Generate a random two-dimensional array of non-repeating numbers, Java

Question

The user will type in the number for i (variant), then the number for j (elements for every variant), and finally the maximum value possible (maxElem). Using the inputed values, the task is to generate nonrepeating random numbers (nonrepeating in a variant, meaning for i, but the numbers may repeat during the entire array).

For example, a successful output giving the input 3 (i), 5 (j), 9 (maxElem), would be:

4|8|1|7|9

3|8|2|4|5

2|6|4|8|5

As you may notice, the number 4 repeats itself during the entire array for 3 times (allowable). But, for i=0, number 4 is unique.

Please, guide me what would be the changes to this code:

static Scanner sc = new Scanner(System.in);

static int maxElem;

public static void main(String[] args) {

    int[][] greatLoto;

    System.out.println("Of how many variants will the ticket consist?  ");
    int variants = sc.nextInt();

    System.out.println("Of how many elements will the variants consist? ");
    int elements = sc.nextInt();

    System.out.println("Which value should be considered the maximum value? ");
    maxElem = sc.nextInt() + 1;

    greatLoto = new int[variants][elements];

    System.out.println("Initial values: ");
    show(greatLoto);

    System.out.println("Modifying values...");
    modified(greatLoto);

    System.out.println("Newest values: ");
    show(greatLoto);

}

private static void show(int[][] greatLoto) {
    for (int i = 0; i < greatLoto.length; i++) {
        for (int j = 0; j < greatLoto[i].length; j++) {
            System.out.print("|" + greatLoto[i][j] + "|");
        }
        System.out.println("");
    }
    System.out.println("");
}

private static void modified(int[][] greatLoto) {
Random r = new Random(System.currentTimeMillis());
    for (int i = 0; i < greatLoto.length; i++) {
        for (int j = 0; j < greatLoto[i].length; j++) {

            while (Arrays.asList(greatLoto[i]).contains(r)) {
               r = new Random(System.currentTimeMillis());
            }
            greatLoto[i][j] = r.nextInt(maxElem);;

        }
        System.out.println("");

    }

}

assylias · Accepted Answer · 2014-09-25 08:51:07Z

3

This is more of a comment but too long: don't use random.next() because it forces you to check for uniqueness. Instead fill a list with the valid values and shuffle it:

List<Integer> values = new ArrayList<> ();
for (int i = 1; i <= max; i++) values.add(i);
Collections.shuffle(values);

Then you can simply iterate over the values and take the j first numbers.

Note that if j is significantly greater than i using the random approach would probably be more efficient.

edited Sep 25, 2014 at 8:51

answered Sep 25, 2014 at 8:38

assylias

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1 Comment

necromancer Over a year ago

you should mention that this is reasonable only if max is pretty small.

necromancer · Accepted Answer · 2014-09-25 08:39:38Z

0

The most minimal change would be:

private static void modified(int[][] greatLoto) {
Random r = new Random(System.currentTimeMillis());
    for (int i = 0; i < greatLoto.length; i++) {
        for (int j = 0; j < greatLoto[i].length; j++) {
            do {
                greatLoto[i][j] = r.nextInt(maxElem);
            } while (Arrays.asList(greatLoto[i]).contains(greatLoto[i][j]));    
        }
        System.out.println("");
    }

}

But there are more elegant (but difficult to code) ways to generate unique random numbers without discarding duplicates.

answered Sep 25, 2014 at 8:39

necromancer

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1 Comment

Robert Over a year ago

It has to be r.nextInt(maxElem)+1; because OP wants to have the maxElem included...

Robert · Accepted Answer · 2014-09-25 08:47:00Z

0

You need three loops:

Loop_1: Builds an array of size j and uses Loop_1B for every field of this array.

Loop_1B: Generate an int with r.nextInt(maxElem)+1; (it has to be +1 because nextInt() is covering the 0 inclusively and the specified value exclusively). Afterwards check if the number is already used in the array, if yes, run this loop again.

Loop_2: Repeats Loop_1 i times.

edited Sep 25, 2014 at 8:47

answered Sep 25, 2014 at 8:35

Robert

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