CODEWITHSUNDEEP
javarandomnumbers
kakabali
kakabali

Reputation: 4033

Creating a 4 digit Random Number using java with no repetition in digits

I wrote a code using java to create a random 4 digit number with no repetition of digits, the code I wrote is given below :-

Random r = new Random();
d1 = r.nextInt(9);
d2 = r.nextInt(9);
d3 = r.nextInt(9);
d4 = r.nextInt(9);
while(d1==d2||d1==d3||d1==d4||d2==d3||d2==d4||d3==d4)
{
    if(d1==d2||d2==d3||d2==d4)
    {
        d2 = r.nextInt(9);
    }
    if(d1==d3||d2==d3||d3==d4)
    {
        d3 = r.nextInt(9);
    }
    if(d1==d4||d2==d4||d3==d4)
    {
        d4 = r.nextInt(9);
    }
}   
System.out.println(d1+""+d2+""+d3+""+d4);


here are the test cases(generated from System.out.println(R1+""+R2+""+R3+""+R4);) are as following :-

 0123 |  OK as required
 1234 |  OK as required
 2123 |  not OK because 2 is present more than one time 
 9870 |  OK as required
 0444 |  not OK because 4 is present more than one time


Now My question here is, that if there is some better way to do this. If I could enhance it in some way?

Upvotes: 8

Views: 22768

Answers (7)

Jeroen Vannevel
Jeroen Vannevel

Reputation: 44448

Create a list of integers from 0 to 9, shuffle it and extract the first 4. 

public static void main(String[] args) {
    List<Integer> numbers = new ArrayList<>();
    for(int i = 0; i < 10; i++){
        numbers.add(i);
    }

    Collections.shuffle(numbers);

    String result = "";
    for(int i = 0; i < 4; i++){
        result += numbers.get(i).toString();
    }
    System.out.println(result);
}

There's some ugly string-to-int conversing going on, but you get the idea. Depending on your use case you can see what is needed.

Upvotes: 18

Hot Licks
Hot Licks

Reputation: 47739

Roughly (not tested): 

int randomNum = r.nextInt(5040);
int firstDigit = randomNum % 10;
randomNum = randomNum / 10;
int secondDigit = randomNum % 9;
randomNum = randomNum / 9;
int thirdDigit = randomNum % 8;
randomNum = randomNum / 8;
int fourthDigit = randomNum % 7;

if (secondDigit == firstDigit) {
  secondDigit++;
}

while ((thirdDigit == firstDigit) || (thirdDigit == secondDigit)) {
  thirdDigit++:
}

while ((fourthDigit == firstDigit) || (fourthDigit == secondDigit) || (fourthDigit == thirdDigit)) {
  fourthDigit++;
}

(After coding this I realized that the increment operations need to be done modulo 10.)

Upvotes: 1

MZ4Code
MZ4Code

Reputation: 76

Here's my approach, even though it uses a lot of string parsing but no data structure:

 static int generateNumber(int length){
            String result = "";
            int random;
            while(true){
                random  = (int) ((Math.random() * (10 )));
                if(result.length() == 0 && random == 0){//when parsed this insures that the number doesn't start with 0
                    random+=1;
                    result+=random;
                }
                else if(!result.contains(Integer.toString(random))){//if my result doesn't contain the new generated digit then I add it to the result
                    result+=Integer.toString(random);
                }
                if(result.length()>=length){//when i reach the number of digits desired i break out of the loop and return the final result
                    break;
                }
            }

            return Integer.parseInt(result);
        }

Upvotes: 4

Juvanis
Juvanis

Reputation: 25950

Here is my solution without using any additional data structure, looping on generating random number till it has unique digits.

int a = 0, b = 0, c = 0, d = 0;
int x = 0;
while (true) {
    x = r.nextInt(9000) + 1000;
    a = x % 10;
    b = (x / 10) % 10;
    c = (x / 100) % 10;
    d = x / 1000;
    if (a == b || a == c || a == d || b == c || b == d || c == d)
        continue;
    else
        break;
}

System.out.println(x);

Upvotes: 1

Algiz
Algiz

Reputation: 1288

Create a list with Integer from 0 to 9 (so 10 items in total)

List<Integer> l = ...
Collections.shuffle(l);
d1 = l.get(0);
d2 = l.get(1);
d3 = l.get(2);
d4 = l.get(3);

Upvotes: 1

Rob
Rob

Reputation: 6497

A couple of approaches:

  1. Use a Set to hold the digits and keep adding random digits until the set has four values in it.

  2. Create an array with values 0-9 in it. Shuffle the array and take the first four values.

If performance matters, you will want to try a couple of different methods and see which is faster.

Upvotes: 5

fred02138
fred02138

Reputation: 3371

Use a Set maybe?

Random r = new Random();
Set<Integer> s = new HashSet<Integer>();
while (s.size() < 4) {
    s.add(r.nextInt(9));
}

Upvotes: 2

Related Questions