# How do I generate random integers within a specific range in Java?

Note that this approach is more biased and less efficient than a `nextInt`

approach, https://stackoverflow.com/a/738651/360211

One standard pattern for accomplishing this is:

```
Min + (int)(Math.random() * ((Max - Min) + 1))
```

The Java Math library function Math.random() generates a double value in the range `[0,1)`

. Notice this range does not include the 1.

In order to get a specific range of values first, you need to multiply by the magnitude of the range of values you want covered.

```
Math.random() * ( Max - Min )
```

This returns a value in the range `[0,Max-Min)`

, where 'Max-Min' is not included.

For example, if you want `[5,10]`

, you need to cover five integer values so you use

```
Math.random() * 5
```

This would return a value in the range `[0,5)`

, where 5 is not included.

Now you need to shift this range up to the range that you are targeting. You do this by adding the Min value.

```
Min + (Math.random() * (Max - Min))
```

You now will get a value in the range `[Min,Max)`

. Following our example, that means `[5,10)`

:

```
5 + (Math.random() * (10 - 5))
```

But, this still doesn't include `Max`

and you are getting a double value. In order to get the `Max`

value included, you need to add 1 to your range parameter `(Max - Min)`

and then truncate the decimal part by casting to an int. This is accomplished via:

```
Min + (int)(Math.random() * ((Max - Min) + 1))
```

And there you have it. A random integer value in the range `[Min,Max]`

, or per the example `[5,10]`

:

```
5 + (int)(Math.random() * ((10 - 5) + 1))
```

How do I generate a random `int`

value in a specific range?

I have tried the following, but those do not work:

**Attempt 1:**

```
randomNum = minimum + (int)(Math.random() * maximum);
// Bug: `randomNum` can be bigger than `maximum`.
```

**Attempt 2:**

```
Random rn = new Random();
int n = maximum - minimum + 1;
int i = rn.nextInt() % n;
randomNum = minimum + i;
// Bug: `randomNum` can be smaller than `minimum`.
```

Just use the Random class:

```
Random ran = new Random();
// Assumes max and min are non-negative.
int randomInt = min + ran.nextInt(max - min + 1);
```

Generate a random number for the difference of min and max by using the nextint(n) method and then add min number to the result:

```
Random rn = new Random();
int result = rn.nextInt(max - min + 1) + min;
System.out.println(result);
```

```
rand.nextInt((max+1) - min) + min;
```

ThreadLocalRandom equivalent of class java.util.Random for multithreaded environment. Generating a random number is carried out locally in each of the threads. So we have a better performance by reducing the conflicts.

```
int rand = ThreadLocalRandom.current().nextInt(x,y);
```

x,y - intervals e.g. (1,10)

You can achieve that concisely in Java 8:

```
Random random = new Random();
int max = 10;
int min = 5;
int totalNumber = 10;
IntStream stream = random.ints(totalNumber, min, max);
stream.forEach(System.out::println);
```

It's better to use SecureRandom rather than just Random.

```
public static int generateRandomInteger(int min, int max) {
SecureRandom rand = new SecureRandom();
rand.setSeed(new Date().getTime());
int randomNum = rand.nextInt((max - min) + 1) + min;
return randomNum;
}
```

Use:

```
minimum + rn.nextInt(maxValue - minvalue + 1)
```

```
public static Random RANDOM = new Random(System.nanoTime());
public static final float random(final float pMin, final float pMax) {
return pMin + RANDOM.nextFloat() * (pMax - pMin);
}
```

```
import java.util.Random;
public class RandomUtil {
// Declare as class variable so that it is not re-seeded every call
private static Random random = new Random();
/**
* Returns a psuedo-random number between min and max (both inclusive)
* @param min Minimim value
* @param max Maximim value. Must be greater than min.
* @return Integer between min and max (both inclusive)
* @see java.util.Random#nextInt(int)
*/
public static int nextInt(int min, int max) {
// nextInt is normally exclusive of the top value,
// so add 1 to make it inclusive
return random.nextInt((max - min) + 1) + min;
}
}
```

```
int random = minimum + Double.valueOf(Math.random()*(maximum-minimun)).intValue();
```

Or take a look to RandomUtils from Apache Commons.

You can edit your second code example to:

```
Random rn = new Random();
int range = maximum - minimum + 1;
int randomNum = rn.nextInt(range) + minimum;
```

```
rand.nextInt((max+1) - min) + min;
```

This is working fine.

This methods might be convenient to use:

This method will return a random number ** between** the provided min and max value:

```
public static int getRandomNumberBetween(int min, int max) {
Random foo = new Random();
int randomNumber = foo.nextInt(max - min) + min;
if (randomNumber == min) {
// Since the random number is between the min and max values, simply add 1
return min + 1;
} else {
return randomNumber;
}
}
```

and this method will return a random number ** from** the provided min and max value (so the generated number could also be the min or max number):

```
public static int getRandomNumberFrom(int min, int max) {
Random foo = new Random();
int randomNumber = foo.nextInt((max + 1) - min) + min;
return randomNumber;
}
```

Forgive me for being fastidious, but the solution suggested by the majority, i.e., `min + rng.nextInt(max - min + 1))`

, seems perilous due to the fact that:

`rng.nextInt(n)`

cannot reach`Integer.MAX_VALUE`

.`(max - min)`

may cause overflow when`min`

is negative.

A foolproof solution would return correct results for any `min <= max`

within [`Integer.MIN_VALUE`

, `Integer.MAX_VALUE`

]. Consider the following naive implementation:

```
int nextIntInRange(int min, int max, Random rng) {
if (min > max) {
throw new IllegalArgumentException("Cannot draw random int from invalid range [" + min + ", " + max + "].");
}
int diff = max - min;
if (diff >= 0 && diff != Integer.MAX_VALUE) {
return (min + rng.nextInt(diff + 1));
}
int i;
do {
i = rng.nextInt();
} while (i < min || i > max);
return i;
}
```

Although inefficient, note that the probability of success in the `while`

loop will always be 50% or higher.