Java - Arrays - checking for duplicates in same input

Question

I'm trying to get the secret code from 1 - 40, not to repeat any numbers. How am I able to compare each of them and not get any duplicates?

I have extensively looked through Java documentation and asked my lectures and I can't get a working answer. I understand the concept, what I'm meant to do, but just can't get the coding right.

Here is my code:

public static void main(String[] args) {
 int[] secret = new int[5];
 //int[] secret = {0,0,0,0,0};
 int[][] num = new int[3][5];
 int correctL1 = 0;
 int correctL2 = 0;
 int correctL3 = 0;
 for (int i = 0; i<5; i++){ // to get secret numbers.
 secret[i] = (int) ((Math.random() * (40 - 1)) + 1);
 }
 System.out.println(Arrays.toString(secret));
}

I have tried putting this into the loop to get another number but it's still giving me duplicates.

if(((secret[i] == secret[0]) || (secret[i] == secret[1]) || 
 (secret[i] == secret[2]) || (secret[i] == secret[3]) ||
 (secret[i] == secret[4])) || (secret[i] != 0)) {
 secret[i] = ((int) ((Math.random() * (40 - 1)) + 1));
}

• Timothy Truckle: since Java is an object oriented language you should get used to OO approaches: Java has different Collection types. they roughly split into Lists, which support and order, null elements and duplicates, and Sets, which support uniqueness of elements. So the easieast approach is to collect the input data in an HashSet.

Answer 1

In Java 8, you can use Random#ints to easily generate an array of distinct random numbers:

int[] secret = new Random().ints(1, 40).distinct().limit(5).toArray();

Ideone Demo

Otherwise, you can generate a Set<Integer> by using a while loop:

Set<Integer> secret = new HashSet<>();
Random gen = new Random();
while (secret.size() < 5) {
 secret.add(gen.nextInt(40 - 1) + 1);
}

Answer 2

Simply use a Set<Integer> to keep the already generated numbers and iterate as long as the generated number is part of the generated numbers.

Set<Integer> existing = new HashSet<>();
for (int i = 0; i<5; i++){ // to get secret numbers.
 // Loop until the generated number is not part of the already generated numbers
 int value;
 do {
 value = (int) ((Math.random() * (40 - 1)) + 1);
 } while (!existing.add(value));
 secret[i] = value;
}

• Romain Linsolas: you can simplify with while(existing.size() < 5) { existing.add(X); }

Answer 3

Why not creating a Set<Integer> (so you will not get any duplicate) and iterate until its size is 5 ?

Answer 4

Just use Set Collection..for better performance you can use below code

Set<Integer> data=new HashSet<Integer>();

If you want to get all elements by ascending order,you can use TreeSet

OR Other Logic is Every collection has contains method which returns boolean for e.g.data.contains(o)

Answer 5

while(true){
 int[] secret = new int[5];
 for (int i = 0; i < 5; i++) { // to get secret numbers.
 secret[i] = (int) ((Math.random() * (40 - 1)) + 1);
 }
 System.out.println("Array:" + Arrays.toString(secret));
 Set<Integer> mySet = new HashSet<Integer>();
 for (int i = 0; i < 5; i++) { // to get secret numbers.
 mySet.add(secret[i]);
 }
 if (mySet.size() == secret.length) {
 System.out.println("OK");
 break;
 } else {
 System.out.println("Not ok");
 }
 }