Question

Does this method check and replace randoms in the array?

public void generateNumbers()
   {

       for(int counter=0; counter< 6; counter++){

           numbers[counter]= 1+ (int)(Math.random()*49);
           for(int y = counter - 1; y >=0 ; y--){
               if(numbers[counter] == numbers[y]){
                   numbers[counter]= 1+ (int)(Math.random()*49);
                }
           }


    }

If there is duplicates I need to replace this with another random number and then check again. I want to check this using a loop if possible and not using collections.

The range of the numbers created by random should be 1 – 49.

Answer 1

You need to “rewind” back to counter-1 on finding the duplicate, so that you start searching for duplicates again:

public void generateNumbers()
{

   for(int counter=0; counter< 6; counter++){

       numbers[counter]= 1+ (int)(Math.random()*49);
       for(int y = counter - 1; y >=0 ; y--){
           if(numbers[counter] == numbers[y]){
               numbers[counter]= 1+ (int)(Math.random()*49);
               y = counter - 1; // <<<===== ADDED
            }
       }


}

Your original method would fail if you start with this array

1 3 7 8

then generate 3, check it against 8 and 7, find duplicate 3, and generate 8 as a replacement. At this point, your algorithm would check 8 against 1, and be satisfied that there’s no duplicate; this would be incorrect.