Iterating through nested arrays in Java

Question

I'm building e-commerce application and i have one case scenario where i need to cross out sizes which are not available in store.

I have fixed arraylist of strings with sample like this: "36", "38", "40", "42" and i have available sizes with sample like this: "36", "38", "40"

Now i need to iterate through first array and cross out those sizes which are not available.

Here is one part of code where i'm doing that:

// tempSizes - available sizes
// mProduct.getSizes() - all sizes
for (String tempSize : tempSizes) {
      for (int i = 0; i < mProduct.getSizes().size(); i++) {
            if (tempSize.equals(mProduct.getSizes().get(i))) {
                 // if size is available
                 sizes.add(new Size(mProduct.getSizes().get(i), true));
            } else {
                 // if size is not available
                 sizes.add(new Size(mProduct.getSizes().get(i), false));
            }
      }
}

Problem here is that nested for loop will be called three times and the result will output with duplicates of sample. If there is an easier way to do this, please let me know, i would appreciate it.

"the result will be three arraylist with one available size" I don't understand: you don't create 3 arraylists here, you add everything into one. — Andy Turner
– Andy Turner, Commented May 28, 2017 at 14:24
If sample is 36, 38, 40 and available sizes are 36, 38 the output will be 36(available), 38, 40, 36, 38(available), 40 and it should be something like this: 36(available), 38(available), 40 — Dusan Dimitrijevic
– Dusan Dimitrijevic, Commented May 28, 2017 at 14:27

Darshan Mehta · Accepted Answer · 2017-05-28 14:22:39Z

3

You can easily do it with ArrayList using one for loop, e.g.:

List<Integer> samples = Arrays.asList(36, 38, 40 ,42);
List<Integer> available = Arrays.asList(36, 38, 40);
List<Integer> unavailable = new ArrayList<>();
for(int size : samples){
    if(!available.contains(size)){
        unavailable.add(size);
    }
}
System.out.println(unavailable);

This will iterate through all the samples, check whether they are available and if not, put them into anoter list. If you have arrays, you can use Arrays.asList() method to convert them into the Lists.

answered May 28, 2017 at 14:22

Darshan Mehta

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5 Comments

Andy Turner Over a year ago

unavailable.addAll(available); unavailable.removeAll(samples);.

Darshan Mehta Over a year ago

@AndyTurner I would rather not change the existing lists/arrays as they might get used further down the line. I agree with the approach though.

Darshan Mehta Over a year ago

@DusanDimitrijevic you can accept the answer if it has solved the problem.

Andy Turner Over a year ago

@DarshanMehta "I'd rather not change the existing lists" This changes the existing lists no more or less than your approach.

Darshan Mehta Over a year ago

@AndyTurner Ah okay, I thought it was samples.removeAll(available), please ignore my previous comment.

Rakesh Soni · Accepted Answer · 2017-05-28 14:33:51Z

1

Try this code

// mProduct.getSizes() - all sizes
for (int i = 0; i < mProduct.getSizes().size(); i++) {


    int prodSize = mProduct.getSizes().get(i);      
    boolean sizeFound = false;

    // tempSizes - available sizes
    for (String tempSize : tempSizes) {

            if (tempSize.equals(mProduct.getSizes().get(i))) {
                 // if size is available
                 sizes.add(new Size(prodSize, true));
                 sizeFound = true;
                 break;

            }

      }

      if(sizeFound == false){
           // if size is not available
           sizes.add(new Size(mProduct.getSizes().get(i), false));
      }

}

edited May 28, 2017 at 14:33

answered May 28, 2017 at 14:26

Rakesh Soni

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2 Comments

Dusan Dimitrijevic Over a year ago

Both answers are helpful, but Darshan's approach i think is a little better because it doesn't requires nested for loop and there is less code.

Rakesh Soni Over a year ago

Yes, agree with you.

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