I'm trying to generate a random distinct 3 or 4 digit number

Question

I'm trying to generate a random distinct 3 or 4 digit number in Java, without using loops, arrays, or methods( except the in built math random method for generating random numbers), just the basic stuff like if statements.

I tried

(int)(Math.random()*9899+100);

But it's only returning 4 digit indistinct numbers. By distinct I mean, no number should repeat itself. Numbers like 232, 7277, 889, 191 aren't allowed. But numbers like 1234, 3456, 8976, 435 are distinct and allowed

Answer 1

The following is not going to generate the numbers fairly but any 4 digit number can be generated, 3 digit numbers are not going to have 0 in them.

// first digit between 0 and 9, if it's 0 we will have a 3 digit number
int first = (int)(Math.random() * 10);  

int second = (int)(Math.random() * 10);
// if it's the same as the first add one to it, use modulo to wrap around
if (first == second)
    second = (second + 1) % 10;

// the same as with second digit just with more checks
int third = (int)(Math.random() * 10);
if (third == first || third == second)
    third = (third + 1) % 10;
// if it's still the same, add one once more, after doing it a second time new digit is guaranteed 
if (third == first || third == second)
    third = (third + 1) % 10;

// the same as with second digit just with more checks
int fourth = (int)(Math.random() * 10);
if (fourth == first || fourth == second || fourth == third)
    fourth = (fourth + 1) % 10;
if (fourth == first || fourth == second || fourth == third)
    fourth = (fourth + 1) % 10;
if (fourth == first || fourth == second || fourth == third)
    fourth = (fourth + 1) % 10;

System.out.println(1000 * first + 100 * second + 10 * third + fourth);

You can add the new digit to itself instead of adding one to have a more uniform distribution, it requires handling of some corner cases.