Pick a random position in the word - IllegalArgumentException?

Question

I have to implement the following instructions (pseudocode):

Read a word.
Repeat word.length() times
   Pick a random position i in the word, but not the last position.
   Pick a random position j > i in the word. (this is a tricky point!)
   Swap the letters at positions j and i.
Print the word.

My code throws - IllegalArgumentException here:

j = i + 1 + generator.nextInt( word.length() - i - 1 );

I stack here and don't know how to circumvent this point.

Code:

public String scramble(String word) {
        Random generator = new Random(42);
        int x, i = 0, j = 0, wordLen = word.length();

        for (x = 0; x < wordLen; x++) {                
            i = generator.nextInt(wordLen);            
            j = i + 1 + this.generator.nextInt( word.length() - i - 1 );
        }

I can't find a good solution for this step:

Pick a random position j > i in the word.

How to solve this issue?

Just as an observation, the line `i = generator.nextInt(wordLen);` is wrong, since that can return `wordlen - 1`, which is the last position (and your directions say explictly not to) — Dennis Meng, Jul 04 '13 at 17:24
http://docs.oracle.com/javase/7/docs/api/java/util/Random.html#nextInt(int) - `IllegalArgumentException - if n is not positive` — Brian Roach, Jul 04 '13 at 17:35
@nazar_art Yes, and you don't want to be able to return `n-1` there... — Dennis Meng, Jul 05 '13 at 00:49

score 1 · Accepted Answer · edited Feb 27 '20 at 17:00

1

The line

generator.nextInt( word.length() - i - 1 );

would throw IllegalArgumentException incase the argument <= 0..

From nextInt Docs

Throws: IllegalArgumentException - if n is not positive

Check the length of a string before calling the line

You could do it this way

for (x = 0; x < wordLen; x++) {
      i = generator.nextInt(wordLen - 1); 
      j = generator.nextInt(wordLen);           
      if (j <= i)
        j = i + generator.nextInt(wordLen - i);
}

edited Feb 27 '20 at 17:00

catch23

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answered Jul 04 '13 at 17:19

Anirudha

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Correction: it throws IllegalArgumentException if the argument is less than **or** **equal** **to** zero (the [documentation](http://docs.oracle.com/javase/7/docs/api/java/util/Random.html) says "if n is not positive"). – ajb Jul 04 '13 at 17:33
@Anirudh How to circumvent this issue? – catch23 Jul 04 '13 at 17:39
@Anirudh keep throwing `IllegalArgumentException` exception here - `j = generator.nextInt(i)+(wordLen-i);` – catch23 Jul 04 '13 at 17:47
@nazar_art as i said i should be greater than 0! – Anirudha Jul 04 '13 at 17:49
@Anirudh it keeps throwing – catch23 Jul 04 '13 at 17:52
@Anirudh What is `temp` here (contains)? – catch23 Jul 05 '13 at 06:52

Kon · Answer 2 · 2013-07-04T17:47:25.760

0

String word = "Example"
int len = word.length(); //len = 7
int placeHolder1 = (int) (Math.random() * length); //generates number between 0 and 6
int placeHolder2 = (int) 
int diff = length - placeHolder1;
placeHolder2 = (int) (Math.random() * diff + (length - diff + 1)); //Generates random between placeHolder1 and the length of the String

This gives you the indices of two characters on the String, and you can swap them however you prefer.

edited Jul 04 '13 at 17:47

answered Jul 04 '13 at 17:22

Kon

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You give some "bull" code: `int len = String.length()` how can it work? – catch23 Jul 04 '13 at 17:30
My mistake, I'm typing on my machine w/o JDK, so did not test run code. I meant int len = word.length(), obviously. – Kon Jul 04 '13 at 17:37

Pick a random position in the word - IllegalArgumentException?

2 Answers2