How to organize Random without repetition in Java?

Question

Java

There are numbers from 0 to 23. How to make it so that when you click on the button, a random number is generated that has not been encountered at least 2 more times?

Answer 1

Mb to make an array with the numbers that fell out, then shuffle array?

Answer 2

spoiler

import java.lang.Math;
import java.util.*;
public class SuperRandom
{
  
  private ArrayList<Integer> numbers;
  private Random rand;
  
  	public static void main(String[] args)
  	{
      		SuperRandom sr=new SuperRandom(10);
      		for(int i=0;i<10;i++)
    		System.out.print(sr.getValue());
  	}
  
  	public SuperRandom(int size){
    rand=new Random();
    numbers=new ArrayList<>();
    		for(int i=0;i<size;i++)
    		numbers.add(i);
 	 }
  
  	public int getValue(){
     		int randomIndex=rand.nextInt(numbers.size());
     		return numbers.remove(randomIndex);
 	}
  
  	public int getSize(){
  		return numbers.size();
 	}
  
}

An alternative solution, just think for yourself how you best handle the end of the values.

Answer 3

Create an array of 46 elements, fill with numbers, mix, then bite out of it one value at a time, as soon as the length becomes zero, refill the array well, or whatever it should be logically after the
upd. If Java is so godless about memory, then instead of biting out, you can have a separate incremental counter

Answer 4

Remember the last 2 generated numbers.
When you press the button, start the generation cycle until the variant differs from both stored ones. Well, shift the saved pair by 1 for each new number found.