I want to create a set of random numbers without duplicates in Java.
For example I have an array to store 10,000 random integers from 0 to 9999.
Here is what I have so far:
import java.util.Random;
public class Sort{
public static void main(String[] args){
int[] nums = new int[10000];
Random randomGenerator = new Random();
for (int i = 0; i < nums.length; ++i){
nums[i] = randomGenerator.nextInt(10000);
}
}
}
But the above code creates duplicates. How can I make sure the random numbers do not repeat?
Integer[] arr = {...};
Collections.shuffle(Arrays.asList(arr));
For example:
public static void main(String[] args) {
Integer[] arr = new Integer[1000];
for (int i = 0; i < arr.length; i++) {
arr[i] = i;
}
Collections.shuffle(Arrays.asList(arr));
System.out.println(Arrays.toString(arr));
}
A simple algorithm that gives you random numbers without duplicates can be found in the book Programming Pearls p. 127.
Attention: The resulting array contains the numbers in order! If you want them in random order, you have to shuffle the array, either with Fisher–Yates shuffle or by using a List and call Collections.shuffle().
The benefit of this algorithm is that you do not need to create an array with all the possible numbers and the runtime complexity is still linear O(n).
public static int[] sampleRandomNumbersWithoutRepetition(int start, int end, int count) {
Random rng = new Random();
int[] result = new int[count];
int cur = 0;
int remaining = end - start;
for (int i = start; i < end && count > 0; i++) {
double probability = rng.nextDouble();
if (probability < ((double) count) / (double) remaining) {
count--;
result[cur++] = i;
}
remaining--;
}
return result;
}
In Java 8, if you want to have a list of non-repeating N random integers in range (a, b), where b is exclusive, you can use something like this:
Random random = new Random();
List<Integer> randomNumbers = random.ints(a, b).distinct().limit(N).boxed().collect(Collectors.toList());
Achintya Jha has the right idea here. Instead of thinking about how to remove duplicates, you remove the ability for duplicates to be created in the first place.
If you want to stick with an array of ints and want to randomize their order (manually, which is quite simple) follow these steps.
create array of size n.
loop through and initialize each value at index i to the value i (or i+1 if you wish to have the numbers 1 to n rather than 0 to n-1).
finally, loop through the array again swapping each value for a value at a random index.
Your code could be modified to look like this:
import java.util.Random;
public class Sort
{
// use a constant rather than having the "magic number" 10000 scattered about
public static final int N = 10000;
public static void main(String[] args)
{
//array to store N random integers (0 - N-1)
int[] nums = new int[N];
// initialize each value at index i to the value i
for (int i = 0; i < nums.length; ++i)
{
nums[i] = i;
}
Random randomGenerator = new Random();
int randomIndex; // the randomly selected index each time through the loop
int randomValue; // the value at nums[randomIndex] each time through the loop
// randomize order of values
for(int i = 0; i < nums.length; ++i)
{
// select a random index
randomIndex = randomGenerator.nextInt(nums.length);
// swap values
randomValue = nums[randomIndex];
nums[randomIndex] = nums[i];
nums[i] = randomValue;
}
}
}
And if I were you I would likely break each of these blocks into separate, smaller methods rather than having one large main method.
Hope this helps.
If you need generate numbers with intervals, it can be just like that:
Integer[] arr = new Integer[((int) (Math.random() * (16 - 30) + 30))];
for (int i = 0; i < arr.length; i++) {
arr[i] = i;
}
Collections.shuffle(Arrays.asList(arr));
System.out.println(Arrays.toString(arr));`
The result:
[1, 10, 2, 4, 9, 8, 7, 13, 18, 17, 5, 21, 12, 16, 23, 20, 6, 0, 22, 14, 24, 15, 3, 11, 19]
Note:
If you need that the zero does not leave you could put an "if"
How about this?
LinkedHashSet<Integer> test = new LinkedHashSet<Integer>();
Random random = new Random();
do{
test.add(random.nextInt(1000) + 1);
}while(test.size() != 1000);
The user can then iterate through the Set using a for loop.
public class RandomNum {
public static void main(String[] args) {
Random rn = new Random();
HashSet<Integer> hSet = new HashSet<>();
while(hSet.size() != 1000) {
hSet.add(rn.nextInt(1000));
}
System.out.println(hSet);
}
}
If you're using JAVA 8 or more than use stream functionality following way,
Stream.generate(() -> (new Random()).nextInt(10000)).distinct().limit(10000);
Here we Go!
public static int getRandomInt(int lower, int upper) {
if(lower > upper) return 0;
if(lower == upper) return lower;
int difference = upper - lower;
int start = getRandomInt();
//nonneg int in the range 0..difference - 1
start = Math.abs(start) % (difference+1);
start += lower;
return start;
}
public static void main(String[] args){
List<Integer> a= new ArrayList();
int i;
int c=0;
for(;;) {
c++;
i= getRandomInt(100, 500000);
if(!(a.contains(i))) {
a.add(i);
if (c == 10000) break;
System.out.println(i);
}
}
for(int rand : a) {
System.out.println(rand);
}
}
Get Random number Returns a random integer x satisfying lower <= x <= upper. If lower > upper, returns 0. #param lower #param upper #return
In the main method I created list then i check if the random number exist on the list if it doesn't exist i will add the random number to the list
It is very slow but straight forward.
A simple stream solution:
new Random().ints(0, 10000)
.distinct()
.limit(10000)
.forEach(System.out::println);
public class Randoms {
static int z, a = 1111, b = 9999, r;
public static void main(String ... args[])
{
rand();
}
public static void rand() {
Random ran = new Random();
for (int i = 1; i == 1; i++) {
z = ran.nextInt(b - a + 1) + a;
System.out.println(z);
randcheck();
}
}
private static void randcheck() {
for (int i = 3; i >= 0; i--) {
if (z != 0) {
r = z % 10;
arr[i] = r;
z = z / 10;
}
}
for (int i = 0; i <= 3; i++) {
for (int j = i + 1; j <= 3; j++) {
if (arr[i] == arr[j]) {
rand();
}
}
}
}
}
HashSet<Integer>hashSet=new HashSet<>();
Random random = new Random();
//now add random number to this set
while(true)
{
hashSet.add(random.nextInt(1000));
if(hashSet.size()==1000)
break;
}
Related
I have some homework where. I have to write a simple program where 1000000 randoms between 1-100 are generated and stored in an array. After the program has to print out the number of occurrences of 25, 50 and 100. I've been trying for loops but no luck. Till now I have this:
package randomnumbers;
import java.util.Random;
public class random {
public static void main(String[] args) {
Random r = new Random();
int number[] = new int[1000000];
for (int count = 0; count < number.length; count++) {
number[count] = 1 + r.nextInt(100);
}
}
}
public static void main(String[] args) {
Random r = new Random();
int low = 1;
int high = 101;
int number[] = new int[1000000];
int count25 = 0;
int count50 = 0;
int count100 = 0;
for (int i = 0; i < number.length; i++) {
int num = r.nextInt(high - low) + low;
if (num == 25) {
count25++;
} else if (num == 50) {
count50++;
} else if (num == 100) {
count100++;
}
number[i] = num;
}
System.out.println(count25);
System.out.println(count50);
System.out.println(count100);
}
Here r.nextInt(high - low) + low generates random value between lower bound(inclusive) and upper bound(exclusive). Hence I have used upper bound as 101.
There are 3 different counters which would increment by one if its value matches the required value(25, 50 and 100).
After the loop is completed the value of the counter can be printed.
You can use IntStream.count() method for this purpose. Your code might look something like this:
public static void main(String[] args) {
// generate array of 1,000,000 elements
int[] arr = IntStream.range(0, 1000000)
// random value from 1 to 100
.map(i -> (int) (1 + Math.random() * 100))
.toArray();
// test output
System.out.println(countOf(arr, 0)); // 0
System.out.println(countOf(arr, 1)); // 10137
System.out.println(countOf(arr, 25)); // 10026
System.out.println(countOf(arr, 50)); // 10103
System.out.println(countOf(arr, 100)); // 10100
System.out.println(countOf(null,100)); // 0
System.out.println(countOf(arr, 101)); // 0
}
private static long countOf(int[] arr, int value) {
if (arr == null)
return 0;
else
return Arrays.stream(arr).filter(i -> i == value).count();
}
See also:
• How to find duplicate elements in array in effective way?
• How to get elements of an array that is not null?
For my program I need to make a 10 element array and the goal is to print out the even numbers from 2 to 20. I have to do this by adding 2 to the beginning element. This is what I have so far. I think I should use a loop as shown but I don't know how to go about adding 2 and printing that out. Thanks!
int[] array = new int[10];
for(int counter=0; counter<array.length; counter++) {
}
if you want program print even number between 2 & 20
for(int i=2;i<=20;i++)
{
if(i%2 == 0)
print(i)
}
Start at 2 and increment by 2 to get the even numbers:
int[] array = new int[10]
for(int counter=2; counter <= 20; counter += 2) {
array[counter/2 - 1] = counter
}
or
int[] array = new int[10]
for(int i=0; i <= 10; i++) {
array[i] = i*2 + 2
}
this is also an option
int i, value;
int nums[] = new int[10];
for (i = 0, value = 2; i < nums.length; value = value + 2, i = i + 1) {
nums[i] = value;
}
for (i = 0; i < nums.length; i++) {
System.out.println(nums[i]);
}
int[] array = new int[10];
for (int i = 0, j = 1; i < array.length && j <= 20; j++) {
if (j % 2 == 0) {
array[i] = j;
i++;
}
}
System.out.println(Arrays.toString(array));
Java 8: int[] array = IntStream.range(1, 11).map(x -> x * 2).toArray();
Or, to just print: IntStream.range(1, 11).map(x -> x * 2).forEach(System.out::println);
From java-9 you can use IntStream.iterate to create int array
int[] arr= IntStream.iterate(2, i->i<=20, i->i+2).toArray();
for Integer array you can use Stream.iterate
Integer[] ary = Stream.iterate(2, i->i<=20, i->i+2).toArray(Integer[]::new);
Dear Alexis,
Below is an example using do-while.
you can simply define a range of expected even numbers using minVal and maxVal.
Program will execute and return list of ordered even numbers for given range. Assuming input values are correct even numbers. You can improve to apply validations.
public class EvenNumberGenerator {
static int minVal=2; //enter valid min value even number
static int maxVal = 20; //enter valid max value even number
public static void main(String[] args) {
List<Integer> evenNumbers = new ArrayList();
do {
if(minVal % 2 == 0) {
evenNumbers.add(minVal);
}
minVal++;
} while (!evenNumbers.contains(maxVal));
System.out.println(evenNumbers);
// evenNumbers.toArray(); in case you need an array
}
}
OUTPUT
[2, 4, 6, 8, 10, 12, 14, 16, 18, 20]
Hope it helps!
Using your code as a start, this will work:
int[] array = new int[10];
for(int counter=0; counter<array.length; counter++) {
array[counter] = (counter + 1) * 2;
}
System.out.println(Arrays.toString(array));
The following will also work with Eclipse Collections:
int[] array = IntInterval.evensFromTo(2, 20).toArray();
System.out.println(Arrays.toString(array));
Note: I am a committer for Eclipse Collections
In this code I have found duplicates from an array and I want to remove them. The output then will be unique generated numbers. I am required to use math.random and modulo. Anyone have any clues? I tried to store them in an array but then the original array has 0's and 0 is part of my domain for the random number generation (from 0 to 52).
public class Decks {
public static void main(String[] args) {
generate();
}
public static void generate() {
int deckOfCard[] = new int[52];
for (int counts = 0; counts < 52; counts++) {
deckOfCard[counts] = (int) (Math.random() * 51);
}
for (int i = 0; i < deckOfCard.length - 1; i++) {
for (int j = i + 1; j < deckOfCard.length; j++) {
if ((deckOfCard[i] == (deckOfCard[j])) && (i != j)) {
System.out.println("DUPLICATE " + deckOfCard[i]);
}
}
}
for (int count = 0; count < deckOfCard.length; count++) {
System.out.print("\t" + deckOfCard[count]);
}
}
Why dont you try using HashSet instead of arrays ? As you know sets only store unique values so you wont have any duplicates.
You must validate the numbers generated during the random number generation like this:
import java.util.Random;
public class Decks {
public static void main(String[] args) {
Random myRandom = new Random();
int[] num = new int[53];
boolean[] check = new boolean[53];
int all = 0;
int ranNum;
while (all < 53) {
ranNum = myRandom.nextInt(53);
if (!check[ranNum]) {
check[ranNum] = true;
num[all] = ranNum;
all++;
}
}
for (int i = 0; i < 53; i++) {
System.out.println(num[i]);
}
}
}
I suggest not including number 0 because it does not exist in a real deck of cards (ACE being the lowest having the number value of 1). I just included it right here because in my understanding, 0 is included in your desired output.
Considering time complexity, you can sort them first, which at best case takes nlogn time, and then use O(1) to find duplicated elements out.
So I created an array with random numbers, i printed and counted the repeated numbers, now I just have to create a new array with the same numbers from the first array but without any repetitions. Can't use ArrayList by the way.
What I have is.
public static void main(String[] args) {
Random generator = new Random();
int aR[]= new int[20];
for(int i=0;i<aR.length;i++){
int number=generator.nextInt(51);
aR[i]=number;
System.out.print(aR[i]+" ");
}
System.out.println();
System.out.println();
int countRep=0;
for(int i=0;i<aR.length;i++){
for(int j=i+1;j<aR.length-1;j++){
if(aR[i]==aR[j]){
countRep++;
System.out.println(aR[i]+" "+aR[j]);
break;
}
}
}
System.out.println();
System.out.println("Repeated numbers: "+countRep);
int newaR[]= new int[aR.length - countRep];
}
Can someone help?
EDIT: Can't really use HashSet either. Also the new array needs to have the correct size.
Using Java 8 and streams you can do the following:
int[] array = new int[1024];
//fill array
int[] arrayWithoutDuplicates = Arrays.stream(array)
.distinct()
.toArray();
This will:
Turn your int[] into an IntStream.
Filter out all duplicates, so retaining distinct elements.
Save it in a new array of type int[].
Try:
Set<Integer> insertedNumbers = new HashSet<>(newaR.length);
int index = 0;
for(int i = 0 ; i < aR.length ; ++i) {
if(!insertedNumbers.contains(aR[i])) {
newaR[index++] = aR[i];
}
insertedNumbers.add(aR[i]);
}
One possible approach is to walk through the array, and for each value, compute the index at which it again occurs in the array (which is -1 if the number does not occur again). The number of values which do not occur again is the number of unique values. Then collect all values from the array for which the corresponding index is -1.
import java.util.Arrays;
import java.util.Random;
public class UniqueIntTest
{
public static void main(String[] args)
{
int array[] = createRandomArray(20, 0, 51);
System.out.println("Array " + Arrays.toString(array));
int result[] = computeUnique(array);
System.out.println("Result " + Arrays.toString(result));
}
private static int[] createRandomArray(int size, int min, int max)
{
Random random = new Random(1);
int array[] = new int[size];
for (int i = 0; i < size; i++)
{
array[i] = min + random.nextInt(max - min);
}
return array;
}
private static int[] computeUnique(int array[])
{
int indices[] = new int[array.length];
int unique = computeIndices(array, indices);
int result[] = new int[unique];
int index = 0;
for (int i = 0; i < array.length; i++)
{
if (indices[i] == -1)
{
result[index] = array[i];
index++;
}
}
return result;
}
private static int computeIndices(int array[], int indices[])
{
int unique = 0;
for (int i = 0; i < array.length; i++)
{
int value = array[i];
int index = indexOf(array, value, i + 1);
if (index == -1)
{
unique++;
}
indices[i] = index;
}
return unique;
}
private static int indexOf(int array[], int value, int offset)
{
for (int i = offset; i < array.length; i++)
{
if (array[i] == value)
{
return i;
}
}
return -1;
}
}
This sounds like a homework question, and if it is, the technique that you should pick up on is to sort the array first.
Once the array is sorted, duplicate entries will be adjacent to each other, so they are trivial to find:
int[] numbers = //obtain this however you normally would
java.util.Arrays.sort(numbers);
//find out how big the array is
int sizeWithoutDuplicates = 1; //there will be at least one entry
int lastValue = numbers[0];
//a number in the array is unique (or a first duplicate)
//if it's not equal to the number before it
for(int i = 1; i < numbers.length; i++) {
if (numbers[i] != lastValue) {
lastValue = i;
sizeWithoutDuplicates++;
}
}
//now we know how many results we have, and we can allocate the result array
int[] result = new int[sizeWithoutDuplicates];
//fill the result array
int positionInResult = 1; //there will be at least one entry
result[0] = numbers[0];
lastValue = numbers[0];
for(int i = 1; i < numbers.length; i++) {
if (numbers[i] != lastValue) {
lastValue = i;
result[positionInResult] = i;
positionInResult++;
}
}
//result contains the unique numbers
Not being able to use a list means that we have to figure out how big the array is going to be in a separate pass — if we could use an ArrayList to collect the results we would have only needed a single loop through the array of numbers.
This approach is faster (O(n log n) vs O (n^2)) than a doubly-nested loop through the array to find duplicates. Using a HashSet would be faster still, at O(n).
I would like to generate 6 numbers inside an array and at the same time, having it compared so it will not be the same or no repeating numbers. For example, I want to generate 1-2-3-4-5-6 in any order, and most importantly without repeating. So what I thought is to compare current array in generated array one by one and if the number repeats, it will re-run the method and randomize a number again so it will avoid repeating of numbers.
Here is my code:
import javax.swing.*;
public class NonRepeat
{
public static void main(String args[])
{
int Array[] = new int [6];
int login = Integer.parseInt(JOptionPane.showInputDialog("ASD"));
while(login != 0)
{
String output="";
for(int index = 0; index<6; index++)
{
Array[index] = numGen();
for(int loop = 0; loop <6 ; loop++)
{
if(Array[index] == Array[loop])
{
Array[index] = numGen();
}
}
}
for(int index = 0; index<6; index++)
{
output += Array[index] + " ";
}
JOptionPane.showMessageDialog(null, output);
}
}
public static int numGen()
{
int random = (int)(1+Math.random()*6);
return random;
}
}
I've been thinking it for 2 hours and still cant generate 6 numbers without repeating.
Hope my question will be answered.
Btw, Im new in codes so please I just want to compare it using for loop or while loop and if else.
You can generate numbers from, say, 1 to 6 (see below for another solution) then do a Collections.shuffle to shuffle your numbers.
final List<Integer> l = new ArrayList<Integer>();
for (int j = 1; j < 7; j++ ) {
l.add( j );
}
Collections.shuffle( l );
By doing this you'll end up with a randomized list of numbers from 1 to 6 without having twice the same number.
If we decompose the solution, first you have this, which really just create a list of six numbers:
final List<Integer> l = new ArrayList<Integer>();
for (int j = 1; j < 7; j++ ) {
l.add( j );
}
So at this point you have the list 1-2-3-4-5-6 you mentioned in your question. You're guaranteed that these numbers are non-repeating.
Then you simply shuffle / randomize that list by swapping each element at least once with another element. This is what the Collections.shuffle method does.
The solutions that you suggested isn't going to be very efficient: depending on how big your list of numbers is and on your range, you may have a very high probability of having duplicate numbers. In that case constantly re-trying to generate a new list will be slow. Moreover any other solution suggesting to check if the list already contains a number to prevent duplicate or to use a set is going to be slow if you have a long list of consecutive number (say a list of 100 000 numbers from 1 to 100 000): you'd constantly be trying to randomly generate numbers which haven't been generated yet and you'd have more and more collisions as your list of numbers grows.
If you do not want to use Collections.shuffle (for example for learning purpose), you may still want to use the same idea: first create your list of numbers by making sure there aren't any duplicates and then do a for loop which randomly swap two elements of your list. You may want to look at the source code of the Collections.shuffle method which does shuffle in a correct manner.
EDIT It's not very clear what the properties of your "random numbers" have to be. If you don't want them incremental from 1 to 6, you could do something like this:
final Random r = new Random();
final List<Integer> l = new ArrayList<Integer>();
for (int j = 0; j < 6; j++ ) {
final int prev = j == 0 ? 0 : l.get(l.size() - 1);
l.add( prev + 1 + r.nextInt(42) );
}
Collections.shuffle( l );
Note that by changing r.nextInt(42) to r.nextInt(1) you'll effectively get non-repeating numbers from 1 to 6.
You have to check if the number already exist, you could easily do that by putting your numbers in a List, so you have access to the method contains. If you insist on using an array then you could make a loop which checks if the number is already in the array.
Using ArrayList:
ArrayList numbers = new ArrayList();
while(numbers.size() < 6) {
int random = numGen(); //this is your method to return a random int
if(!numbers.contains(random))
numbers.add(random);
}
Using array:
int[] numbers = new int[6];
for (int i = 0; i < numbers.length; i++) {
int random = 0;
/*
* This line executes an empty while until numGen returns a number
* that is not in the array numbers yet, and assigns it to random
*/
while (contains(numbers, random = numGen()))
;
numbers[i] = random;
}
And add this method somewhere as its used in the snippet above
private static boolean contains(int[] numbers, int num) {
for (int i = 0; i < numbers.length; i++) {
if (numbers[i] == num) {
return true;
}
}
return false;
}
Here is the solution according to your code -
You just need to change the numGen method -
public static int numGen(int Array[])
{
int random = (int)(1+Math.random()*6);
for(int loop = 0; loop <Array.length ; loop++)
{
if(Array[loop] == random)
{
return numGen(Array);
}
}
return random;
}
Complete code is -
import javax.swing.*;
public class NonRepeat
{
public static void main(String args[])
{
int login = Integer.parseInt(JOptionPane.showInputDialog("ASD"));
while(login != 0)
{
int Array[] = new int [6];
String output="";
for(int index = 0; index<6; index++)
{
Array[index] = numGen(Array);
}
for(int index = 0; index<6; index++)
{
output += Array[index] + " ";
}
JOptionPane.showMessageDialog(null, output);
}
}
public static int numGen(int Array[])
{
int random = (int)(1+Math.random()*6);
for(int loop = 0; loop <Array.length ; loop++)
{
if(Array[loop] == random)
{
return numGen(Array);
}
}
return random;
}
}
Use List instead of array and List#contains to check if number is repeated.
you can use a boolean in a while loop to identify duplicates and regenerate
int[] array = new int[10]; // array of length 10
Random rand = new Random();
for (int i = 0 ; i < array.length ; i ++ ) {
array[i] = rand.nextInt(20)+1; // random 1-20
boolean found = true;
while (found) {
found = false;
// if we do not find true throughout the loop it will break (no duplicates)
int check = array[i]; // check for duplicate
for (int j = 0 ; j < i ; j ++) {
if ( array[j] == check ) {
found = true; // found duplicate
}
}
if (found) {
array[i] = rand.nextInt(20)+1 ; // replace
}
}
}
System.out.println(Arrays.toString(array));
You may use java.util.Random. And please specify if you want any random number or just the number 1,2,3,4,5,6. If you wish random numbers then , this is a basic code:
import java.util.*;
public class randomnumber
{
public static void main(String[] args)
{
Random abc = new Random();
int[] a = new int[6];
int limit = 100,c=0;
int chk = 0;
boolean y = true;
for(;c < 6;)
{
int x = abc.nextInt(limit+1);
for(int i = 0;i<a.length;i++)
{
if(x==a[i])
{
y=false;
break;
}
}
if(y)
{
if(c!=0)if(x == (a[c-1]+1))continue;
a[c]=x;
c++;
}
}
for (Integer number : a)
{
System.out.println(number);
}
}
}
if you don't understand the last for loop , please tell , i will update it.
Use List and .contains(Object obj) method.
So you can verify if list has the random number add before.
update - based on time you can lost stuck in random loop.
List<Integer> list = new ArrayList<Integer>();
int x = 1;
while(x < 7){
list.add(x);
x++;
}
Collections.shuffle(list);
for (Integer number : list) {
System.out.println(number);
}
http://docs.oracle.com/javase/7/docs/api/java/util/List.html#contains(java.lang.Object)