Generate random numbers without duplicates using arrays only

Question

I want to fill an array of size X with random integers from 0 to X with no duplicates. The catch is I must only use arrays to store the collections of int, no ArrayLists. How do I go about implementing this?

I don't understand why I can't seem to get this. But this is my most recent bit of code that fills the list but allows for duplicates.

System.out.print("Zero up to but excluding ");
int limit = scanner.nextInt();

// create index the size of the limit
int [] index = new int[limit];

for(int fill=0;fill<limit;fill+=1){
    index[fill] = (limit);
}

int randomNumber = 0;
Random rand = new Random();
int [] randoms = new int[limit];
boolean flag = true;

// CODE TO NOT PRINT DOUBLES
for (int z=0;z<limit;z+=1){
    randomNumber = rand.nextInt(limit);
    int i=0;
    while (i<limit){
        if (index[i] == randomNumber){
            flag = true;
        }
        else {
            flag = false;
            break;
        }
        i+=1;
    }
    if (flag == false){
        randoms[z] = randomNumber;
        index[z] = randomNumber;
    }
}
System.out.println("Randoms: "+java.util.Arrays.toString(randoms));

The standard way to do this is to use an ArrayList, fill it with the numbers 0..X-1, then shuffle it. So, with arrays, do the same. — Andy Turner
– Andy Turner, Commented Jul 23, 2016 at 22:56

CMPS · Accepted Answer · 2016-07-23 23:09:07Z

6

Here's one way to do it:

Create an array of length N
Fill it from 0 to N-1
Run a for loop and swap randomly 2 indices

Code:

// Step 1
int N = 10;
int[] array = new int[N];

// Step 2
for(int i=0; i < N; i++)
   array[i] = i;

// Step 3
for(int i=0; i < N; i++) {
   int randIndex = (int) (Math.random() * N);
   int tmp = array[i];
   array[i] = array[randIndex];
   array[randIndex] = tmp;
}

edited Jul 23, 2016 at 23:09

answered Jul 23, 2016 at 22:57

CMPS

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Comments

Roland Kwee · Accepted Answer · 2016-07-23 22:57:46Z

0

Why not rephrase the problem to shuffling an array of integers. First fill the array monotonically with the numbers 0 to X. Then use the Random() function to select one of the X numbers to exchange with the number in position 0. Repeat as many times as you may like. Done.

answered Jul 23, 2016 at 22:57

Roland Kwee

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Comments

beatrice · Accepted Answer · 2016-07-23 23:04:29Z

0

Here is your bug:

  while (i<limit){
    if (index[i] == randomNumber){
      flag = true;
    }
    else {flag = false;break;}  <--- rest of the array is skipped
    i+=1;
  }

after you generated a new number, you start to check for equality , however once you find that randomNumber!=index[i] (else statement) you break out of the while. look this: actual array is 3,4,5,1 your new number is 5, you compare it to 3 just to find out that they different so flag is set to false and break out happens.

answered Jul 23, 2016 at 23:04

beatrice

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1 Comment

beatrice Over a year ago

remove your break and put it after flag=true if (index[i] == randomNumber){ flag = true;break; } else { flag = false; }

Chris Gong · Accepted Answer · 2016-07-23 23:43:14Z

Consider using another array filled with elements in order from 0 to X. Then, with this array, shuffle the elements around. How do you go about this? Use a loop to traverse through every single element of the array, and for each iteration, choose a random number from 0 to array.length - 1 and switch the elements at the index you're currently on and the random index. This is how it would look like,

In your main, you would have an array initialized by doing this,

int[] arr = new int[10];//10 can be interchangeable with any other number
for(int i = 0; i < arr.length; i++){
    arr[i] = i;
}
shuffleArray(arr);

And the shuffle method would look like this,

public int[] shuffleArray(int[] arr){
    Random rand = new Random();
    for(int i = 0; i < arr.length; i++){
        int r = rand.nextInt(arr.length);//generate a random number from 0 to X
        int k = arr[i];
        arr[i] = arr[r];
        arr[r] = k;
    }
}

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