How to generate random numbers from 0 to 100 without duplicates when user enters the size of the array in Java?

user_15868797 0
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Write a program to generate any number of random integers in 0 to 100 range. Your program should get the size as a parameter and return the numbers as an array.

This the question I got and below is the code I tried, but I get duplicates with this code. How to generate random numbers without any duplicates?

    import java.util.Arrays;
    import java.util.Random;
    import java.util.Scanner;
    
    public class Main {
    
        public static void main(String[] args) {
            Scanner sc = new Scanner(System.in);
            
            System.out.println("size:\t");
            int n = sc.nextInt();
    
            int[] arr = new int[n];
    
            Random rand = new Random(); //instance of a random class.
            int upperbound = 101; //generate random values from 0-100
    
            for(int i=0;i<n;i++) {
                arr[i] = rand.nextInt(upperbound);
            }
            System.out.println("Random Numbers: "+ Arrays.toString(arr));
            
        }
    }
linear-search data-structures sorting algorithm java
5 Answers

You can create a list that contains all values from 0 to 100, then shuffle it and extract the first n elements of that list like this:

public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        System.out.println("size:\t");
        int n = sc.nextInt();

        int bound = 100;
        
        List<Integer> items = IntStream.range(0, bound + 1)
                .boxed()
                .collect(Collectors.toList());

        Collections.shuffle(items);

        int[] arr = new int[Math.min(n, bound + 1)];
        for(int i = 0; i < arr.length; i++){
            arr[i] = items.get(i);
        }

        System.out.println("Random Numbers: "+ Arrays.toString(arr));
    }
user_40118900

You can use a HashSet and keep adding all the new entries to it and while generating a new random integer, check if it was already generated (if the number is present in the unique set).
If the number was already generated, then keep generating new number until a unique number is found.

        Set<Integer> unique = new HashSet<Integer>();
        for (int i=0; i<100; i++)
        {
            int number = rand.nextInt(upperbound);
            while (unique.contains(number)) // `number` is a duplicate
                number = rand.nextInt(upperbound);
            arr[i] = number;
            unique.add(number); // adding to the set
        }
user_138824250

You can create a list of 1...100 (let's call in "l") , and an array of size 100 (let's call it "a").

Then, create a for loop and generate a number between 0 to l.size() - 1.

First iteration generation (let's call it index0) will be the index in l that will be taken as the first object in the array. So call l.remove(index0), and use the retrieved object in a[0].

Second iteration generation, between 0 to l.size()-1, which means 0 to 98. That will be index1. call l.remove(index1), the retrieved object will be used to the second index in a[1].

Do it until l is empty.

user_127859940

this code work as well

public void solution(int size){


    int upperbound = 101;
    int arr[]=new int[upperbound];

    /**
     * initial a number arr store 0-100
     */
    for (int i = 0; i <upperbound ; i++) {
        arr[i]=i;
    }

    SecureRandom secureRandom=new SecureRandom();
    for (int i = 0; i <size ; i++) {

        /**
         * if generate a index, print number and swap now index number and tail
         */
        int index=secureRandom.nextInt(upperbound-i);
        int tmp=arr[index];
        int last=arr[arr.length-1-i];

        System.out.println(tmp);
        arr[index]=last;
    }

}
user_83171120

Instead of using an Array you could use an ArrayList and add the value to the ArrayList only if it has not been used before.

ArrayList<Integer> numbers = new ArrayList<Integer>();
    Random random = new Random();
if (!numbers.contains(randomNumber)) {
    numbers.add(randomNumber);
}

After this you could directly the print the ArrayList or convert it into an array and then print it out.

Integer[] arr = new Integer[number.size()];
    // ArrayList to Array Conversion
    for (int i = 0; i < number.size(); i++) {
        arr[i] = number.get(i);
user_162244010
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