How to conduct permutation using an arraylist

Question

I am trying to compare the first element in an arraylist to the rest of the elements.

Then comparing the second element of the arraylist to the rest of the elements and so on until
the end of the arraylist.

The code is as below:

ArrayList<String> a = new ArrayList<String>();

    a.add("G1");
    a.add("G2");
    a.add("G3");
    a.add("G1");

    System.out.println(a.size());

    for(int i = 0; i<a.size(); i++){

        for(int j = 0; j<a.size(); j++){

        if(a.get(i) == a.get(j))
            System.out.println("Element: " + a.get(i)+ " at " + i + " and " + "Element: "+ a.get(j)+ " at " + j);

        }





    }

The problem is that it would check element 0 against 1 then 2 so on but after this it would check element 1 against 0 then 1 and so on it should not compare the same elements twice. — Sandeep Johal
– Sandeep Johal, Commented Jan 11, 2013 at 5:30
I already figured that out. Take a look at my answer and let me know if this helps. — Smit
– Smit, Commented Jan 11, 2013 at 5:32

Amir Afghani · Accepted Answer · 2013-01-11 05:27:17Z

1

== is reference equality (e.g do these two objects point to the same location in memory). For object equality, use .equals() instead.

answered Jan 11, 2013 at 5:27

Amir Afghani

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Comments

Archit Saxena · Accepted Answer · 2013-10-05 17:21:24Z

1

Use this code, instead of yours.

for(int i = 0; i<a.size()-1; i++){

    for(int j = i+1; j<a.size(); j++){

    if(a.get(i).equals(a.get(j)))
        System.out.println("Element: " + a.get(i)+ " at " + i + " and " + "Element: "+ a.get(j)+ " at " + j);

    }

Hope that helps.. :)

edited Oct 5, 2013 at 17:21

answered Jan 11, 2013 at 5:44

Archit Saxena

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3 Comments

nhydock Over a year ago

Additionally each loop through will end 1 short with that a.size()-1 on the interior loop. a.size()-1 is acceptable on the outer loop so then it doesn't jump through one extra time at the end when there's only one entry to compare with.

Archit Saxena Over a year ago

oh, my bad... replace == with .equals and the i loop should end at a.size()-1 not the j loop.

Smit Over a year ago

@ArchitSaxena Better you edit this in your answer. Make it complete.

Smit · Accepted Answer · 2013-01-11 17:03:03Z

0

Use

 if((a.get(i)).equals(a.get(j)))

instead of

 if(a.get(i) == a.get(j))

Morover start your initialization of j = i+1 You have already checked these previous string so no need to start it again.

EDIT

You have to limit your outer loop to i<a.size()-1 so that it wont check very last element with itself.

I hope this helps. if you need anymore help just ask.

edited Jan 11, 2013 at 17:03

answered Jan 11, 2013 at 5:30

Smit

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3 Comments

Sandeep Johal Over a year ago

Thanks, another issue is that every element must only be checked once any idea on how I could achieve this ? for example if the contents of the array were a.add("G1");//0 a.add("G2");//1 a.add("G3");//2 a.add("G1");//3 a.add("G2");//4 a.add("G2");//5 a.add("G3");//6 }

Smit Over a year ago

@SandeepJohal I updated the answer. The issue you are referring to already stated in answer by others.

Smit Over a year ago

@SandeepJohal I think I misunderstood your question. I didnt get it. Can you make that edit in your question and let me know when you do it.

nhydock · Accepted Answer · 2013-01-11 05:51:14Z

In your case, since you're using Strings, instead of using ==, which is referential equality, as in are the memory addresses the same, you want to use .equals() which will compare the two strings by their actual values.

As for your for-loop, you can make it slightly more efficient by doing this.

for (int i = 0; i < a.size()-1; i++)
{
    for (int j = i+1; j < a.size(); j++)
    {
        if(a.get(i).equals(a.get(j)))
        {
            System.out.println("Element: " + a.get(i)+ " at " + i + " and " + "Element: "+ a.get(j)+ " at " + j);
        }

    }
}

Since there are times when you already compare a[i] with a[j], the result will be the same if you do a[j] equals a[i], so you can just skip over them. This also saves you from checking a[i] equals a[j] when j and i are the same.

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