integer> aList = new ArrayList<>(Arrays.asList(array1));
    List<Integer> bList = new ArrayList<>(Arrays.asList(array2));
    List<Integer> union = new ArrayList(aList);
    union.addAll(bList);
    List<Integer> intersection = new ArrayList(aList);
    intersection.retainAll(bList);
    List<Integer> symmetricDifference = new ArrayList(union);
    symmetricDifference.removeAll(intersection);
    YES.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {

       //here i want to show common number that is (3,4,5)
      System.out.println("common number: " + intersection);

        }
    });

    NO.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {

      //here i want to show number  no matched in both Textview
        System.out.println("**symmetricDifference: " + symmetricDifference+"**");

       }
    });

Prints:

 common number: [3, 4, 5]
 **symmetricDifference: [1, 2, 6, 7]**

How find common number from two textview

3 Answers3