Create 2 arrays - 1st has random integers, 2nd has unique random integers

Question

I am working on a school homework problem. I need to create 2 int[] arrays. The first array int[10] is filled with random integers. The second array has the same numbers as in the first array, but without any duplicates.

For example, assume my first array is 1,2,2,3,1,5,5,7,9,9. My second array would then be 1,2,3,5,7,9.

Could someone please point me in the right direction to solving this problem.

For questions about homework, please always add the homework tag. — James Montagne, Apr 18 '12 at 13:25

score 4 · Answer 1 · answered Apr 18 '12 at 13:31

4

Put the numbers into a Set. Then retrieve numbers from the Set. Simple! Duplicates will automatically be removed!

answered Apr 18 '12 at 13:31

CodeBlue

14,631
33
94
132

But would this be too complicated for homework? The question makes it look as though the poster is only new to Java and is probably learning about arrays, so using a `Set` might be too technical? – wattostudios Apr 18 '12 at 13:35
Ya, possibly. But he could get bonus points for using a Set! :D – CodeBlue Apr 18 '12 at 13:36
No you are correct, I'll probably try do it both ways because we haven't learned Set yet. I'm talking about FOR/WHILE loops and only just now started with arrays level of knowledge I guess... The "tip" we received that might be helpful is using a boolean array that determines if a number is being duplicated using some sort of flag. Not sure how to implement that.... – Alon Minski Apr 18 '12 at 13:48
@user1341419 Please see my second answer. The function I have defined will help you with the boolean array part. Honestly though, you don't need a boolean array at all. You could just check the boolean value from the function and make a decision whether you want to put the number in the new array or not. – CodeBlue Apr 18 '12 at 13:51

score 2 · Answer 2 · answered Apr 18 '12 at 13:29

I would do the following (assuming that it is homework and you shouldn't be doing anything too complicated)...

Sort the array using java.util.Arrays.sort(myArray); - this will order the numbers, and make sure that all repeating numbers are next to each other.
Loop through the array and keep a count of the number of unique numbers (ie compare the current number to the next number - if they're different, increment the counter by 1)
Create your second int[] array to the correct size (from point 2)
Repeat the same process as point 2, but fill your new array with the unique numbers, rather than incrementing a counter.

This should be enough to get you moving in the right direction. When you have some code, if you still have questions, come back to us and ask.

Excellent that helped! I'll get back to you with the code! Thanks :) — Alon Minski, Apr 18 '12 at 13:41
Glad to help. Please remember to 'tick' an accepted answer if your question has been solved. — wattostudios, Apr 18 '12 at 13:43

score 1 · Accepted Answer · answered Apr 18 '12 at 13:43

1

I recommend using a Set , but here's a way to do it without using a Set. (Note: This will work, but don't ask me about the efficiency of this!)

Have a function like this -

   public static boolean isNumberInArray(int[] array, int number) 
      {
          for(int i=0; i<array.length; i++)
           {
                if(number == array[i])
                   return true;
           }
          return false;
      }

Now use this function before you make an insert into the new array. I leave you to figure out that part. It's homework after all!

answered Apr 18 '12 at 13:43

CodeBlue

14,631
33
94
132

Sweet! I think that THAT is what my teacher was looking for. I got it from here :) Thanks! – Alon Minski Apr 18 '12 at 14:06
where do I post the solution I got at the end? – Alon Minski Apr 24 '12 at 14:32

score 0 · Answer 4 · answered Apr 18 '12 at 13:34

0

Hints(WATTO explains it better):

a = sorted first array
lastItem = a[0]
append lastItem into new array
for i in 1 to length(a):
   if a[i] != lastItem:
      append a[i] into new array
      lastItem = a[i]

answered Apr 18 '12 at 13:34

mshsayem

17,557
11
61
69

Tudor · Answer 5 · 2012-04-18T13:48:07.373

@WATTO Studios has a good approach. Sorting is always useful when duplicates are involved.

I will suggest an alternative method using hash tables:

Create a hashing structure with an integer as key (the number in the original array) and a counter as a value.
Go through the original array and for each number encountered increment it's corresponding counter value in the hash table.
Go through the original array again. For each number check back the hash table. If the counter associated is greater than 1, remove the value and decrement the counter.

Let's see a practical case:

4 5 6 4 1 1 3

First pass will create the following table:

1 -> 2
3 -> 1
4 -> 2
5 -> 1
6 -> 1

Second pass step by step:

4 5 6 4 1 1 3
^

4 has a counter of 2 -> remove and decrement:

1 -> 2
3 -> 1
4 -> 1
5 -> 1
6 -> 1

5 6 4 1 1 3
^

5 has a counter of 1 -> ignore 
6 has a counter of 1 -> ignore 
4 has a counter of 1 -> ignore 
1 has a counter of 2 -> remove and decrement

1 -> 1
3 -> 1
4 -> 1
5 -> 1
6 -> 1

5 6 4 1 3
      ^    
1 has a counter of 1 -> ignore
3 has a counter of 1 -> ignore

Final array:

5 6 4 1 3

There are, of course, more efficient ways to handle the removal (since using an array implies shifting), like inserting the items into a linked list for example. I'll let you decide that. :)

Edit: An even faster approach, requiring a single pass:

Use the same hashing structure as above.
Go through the original array. For each item check the table. If the associated counter is 0, increment it to 1. If it's already 1, remove the item.

Create 2 arrays - 1st has random integers, 2nd has unique random integers

5 Answers5