Random numbers in array

Question

I need to fill an array A with 5 random numbers from a possible 50. Duplicates are allowed. I then need to fill the same array a second time but with unique numbers that did not appear in the first or second filling of the array. I am using the code below to generate the array the second time. I am stuck on how to search the first array for duplicates at the same time. Any help appreciated! Thanks

boolean drawn;
 for (int i=0; i<A.length; i++) {
 do {
 drawn = false;
 A[i] = 1 + (int)(Math.random() * 50);
 for (int j=0; j<i; j++)
 if (A[i] == A[j]) 
 drawn = true; 
 } while (drawn);

• Caffeinated: I'd draw the logic on paper. You can use if-then logic, though I think there are many ways to skin this cat
• Jesse Webb: If you used a List instead of an array, you could leverage the contains() method.

Answer 1

Use a List<> to keep track of all allowed numbers, and as you use them, remove them from the list. Then, for your 'second pass', instead of always getting a random number * 50, get a random number * list size(). Then the actual number you use is list.get(location) instead of just number.

(your list will start out as 1,2,3,4,5, but then when 3 is used, the list will become 1,2,4,5, ..., So that when you get the next "random()" number of 3, that is actually a 4.)

Answer 2

I would do this:

List<Integer> choices = new ArrayList<Integer>(); 
Random r = new Random();
for(int i = 0; i <N;i++)
{ 
 choices.add(r.nextInt(N-1)+1;
} 
Set<Integer> uniques = new HashSet<Integer>(); 
Collections.shuffle(uniques); 
//remove first 5 entries from uniques 
//repeat for second pass 

This of course has the possibility (as minimal as it is) that there will be a set remaining of less than size 5.

Answer 3

You could create a list containing your 50 numbers, and use a Random to get 5 indices between 0 and 50, to create your first random array.

For the second pass, create a copy of the first list, but remove the numbers that are in the first random array. Then shuffle this second list (using Collections.shuffle()), and take the 5 first elements.

Answer 4

You have to use one extra array each time you want to fill the array as you need to store the previous array elements. Something like:

boolean drawn;
System.arraycopy( A, 0, B, 0, A.length );
for (int i=0; i<A.length; i++) {
 do {
 drawn = false;
 A[i] = 1 + (int)(Math.random() * 50);
 for (int j=0; j<A.length; j++)
 if (A[i] == B[j]) 
 drawn = true; 
 for (int j=0; j<i; j++)
 if (A[i] == A[j]) 
 drawn = true; 
 } while (drawn);

`

• Rohit Jain: @user1930614.. Don't just copy the code. Try to understand what exactly is happening. That will help you in long run.

Answer 5

Store your 50 random number candidates in an array random of size 50

Keep a counter initially set to random.length.So, herecounter will be initialized to 49.

Then Generate a number r from 0 to counter-1 and choose random[r] as your number.

Now go ahead and swap random[r] with random[counter-1]. Decrement counter so that next time you will only be searching the first 49 elements of the array, and continue this process whenever you pick a number.

Now simply If you want a number that you have not chosen before, generate r from 0 to counter-1 so that random[r] comes from the unique elements in the array which will be in the beginning.

 int[] randoms= new int[50];
 .
 . //populate your array with the candidate numbers
 .
 int counter= randoms.length;
 Random rand = new Random();
 public int getUnique(){
 //Get a random number in the range 0 to counter-1
 int r = rand.nextInt(counter);
 int myElement = randoms[r];
 randoms[r] = randoms[counter-1];
 randoms[counter-1]= myElement;
 counter--;
 return myElement; 
 }

And whenever you want to allow duplicates, generate an r from 0 to random.length -1 and choose random[r] as your number. Notice that we still have to keep track of the numbers used so that getUnique() still works:

 public int getAny(){
 //Get a random number in the range 0 to random.length-1
 int r = rand.nextInt(random.length);
 int myElement = randoms[r];
 if(r < counter){
 randoms[r] = randoms[counter-1];
 randoms[counter-1]= myElement;
 counter--;
 }
 return myElement; 
 }