Develop Reference

How to increace propability of a specific number in Random() in java [duplicate]

I want to choose a random item from a set, but the chance of choosing any item should be proportional to the associated weight
Example inputs:
item weight
---- ------
sword of misery 10
shield of happy 5
potion of dying 6
triple-edged sword 1
So, if I have 4 possible items, the chance of getting any one item without weights would be 1 in 4.
In this case, a user should be 10 times more likely to get the sword of misery than the triple-edged sword.
How do I make a weighted random selection in Java?

I would use a NavigableMap
public class RandomCollection<E> {
private final NavigableMap<Double, E> map = new TreeMap<Double, E>();
private final Random random;
private double total = 0;
public RandomCollection() {
this(new Random());
}
public RandomCollection(Random random) {
this.random = random;
}
public RandomCollection<E> add(double weight, E result) {
if (weight <= 0) return this;
total += weight;
map.put(total, result);
return this;
}
public E next() {
double value = random.nextDouble() * total;
return map.higherEntry(value).getValue();
}
}
Say I have a list of animals dog, cat, horse with probabilities as 40%, 35%, 25% respectively
RandomCollection<String> rc = new RandomCollection<>()
.add(40, "dog").add(35, "cat").add(25, "horse");
for (int i = 0; i < 10; i++) {
System.out.println(rc.next());
}

There is now a class for this in Apache Commons: EnumeratedDistribution
Item selectedItem = new EnumeratedDistribution<>(itemWeights).sample();
where itemWeights is a List<Pair<Item, Double>>, like (assuming Item interface in Arne's answer):
final List<Pair<Item, Double>> itemWeights = Collections.newArrayList();
for (Item i: itemSet) {
itemWeights.add(new Pair(i, i.getWeight()));
}
or in Java 8:
itemSet.stream().map(i -> new Pair(i, i.getWeight())).collect(toList());
Note: Pair here needs to be org.apache.commons.math3.util.Pair, not org.apache.commons.lang3.tuple.Pair.

You will not find a framework for this kind of problem, as the requested functionality is nothing more then a simple function. Do something like this:
interface Item {
double getWeight();
}
class RandomItemChooser {
public Item chooseOnWeight(List<Item> items) {
double completeWeight = 0.0;
for (Item item : items)
completeWeight += item.getWeight();
double r = Math.random() * completeWeight;
double countWeight = 0.0;
for (Item item : items) {
countWeight += item.getWeight();
if (countWeight >= r)
return item;
}
throw new RuntimeException("Should never be shown.");
}
}

139
There is a straightforward algorithm for picking an item at random, where items have individual weights:
calculate the sum of all the weights
pick a random number that is 0 or greater and is less than the sum of the weights
go through the items one at a time, subtracting their weight from your random number until you get the item where the random number is less than that item's weight

Use an alias method
If you're gonna roll a lot of times (as in a game), you should use an alias method.
The code below is rather long implementation of such an alias method, indeed. But this is because of the initialization part. The retrieval of elements is very fast (see the next and the applyAsInt methods they don't loop).
Usage
Set<Item> items = ... ;
ToDoubleFunction<Item> weighter = ... ;
Random random = new Random();
RandomSelector<T> selector = RandomSelector.weighted(items, weighter);
Item drop = selector.next(random);
Implementation
This implementation:
uses Java 8;
is designed to be as fast as possible (well, at least, I tried to do so using micro-benchmarking);
is totally thread-safe (keep one Random in each thread for maximum performance, use ThreadLocalRandom?);
fetches elements in O(1), unlike what you mostly find on the internet or on StackOverflow, where naive implementations run in O(n) or O(log(n));
keeps the items independant from their weight, so an item can be assigned various weights in different contexts.
Anyways, here's the code. (Note that I maintain an up to date version of this class.)
import static java.util.Objects.requireNonNull;
import java.util.*;
import java.util.function.*;
public final class RandomSelector<T> {
public static <T> RandomSelector<T> weighted(Set<T> elements, ToDoubleFunction<? super T> weighter)
throws IllegalArgumentException {
requireNonNull(elements, "elements must not be null");
requireNonNull(weighter, "weighter must not be null");
if (elements.isEmpty()) { throw new IllegalArgumentException("elements must not be empty"); }
// Array is faster than anything. Use that.
int size = elements.size();
T[] elementArray = elements.toArray((T[]) new Object[size]);
double totalWeight = 0d;
double[] discreteProbabilities = new double[size];
// Retrieve the probabilities
for (int i = 0; i < size; i++) {
double weight = weighter.applyAsDouble(elementArray[i]);
if (weight < 0.0d) { throw new IllegalArgumentException("weighter may not return a negative number"); }
discreteProbabilities[i] = weight;
totalWeight += weight;
}
if (totalWeight == 0.0d) { throw new IllegalArgumentException("the total weight of elements must be greater than 0"); }
// Normalize the probabilities
for (int i = 0; i < size; i++) {
discreteProbabilities[i] /= totalWeight;
}
return new RandomSelector<>(elementArray, new RandomWeightedSelection(discreteProbabilities));
}
private final T[] elements;
private final ToIntFunction<Random> selection;
private RandomSelector(T[] elements, ToIntFunction<Random> selection) {
this.elements = elements;
this.selection = selection;
}
public T next(Random random) {
return elements[selection.applyAsInt(random)];
}
private static class RandomWeightedSelection implements ToIntFunction<Random> {
// Alias method implementation O(1)
// using Vose's algorithm to initialize O(n)
private final double[] probabilities;
private final int[] alias;
RandomWeightedSelection(double[] probabilities) {
int size = probabilities.length;
double average = 1.0d / size;
int[] small = new int[size];
int smallSize = 0;
int[] large = new int[size];
int largeSize = 0;
// Describe a column as either small (below average) or large (above average).
for (int i = 0; i < size; i++) {
if (probabilities[i] < average) {
small[smallSize++] = i;
} else {
large[largeSize++] = i;
}
}
// For each column, saturate a small probability to average with a large probability.
while (largeSize != 0 && smallSize != 0) {
int less = small[--smallSize];
int more = large[--largeSize];
probabilities[less] = probabilities[less] * size;
alias[less] = more;
probabilities[more] += probabilities[less] - average;
if (probabilities[more] < average) {
small[smallSize++] = more;
} else {
large[largeSize++] = more;
}
}
// Flush unused columns.
while (smallSize != 0) {
probabilities[small[--smallSize]] = 1.0d;
}
while (largeSize != 0) {
probabilities[large[--largeSize]] = 1.0d;
}
}
#Override public int applyAsInt(Random random) {
// Call random once to decide which column will be used.
int column = random.nextInt(probabilities.length);
// Call random a second time to decide which will be used: the column or the alias.
if (random.nextDouble() < probabilities[column]) {
return column;
} else {
return alias[column];
}
}
}
}

public class RandomCollection<E> {
private final NavigableMap<Double, E> map = new TreeMap<Double, E>();
private double total = 0;
public void add(double weight, E result) {
if (weight <= 0 || map.containsValue(result))
return;
total += weight;
map.put(total, result);
}
public E next() {
double value = ThreadLocalRandom.current().nextDouble() * total;
return map.ceilingEntry(value).getValue();
}
}

A simple (even naive?), but (as I believe) straightforward method:
/**
* Draws an integer between a given range (excluding the upper limit).
* <p>
* Simulates Python's randint method.
*
* #param min: the smallest value to be drawed.
* #param max: the biggest value to be drawed.
* #return The value drawn.
*/
public static int randomInt(int min, int max)
{return (int) (min + Math.random()*max);}
/**
* Tests wether a given matrix has all its inner vectors
* has the same passed and expected lenght.
* #param matrix: the matrix from which the vectors length will be measured.
* #param expectedLenght: the length each vector should have.
* #return false if at least one vector has a different length.
*/
public static boolean haveAllVectorsEqualLength(int[][] matrix, int expectedLenght){
for(int[] vector: matrix){if (vector.length != expectedLenght) {return false;}}
return true;
}
/**
* Draws an integer between a given range
* by weighted values.
*
* #param ticketBlock: matrix with limits and weights for the drawing. All its
* vectors should have lenght two. The weights, instead of percentages, should be
* measured as integers, according to how rare each one should be draw, the rarest
* receiving the smallest value.
* #return The value drawn.
*/
public static int weightedRandomInt(int[][] ticketBlock) throws RuntimeException {
boolean theVectorsHaventAllLengthTwo = !(haveAllVectorsEqualLength(ticketBlock, 2));
if (theVectorsHaventAllLengthTwo)
{throw new RuntimeException("The given matrix has, at least, one vector with length lower or higher than two.");}
// Need to test for duplicates or null values in ticketBlock!
// Raffle urn building:
int raffleUrnSize = 0, urnIndex = 0, blockIndex = 0, repetitionCount = 0;
for(int[] ticket: ticketBlock){raffleUrnSize += ticket[1];}
int[] raffleUrn = new int[raffleUrnSize];
// Raffle urn filling:
while (urnIndex < raffleUrn.length){
do {
raffleUrn[urnIndex] = ticketBlock[blockIndex][0];
urnIndex++; repetitionCount++;
} while (repetitionCount < ticketBlock[blockIndex][1]);
repetitionCount = 0; blockIndex++;
}
return raffleUrn[randomInt(0, raffleUrn.length)];
}

What is the best way to check if ALL values in a range exist in an array? (java)

I have the task of determining whether each value from 1, 2, 3... n is in an unordered int array. I'm not sure if this is the most efficient way to go about this, but I created an int[] called range that just has all the numbers from 1-n in order at range[i] (range[0]=1, range[1]=2, ect). Then I tried to use the containsAll method to check if my array of given numbers contains all of the numbers in the range array. However, when I test this it returns false. What's wrong with my code, and what would be a more efficient way to solve this problem?
public static boolean hasRange(int [] givenNums, int[] range) {
boolean result = true;
int n = range.length;
for (int i = 1; i <= n; i++) {
if (Arrays.asList(givenNums).containsAll(Arrays.asList(range)) == false) {
result = false;
}
}
return result;
}
(I'm pretty sure I'm supposed to do this manually rather than using the containsAll method, so if anyone knows how to solve it that way it would be especially helpful!)
Here's where this method is implicated for anyone who is curious:
public static void checkMatrix(int[][] intMatrix) {
File numberFile = new File("valid3x3") ;
intMatrix= readMatrix(numberFile);
int nSquared = sideLength * sideLength;
int[] values = new int[nSquared];
int[] range = new int[nSquared];
int valCount = 0;
for (int i = 0; i<sideLength; i++) {
for (int j=0; j<sideLength; j++) {
values[valCount] = intMatrix[i][j];
valCount++;
}
}
for (int i=0; i<range.length; i++) {
range[i] = i+1;
}
Boolean valuesThere = hasRange(values, range);
valuesThere is false when printed.

First style:
if (condition == false) // Works, but at the end you have if (true == false) or such
if (!condition) // Better: not condition
// Do proper usage, if you have a parameter, do not read it in the method.
File numberFile = new File("valid3x3") ;
intMatrix = readMatrix(numberFile);
checkMatrix(intMatrix);
public static void checkMatrix(int[][] intMatrix) {
int nSquared = sideLength * sideLength;
int[] values = new int[nSquared];
Then the problem. It is laudable to see that a List or even better a Set approach is the exact abstraction level: going into detail not sensible. Here however just that is wanted.
To know whether every element in a range [1, ..., n] is present.
You could walk through the given numbers,
and for every number look whether it new in the range, mark it as no longer new,
and if n new numbers are reached: return true.
int newRangeNumbers = 0;
boolean[] foundRangeNumbers = new boolean[n]; // Automatically false
Think of better names.

You say you have a one dimensional array right?
Good. Then I think you are thinking to complicated.
I try to explain you another way to check if all numbers in an array are in number order.
For instance you have the array with following values:
int[] array = {9,4,6,7,8,1,2,3,5,8};
First of all you can order the Array simpel with
Arrays.sort(array);
After you've done this you can loop through the array and compare with the index like (in a method):
for(int i = array[0];i < array.length; i++){
if(array[i] != i) return false;

One way to solve this is to first sort the unsorted int array like you said then run a binary search to look for all values from 1...n. Sorry I'm not familiar with Java so I wrote in pseudocode. Instead of a linear search which takes O(N), binary search runs in O(logN) so is much quicker. But precondition is the array you are searching through must be sorted.
//pseudocode
int range[N] = {1...n};
cnt = 0;
while(i<-inputStream)
int unsortedArray[cnt]=i
cnt++;
sort(unsortedArray);
for(i from 0 to N-1)
{
bool res = binarySearch(unsortedArray, range[i]);
if(!res)
return false;
}
return true;

What I comprehended from your description is that the array is not necessarily sorted (in order). So, we can try using linear search method.
public static void main(String[] args){
boolean result = true;
int[] range <- Contains all the numbers
int[] givenNums <- Contains the numbers to check
for(int i=0; i<givenNums.length; i++){
if(!has(range, givenNums[i])){
result = false;
break;
}
}
System.out.println(result==false?"All elements do not exist":"All elements exist");
}
private static boolean has(int[] range, int n){
//we do linear search here
for(int i:range){
if(i == n)
return true;
}
return false;
}
This code displays whether all the elements in array givenNums exist in the array range.

Arrays.asList(givenNums).
This does not do what you think. It returns a List<int[]> with a single element, it does not box the values in givenNums to Integer and return a List<Integer>. This explains why your approach does not work.
Using Java 8 streams, assuming you don't want to permanently sort givens. Eliminate the copyOf() if you don't care:
int[] sorted = Arrays.copyOf(givens,givens.length);
Arrays.sort(sorted);
boolean result = Arrays.stream(range).allMatch(t -> Arrays.binarySearch(sorted, t) >= 0);

public static boolean hasRange(int [] givenNums, int[] range) {
Set result = new HashSet();
for (int givenNum : givenNums) {
result.add(givenNum);
}
for (int num : range) {
result.add(num);
}
return result.size() == givenNums.length;
}

The problem with your code is that the function hasRange takes two primitive int array and when you pass primitive int array to Arrays.asList it will return a List containing a single element of type int[]. In this containsAll will not check actual elements rather it will compare primitive array object references.
Solution is either you create an Integer[] and then use Arrays.asList or if that's not possible then convert the int[] to Integer[].
public static boolean hasRange(Integer[] givenNums, Integer[] range) {
return Arrays.asList(givenNums).containsAll(Arrays.asList(range));
}
Check here for sample code and output.
If you are using ApacheCommonsLang library you can directly convert int[] to Integer[].
Integer[] newRangeArray = ArrayUtils.toObject(range);

A mathematical approach: if you know the max value (or search the max value) check the sum. Because the sum for the numbers 1,2,3,...,n is always equal to n*(n+1)/2. So if the sum is equal to that expression all values are in your array and if not some values are missing. Example
public class NewClass12 {
static int [] arr = {1,5,2,3,4,7,9,8};
public static void main(String [] args){
System.out.println(containsAllValues(arr, highestValue(arr)));
}
public static boolean containsAllValues(int[] arr, int n){
int sum = 0;
for(int k = 0; k<arr.length;k++){
sum +=arr[k];
}
return (sum == n*(n+1)/2);
}
public static int highestValue(int[]arr){
int highest = arr[0];
for(int i = 0; i < arr.length; i++) {
if(highest<arr[i]) highest = arr[i];
}
return highest;
}
}
according to this your method could look like this
public static boolen hasRange (int [] arr){
int highest = arr[0];
int sum = 0;
for(int i = 0; i < arr.length; i++) {
if(highest<arr[i]) highest = arr[i];
}
for(int k = 0; k<arr.length;k++){
sum +=arr[k];
}
return (sum == highest *(highest +1)/2);
}

How does this custom ArrayList alter its size? [duplicate]

I have searched for a way to resize an array in Java, but I could not find ways of resizing the array while keeping the current elements.
I found for example code like int[] newImage = new int[newWidth];, but this deletes the elements stored before.
My code would basically do this: whenever a new element is added, the array largens by 1. I think this could be done with dynamic programming, but I'm, not sure how to implement it.

You can't resize an array in Java. You'd need to either:
Create a new array of the desired size, and copy the contents from the original array to the new array, using java.lang.System.arraycopy(...);
Use the java.util.ArrayList<T> class, which does this for you when you need to make the array bigger. It nicely encapsulates what you describe in your question.
Use java.util.Arrays.copyOf(...) methods which returns a bigger array, with the contents of the original array.

Not nice, but works:
int[] a = {1, 2, 3};
// make a one bigger
a = Arrays.copyOf(a, a.length + 1);
for (int i : a)
System.out.println(i);
as stated before, go with ArrayList

Here are a couple of ways to do it.
Method 1: System.arraycopy():
Copies an array from the specified source array, beginning at the specified position, to the specified position of the destination array. A subsequence of array components are copied from the source array referenced by src to the destination array referenced by dest. The number of components copied is equal to the length argument. The components at positions srcPos through srcPos+length-1 in the source array are copied into positions destPos through destPos+length-1, respectively, of the destination array.
Object[] originalArray = new Object[5];
Object[] largerArray = new Object[10];
System.arraycopy(originalArray, 0, largerArray, 0, originalArray.length);
Method 2: Arrays.copyOf():
Copies the specified array, truncating or padding with nulls (if necessary) so the copy has the specified length. For all indices that are valid in both the original array and the copy, the two arrays will contain identical values. For any indices that are valid in the copy but not the original, the copy will contain null. Such indices will exist if and only if the specified length is greater than that of the original array. The resulting array is of exactly the same class as the original array.
Object[] originalArray = new Object[5];
Object[] largerArray = Arrays.copyOf(originalArray, 10);
Note that this method usually uses System.arraycopy() behind the scenes.
Method 3: ArrayList:
Resizable-array implementation of the List interface. Implements all optional list operations, and permits all elements, including null. In addition to implementing the List interface, this class provides methods to manipulate the size of the array that is used internally to store the list. (This class is roughly equivalent to Vector, except that it is unsynchronized.)
ArrayList functions similarly to an array, except it automatically expands when you add more elements than it can contain. It's backed by an array, and uses Arrays.copyOf.
ArrayList<Object> list = new ArrayList<>();
// This will add the element, resizing the ArrayList if necessary.
list.add(new Object());

You could just use ArrayList which does the job for you.

It is not possible to change the Array Size.
But you can copy the element of one array into another array by creating an Array of bigger size.
It is recommended to create Array of double size if Array is full and Reduce Array to halve if Array is one-half full
public class ResizingArrayStack1 {
private String[] s;
private int size = 0;
private int index = 0;
public void ResizingArrayStack1(int size) {
this.size = size;
s = new String[size];
}
public void push(String element) {
if (index == s.length) {
resize(2 * s.length);
}
s[index] = element;
index++;
}
private void resize(int capacity) {
String[] copy = new String[capacity];
for (int i = 0; i < s.length; i++) {
copy[i] = s[i];
s = copy;
}
}
public static void main(String[] args) {
ResizingArrayStack1 rs = new ResizingArrayStack1();
rs.push("a");
rs.push("b");
rs.push("c");
rs.push("d");
}
}

You could use a ArrayList instead of array. So that you can add n number of elements
List<Integer> myVar = new ArrayList<Integer>();

Standard class java.util.ArrayList is resizable array, growing when new elements added.

You can't resize an array, but you can redefine it keeping old values or use a java.util.List
Here follows two solutions but catch the performance differences running the code below
Java Lists are 450 times faster but 20 times heavier in memory!
testAddByteToArray1 nanoAvg:970355051 memAvg:100000
testAddByteToList1 nanoAvg:1923106 memAvg:2026856
testAddByteToArray1 nanoAvg:919582271 memAvg:100000
testAddByteToList1 nanoAvg:1922660 memAvg:2026856
testAddByteToArray1 nanoAvg:917727475 memAvg:100000
testAddByteToList1 nanoAvg:1904896 memAvg:2026856
testAddByteToArray1 nanoAvg:918483397 memAvg:100000
testAddByteToList1 nanoAvg:1907243 memAvg:2026856
import java.util.ArrayList;
import java.util.List;
public class Test {
public static byte[] byteArray = new byte[0];
public static List<Byte> byteList = new ArrayList<>();
public static List<Double> nanoAvg = new ArrayList<>();
public static List<Double> memAvg = new ArrayList<>();
public static void addByteToArray1() {
// >>> SOLUTION ONE <<<
byte[] a = new byte[byteArray.length + 1];
System.arraycopy(byteArray, 0, a, 0, byteArray.length);
byteArray = a;
//byteArray = Arrays.copyOf(byteArray, byteArray.length + 1); // the same as System.arraycopy()
}
public static void addByteToList1() {
// >>> SOLUTION TWO <<<
byteList.add(new Byte((byte) 0));
}
public static void testAddByteToList1() throws InterruptedException {
System.gc();
long m1 = getMemory();
long n1 = System.nanoTime();
for (int i = 0; i < 100000; i++) {
addByteToList1();
}
long n2 = System.nanoTime();
System.gc();
long m2 = getMemory();
byteList = new ArrayList<>();
nanoAvg.add(new Double(n2 - n1));
memAvg.add(new Double(m2 - m1));
}
public static void testAddByteToArray1() throws InterruptedException {
System.gc();
long m1 = getMemory();
long n1 = System.nanoTime();
for (int i = 0; i < 100000; i++) {
addByteToArray1();
}
long n2 = System.nanoTime();
System.gc();
long m2 = getMemory();
byteArray = new byte[0];
nanoAvg.add(new Double(n2 - n1));
memAvg.add(new Double(m2 - m1));
}
public static void resetMem() {
nanoAvg = new ArrayList<>();
memAvg = new ArrayList<>();
}
public static Double getAvg(List<Double> dl) {
double max = Collections.max(dl);
double min = Collections.min(dl);
double avg = 0;
boolean found = false;
for (Double aDouble : dl) {
if (aDouble < max && aDouble > min) {
if (avg == 0) {
avg = aDouble;
} else {
avg = (avg + aDouble) / 2d;
}
found = true;
}
}
if (!found) {
return getPopularElement(dl);
}
return avg;
}
public static double getPopularElement(List<Double> a) {
int count = 1, tempCount;
double popular = a.get(0);
double temp = 0;
for (int i = 0; i < (a.size() - 1); i++) {
temp = a.get(i);
tempCount = 0;
for (int j = 1; j < a.size(); j++) {
if (temp == a.get(j))
tempCount++;
}
if (tempCount > count) {
popular = temp;
count = tempCount;
}
}
return popular;
}
public static void testCompare() throws InterruptedException {
for (int j = 0; j < 4; j++) {
for (int i = 0; i < 20; i++) {
testAddByteToArray1();
}
System.out.println("testAddByteToArray1\tnanoAvg:" + getAvg(nanoAvg).longValue() + "\tmemAvg:" + getAvg(memAvg).longValue());
resetMem();
for (int i = 0; i < 20; i++) {
testAddByteToList1();
}
System.out.println("testAddByteToList1\tnanoAvg:" + getAvg(nanoAvg).longValue() + "\t\tmemAvg:" + getAvg(memAvg).longValue());
resetMem();
}
}
private static long getMemory() {
Runtime runtime = Runtime.getRuntime();
return runtime.totalMemory() - runtime.freeMemory();
}
public static void main(String[] args) throws InterruptedException {
testCompare();
}
}

You can try below solution inside some class:
int[] a = {10, 20, 30, 40, 50, 61};
// private visibility - or change it as needed
private void resizeArray(int newLength) {
a = Arrays.copyOf(a, a.length + newLength);
System.out.println("New length: " + a.length);
}

It is not possible to resize an array. However, it is possible change the size of an array through copying the original array to the newly sized one and keep the current elements. The array can also be reduced in size by removing an element and resizing.
import java.util.Arrays
public class ResizingArray {
public static void main(String[] args) {
String[] stringArray = new String[2] //A string array with 2 strings
stringArray[0] = "string1";
stringArray[1] = "string2";
// increase size and add string to array by copying to a temporary array
String[] tempStringArray = Arrays.copyOf(stringArray, stringArray.length + 1);
// Add in the new string
tempStringArray[2] = "string3";
// Copy temp array to original array
stringArray = tempStringArray;
// decrease size by removing certain string from array (string1 for example)
for(int i = 0; i < stringArray.length; i++) {
if(stringArray[i] == string1) {
stringArray[i] = stringArray[stringArray.length - 1];
// This replaces the string to be removed with the last string in the array
// When the array is resized by -1, The last string is removed
// Which is why we copied the last string to the position of the string we wanted to remove
String[] tempStringArray2 = Arrays.copyOf(arrayString, arrayString.length - 1);
// Set the original array to the new array
stringArray = tempStringArray2;
}
}
}
}

Sorry, but at this time is not possible resize arrays, and may be never will be.
So my recommendation, is to think more to find a solution that allow you get from the beginning of the process, the size of the arrays that you will requiere. This often will implicate that your code need a little more time (lines) to run, but you will save a lot of memory resources.

We can't do that using array datatype. Instead use a growable array which is arrayList in Java.

How to create a number generator that will only pick a number 1 time?

I am creating a concentration game.
I have an buffered image array where I load in a 25 image sprite sheet.
public static BufferedImage[] card = new BufferedImage[25];
0 index being the card back. and 1 - 24 being the values for the face of the cards to check against if the cards match.
What I am tying to do is this I will have 4 difficulties Easy, Normal, Hard, and Extreme. Each difficulty will have a certain amount of cards it will need to draw and then double the ones it chosen. for example the default level will be NORMAL which is 12 matches so it need to randomly choose 12 unique cards from the Buffered Image array and then double each value so it will only have 2 of each cards and then shuffle the results.
This is what I got so far but it always seems to have duplicates about 99% of the time.
//generate cards
Random r = new Random();
int j = 0;
int[] rowOne = new int[12];
int[] rowTwo = new int[12];
boolean[] rowOneBool = new boolean[12];
for(int i = 0; i < rowOneBool.length; i++)
rowOneBool[i] = false;
for(int i = 0; i < rowOne.length; i++){
int typeId = r.nextInt(12)+1;
while(rowOneBool[typeId]){
typeId = r.nextInt(12)+1;
if(rowOneBool[typeId] == false);
}
rowOne[i] = typeId;
j=0;
}
the 3 amounts I will be needing to generate is Easy 6, Normal 12, and Hard 18 extreme will use all of the images except index 0 which is the back of the cards.

This is more or less in the nature of random numbers. Sometimes they are duplicates. You can easily factor that in though if you want them to be more unique. Just discard the number and generate again if it's not unique.
Here's a simple method to generate unique random numbers with a specified allowance of duplicates:
public static void main(String[] args) {
int[] randoms = uniqueRandoms(new int[16], 1, 25, 3);
for (int r : randoms) System.out.println(r);
}
public static int[] uniqueRandoms(int[] randoms, int lo, int hi, int allowance) {
// should do some error checking up here
int range = hi - lo, duplicates = 0;
Random gen = new Random();
for (int i = 0, k; i < randoms.length; i++) {
randoms[i] = gen.nextInt(range) + lo;
for (k = 0; k < i; k++) {
if (randoms[i] == randoms[k]) {
if (duplicates < allowance) {
duplicates++;
} else {
i--;
}
break;
}
}
}
return randoms;
}
Edit: Tested and corrected. Now it works. : )

From what I understand from your question, the answer should look something like this:
Have 2 classes, one called Randp and the other called Main. Run Main, and edit the code to suit your needs.
package randp;
public class Main {
public static void main(String[] args) {
Randp randp = new Randp(10);
for (int i = 0; i < 10; i++) {
System.out.print(randp.nextInt());
}
}
}
package randp;
public class Randp {
private int numsLeft;
private int MAX_VALUE;
int[] chooser;
public Randp(int startCounter) {
MAX_VALUE = startCounter; //set the amount we go up to
numsLeft = startCounter;
chooser = new int[MAX_VALUE];
for (int i = 1; i <= chooser.length; i++) {
chooser[i-1] = i; //fill the array up
}
}
public int nextInt() {
if(numsLeft == 0){
return 0; //nothing left in the array
}
int a = chooser[(int)(Math.random() * MAX_VALUE)]; //picking a random index
if(a == 0) {
return this.nextInt(); //we hit an index that's been used already, pick another one!
}
chooser[a-1] = 0; //don't want to use it again
numsLeft--; //keep track of the numbers
return a;
}
}

This is how I would handle it. You would move your BufferedImage objects to a List, although I would consider creating an object for the 'cards' you're using...
int removalAmount = 3; //Remove 3 cards at random... Use a switch to change this based upon difficulty or whatever...
List<BufferedImage> list = new ArrayList<BufferedImage>();
list.addAll(Arrays.asList(card)); // Add the cards to the list, from your array.
Collections.shuffle(list);
for (int i = 0; i < removalAmount; i++) {
list.remove(list.size() - 1);
}
list.addAll(list);
Collections.shuffle(list);
for (BufferedImage specificCard : list) {
//Do something
}

Ok, I said I'd give you something better, and I will. First, let's improve Jeeter's solution.
It has a bug. Because it relies on 0 to be the "used" indicator, it won't actually produce index 0 until the end, which is not random.
It fills an array with indices, then uses 0 as effectively a boolean value, which is redundant. If a value at an index is not 0 we already know what it is, it's the same as the index we used to get to it. It just hides the true nature of algorithm and makes it unnecessarily complex.
It uses recursion when it doesn't need to. Sure, you can argue that this improves code clarity, but then you risk running into a StackOverflowException for too many recursive calls.
Thus, I present an improved version of the algorithm:
class Randp {
private int MAX_VALUE;
private int numsLeft;
private boolean[] used;
public Randp(int startCounter) {
MAX_VALUE = startCounter;
numsLeft = startCounter;
// All false by default.
used = new boolean[MAX_VALUE];
}
public int nextInt() {
if (numsLeft <= 0)
return 0;
numsLeft--;
int index;
do
{
index = (int)(Math.random() * MAX_VALUE);
} while (used[index]);
return index;
}
}
I believe this is much easier to understand, but now it becomes clear the algorithm is not great. It might take a long time to find an unused index, especially when we wanted a lot of values and there's only a few left. We need to fundamentally change the way we approach this. It'd be better to generate the values randomly from the beginning:
class Randp {
private ArrayList<Integer> chooser = new ArrayList<Integer>();
private int count = 0;
public Randp(int startCounter) {
for (int i = 0; i < startCounter; i++)
chooser.add(i);
Collections.shuffle(chooser);
}
public int nextInt() {
if (count >= chooser.size())
return 0;
return chooser.get(count++);
}
}
This is the most efficient and extremely simple since we made use of existing classes and methods.

Complexity and Efficiency in Algorithm for: a[j]-a[i] i>=j

I'm looking to make this much quicker. I've contemplated using a tree, but I'm not sure if that would actually help much.
I feel like the problem is for most cases you don't need to calculate all the possible maximums only a hand full, but I'm not sure where to draw the line
Thanks so much for the input,
Jasper
public class SpecialMax {
//initialized to the lowest possible value of j;
public static int jdex = 0;
//initialized to the highest possible value of i;
public static int idex;
//will hold possible maximums
public static Stack<Integer> possibleMaxs = new Stack<Integer> ();
public static int calculate (int[] a){
if (isPositive(a)){
int size = a.length;
int counterJ;
counterJ = size-1;
//find and return an ordered version of a
int [] ordered = orderBySize (a);
while (counterJ>0){
/* The first time this function is called, the Jvalue will be
* the largest it can be, similarly, the Ivalue that is found
* is the smallest
*/
int jVal = ordered[counterJ];
int iVal = test (a, jVal);
possibleMaxs.push(jVal-iVal);
counterJ--;
}
int answer = possibleMaxs.pop();
while (!possibleMaxs.empty()){
if (answer<possibleMaxs.peek()){
answer = possibleMaxs.pop();
} else {
possibleMaxs.pop();
}
}
System.out.println("The maximum of a[j]-a[i] with j>=i is: ");
return answer;
} else {
System.out.println ("Invalid input, array must be positive");
return 0; //error
}
}
//Check to make sure the array contains positive numbers
public static boolean isPositive(int[] a){
boolean positive = true;
int size = a.length;
for (int i=0; i<size; i++){
if (a[i]<0){
positive = false;
break;
}
}
return positive;
}
public static int[] orderBySize (int[] a){
//orders the array into ascending order
int [] answer = a.clone();
Arrays.sort(answer);
return answer;
}
/*Test returns an Ival to match the input Jval it accounts for
* the fact that jdex<idex.
*/
public static int test (int[] a, int jVal){
int size = a.length;
//initialized to highest possible value
int tempMin = jVal;
//keeps a running tally
Stack<Integer> mIndices = new Stack<Integer> ();
//finds the index of the jVal being tested
for (int i=0; i<size; i++) {
if (jVal==a[i]){
//finds the highest index for instance
if (jdex<i){
jdex = i;
}
}
}
//look for the optimal minimal below jdex;
for (int i=0; i<jdex; i++){
if (a[i]<tempMin){
tempMin = a[i];
mIndices.push(i);
}
}
//returns the index of the last min
if (!mIndices.empty()){
idex = mIndices.pop();
}
return tempMin;
}
}

It can be done in linear time and linear memory. The idea is: find the minimum over each suffix of the array and maximum over each prefix, then find the point where the difference between the two is the highest. You'll also have to store the index on which the maximum/minimum for each prefix is reached if you need the indices, rather than just the difference value.

Pre-sorting a[] makes the procedure complicated and impairs performance. It is not necessary, so we leave a[] unsorted.
Then (EDITED, because I had read j>=i in the body of your code, rather than i>=j in the problem description/title, which I now assume is what is required (I didn't go over your coding details); The two varieties can easily be derived from each other anyway.)
// initialize result(indices)
int iFound = 0;
int jFound = 0;
// initialize a candidate that MAY replace jFound
int jAlternative = -1; // -1 signals: no candidate currently available
// process the (remaining) elements of the array - skip #0: we've already handled that one at the initialization
for (int i=1; i<size; i++)
{
// if we have an alternative, see if that combines with the current element to a higher "max".
if ((jAlternative != -1) && (a[jAlternative]-a[i] > a[jFound]-a[iFound]))
{
jFound = jAlternative;
iFound = i;
jAlternative = -1;
}
else if (a[i] < a[iFound]) // then we can set a[iFound] lower, thereby increasing "max"
{
iFound = i;
}
else if (a[i] > a[jFound])
{ // we cannot directly replace jFound, because of the condition iFound>=jFound,
// but when we later may find a lower a[i], then it can jump in:
// set it as a waiting candidate (replacing an existing one if the new one is more promising).
if ((jAlternative = -1) || (a[i] > a[jAlternative]))
{
jAlternative = i;
}
}
}
double result = a[jFound] - a[iFound];