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Generate random numbers without duplicates using arrays only

I want to fill an array of size X with random integers from 0 to X with no duplicates. The catch is I must only use arrays to store the collections of int, no ArrayLists. How do I go about implementing this?
I don't understand why I can't seem to get this. But this is my most recent bit of code that fills the list but allows for duplicates.
System.out.print("Zero up to but excluding ");
int limit = scanner.nextInt();
// create index the size of the limit
int [] index = new int[limit];
for(int fill=0;fill<limit;fill+=1){
index[fill] = (limit);
}
int randomNumber = 0;
Random rand = new Random();
int [] randoms = new int[limit];
boolean flag = true;
// CODE TO NOT PRINT DOUBLES
for (int z=0;z<limit;z+=1){
randomNumber = rand.nextInt(limit);
int i=0;
while (i<limit){
if (index[i] == randomNumber){
flag = true;
}
else {
flag = false;
break;
}
i+=1;
}
if (flag == false){
randoms[z] = randomNumber;
index[z] = randomNumber;
}
}
System.out.println("Randoms: "+java.util.Arrays.toString(randoms));

Here's one way to do it:
Create an array of length N
Fill it from 0 to N-1
Run a for loop and swap randomly 2 indices
Code:
// Step 1
int N = 10;
int[] array = new int[N];
// Step 2
for(int i=0; i < N; i++)
array[i] = i;
// Step 3
for(int i=0; i < N; i++) {
int randIndex = (int) (Math.random() * N);
int tmp = array[i];
array[i] = array[randIndex];
array[randIndex] = tmp;
}

Why not rephrase the problem to shuffling an array of integers. First fill the array monotonically with the numbers 0 to X. Then use the Random() function to select one of the X numbers to exchange with the number in position 0. Repeat as many times as you may like. Done.

Here is your bug:
while (i<limit){
if (index[i] == randomNumber){
flag = true;
}
else {flag = false;break;} <--- rest of the array is skipped
i+=1;
}
after you generated a new number, you start to check for equality , however once you find that randomNumber!=index[i] (else statement) you break out of the while. look this: actual array is 3,4,5,1 your new number is 5, you compare it to 3 just to find out that they different so flag is set to false and break out happens.

Consider using another array filled with elements in order from 0 to X. Then, with this array, shuffle the elements around. How do you go about this? Use a loop to traverse through every single element of the array, and for each iteration, choose a random number from 0 to array.length - 1 and switch the elements at the index you're currently on and the random index. This is how it would look like,
In your main, you would have an array initialized by doing this,
int[] arr = new int[10];//10 can be interchangeable with any other number
for(int i = 0; i < arr.length; i++){
arr[i] = i;
}
shuffleArray(arr);
And the shuffle method would look like this,
public int[] shuffleArray(int[] arr){
Random rand = new Random();
for(int i = 0; i < arr.length; i++){
int r = rand.nextInt(arr.length);//generate a random number from 0 to X
int k = arr[i];
arr[i] = arr[r];
arr[r] = k;
}
}

Weighted random numbers in 2D Array - Processing

I would like to fill a 3x3 2D array with values 1,2,3.
I need each number to appear for a given times.
For example:
1 to appear 2 times
2 to appear 4 times
3 to appear 3 times
What I need is to store this numbers to array in a random position.
For Example:
1,2,2
3,2,2
1,3,3
I already did this in a simple way using only 2 different numbers controlled by a counter. So I loop through the 2D array and applying random values of number 1 and number 2.
I'm checking if the value is 1 and add it in the counter and the same with number 2. if one of the counter exceeds the number I have set as the maximum appear times then it continues and applies the other value.
Is there any better approach to fill the 3 numbers in random array position?
See code below:
int [][] array = new int [3][3];
int counter1 =0;
int counter2 =0;
for (int i=0; i<3; i++) {
for (int j=0; j<3; j++) {
array[i][j] = (int)random(1, 3); //1,2
if (arrray[i][j]==1) {
counter1++;
} else if (array[i][j] ==2) {
counter2++;
}
//if it is more than 5 times in the array put only the other value
if (counter1>5) {
array[i][j] = 2;
}
//if it is more than 4 times in the array put only the other value
else if (counter2>4) {
array[i][j] = 1;
}
}
}
I finally did this according to this discussion:
How can I generate a random number within a range but exclude some?, with 1D array for tesing, but it does not always works.
Please see attached code:
int []values = new int[28];
int counter1=0;
int counter2=0;
int counter3=0;
for (int i=0; i<values.length; i++) {
if (counter1==14) {
ex = append(ex, 5);
}
if (counter2==4) {
ex =append(ex, 6);
}
if (counter3==10) {
ex =append(ex, 7);
}
values[i] = getRandomWithExclusion(5, 8, ex);
if (values[i]==5) {
counter1++;
} else if (values[i] ==6) {
counter2++;
} else if (values[i] ==7) {
counter3++;
}
}
int getRandomWithExclusion(int start, int end, int []exclude) {
int rand = 0;
do {
rand = (int) random(start, end);
}
while (Arrays.binarySearch (exclude, rand) >= 0);
return rand;
}
I would like to fill the 1D array with values of 5,6 or 7. Each one a specific number. Number 5 can be added 14 times. Number 6 can be added 4 times. Number 7 can be added 10 times.
The above code works most of the times, however somethimes it does not. Please let me know if you have any ideas

This is the Octave/Matlab code for your problem.
n=3;
N=n*n;
count = [1 2; 2 4; 3 3];
if sum(count(:,2)) ~= N
error('invalid input');
end
m = zeros(n, n);
for i = 1:size(count,1)
for j = 1:count(i,2)
r = randi(N);
while m(r) ~= 0
r = randi(N);
end
m(r) = count(i,1);
end
end
disp(m);
Please note that when you address a 2D array using only one index, Matlab/Octave would use Column-major order.

There are a ton of ways to do this. Since you're using processing, one way is to create an IntList from all of the numbers you want to add to your array, shuffle it, and then add them to your array. Something like this:
IntList list = new IntList();
for(int i = 1; i <= 3; i++){ //add numbers 1 through 3
for(int j = 0; j < 3; j++){ add each 3 times
list.append(i);
}
}
list.shuffle();
for (int i=0; i<3; i++) {
for (int j=0; j<3; j++) {
array[i][j] = list.remove(0);
}
}
You could also go the other way: create an ArrayList of locations in your array, shuffle them, and then add your ints to those locations.

How to convert a one-dimensional array to two-dimensional array in Java

I would like to read a file which is in the form of a matrix, so i tried reading a file put it in String arraylist and then converted to integer array. Now I need a 2D integer array. Can anyone help? Is there any better way to do this.
public class readMat {
private static ArrayList<String> list = new ArrayList<String>();
public static void main (String[] args)
{
// read file and put in arraylist
try
{
Scanner s = new Scanner(new File("link_info_test.txt"));
while (s.hasNext())
{
list.add(s.next());
}
}
catch (Exception e)
{
e.printStackTrace();
}
String[] stockArr = new String[list.size()];
stockArr = list.toArray(stockArr);
int[] sum= Convert(stockArr);
}
// convert string arraylist to integer 1 dimensional array private static int[] Convert(String[] stockArr)
{
if (list != null)
{
int intarray[] = new int[stockArr.length];
for (int i = 0; i < stockArr.length; i++)
{
intarray[i] = Integer.parseInt(stockArr[i]);
}
return intarray;
}
return null;
}
}

Let's say you have temperature data for each day of the week for 10 weeks (that is 70 pieces of data). You want to convert it to a 2d array with rows representing weeks and columns representing days. Here you go:
int temp[70] = {45, 43, 54, ........}
int twoD[30][7]
for(int i=0; i < 70; i++) {
twoD[i / 7][i % 7] = temp[i]
}
That's it.

After the line
int[] sum= Convert(stockArr);
you have your entire file in a 1D array of integers. At this point, you have to determine the width and height of the 2D array.
Let's say you want the 2D array to have 3 rows and 4 columns as an example. Do this:
int[][] int_table = new int[3][4];
for(int j = 0; j < 3; j++)
{
for(int i = 0; i < 4; i++)
{
int_table[j][i] = sum[j * 4 + i];
}
}
The equation I'm using within sum's index is a conversion function that goes from 1D to 2D coordinates. Starting at both j and i being equal to 0, sum[j * 4 + i] = sum[0 * 4 + 0] = sum[0]. The variable i would increment by one at the next step, and we would have sum[0 * 4 + 1] = sum[1]. At the end of the row, i would reset to 0 and j would increment by 1. At that point, we would have sum[1 * 4 + 0] = sum[4], or sum's fifth element. This makes sense if you consider the first four elements as those of the first row. now that we're on a new row, we can fill it with the next four. The "four" that I've been mentioning is the width of the row that we defined earlier while declaring the 2D array.
Keep in mind that the 2D array's width and height can't multiply together to be larger than the total number of integers in the 1D array. You'll get an IndexOutOfBoundsException if you try to read beyond that size.

Assuming that each entry of your String array consists of some integers, separated by some delimiter (comma, dot, hyphen, etc), then you can use the String.split() method. For instance, if your delimiter is a comma, then you would do something like this:
String Integer1;
String Integer2;
String[] TotalString;
TotalString = stockArr[i].Split(",");
Integer1 = TotalString[0];
Integer2 = TotalString[1];
Then just parse the Strings into integers and put them into your array.

If i understood, your question correctly you want to convert 1D array to 2D array ... You can use following method
public static int[][] convertArrayTo2DArray(final int[] _1darray) {
int[][] _2dArray = null;
int size = _1darray.length / 2;
if (_1darray.length % 2 == 0) {
_2dArray = new int[2][size];
} else {
_2dArray = new int[3][size];
}
int index = 0;
outter: for (int i = 0; i < _2dArray.length; i++) {
for (int j = 0; j < _2dArray[i].length; j++) {
if (index == _1darray.length) {
break outter;
}
_2dArray[i][j] = _1darray[index];
index++;
}
}
return _2dArray;
}

Find the most frequent value in an array of double in Java (without hashmaps or sorting)

Write a full Java program that does the following:
Creates an array of 100 double.
Reads in an unknown number of doubles from a file named values.txt .
There will be at least 2 distinct values, and no more than 100 distinct values in the file. The values will be in unsorted order. Values will be no smaller than 0, and no larger than 99.
Outputs the most frequently occurring value in the file.
Outputs the least frequently occurring value in the file. The value must occur at least once in order to be output.
Outputs the average of all array values.
You must create and use separate methods for each of the items #2-5.
This is what I have so far. I cannot for the life of me figure out how to get this right:
import java.util.*;
import java.io.*;
public class arrayProgram2 {
static Scanner console = new Scanner(System.in);
static final int ARRAY_SIZE = 100;
static int numOfElements = 0;
public static void main(String[] args) throws FileNotFoundException {
Scanner inFile = new Scanner(new FileReader("values.txt"));
double[] Arr1 = new double[ARRAY_SIZE];
while (inFile.hasNext()) {
Arr1[numOfElements] = inFile.nextDouble();
numOfElements++;
}
System.out.println("There are " + numOfElements + " values.");
System.out.printf("The average of the values is %.2f%n", avgArray(Arr1));
System.out.println("The sum is " + sumArray(Arr1));
inFile.close();
} //end main
//Method to calculate the sum
public static double sumArray(double[] list) {
double sum = 0;
for (int index = 0; index < numOfElements; index++) {
sum = sum + list[index];
}
return sum;
}
//Method to calculate the average
public static double avgArray(double[] list) {
double sum = 0;
double average = 0;
for (int index = 0; index < numOfElements; index++) {
sum = sum + list[index];
}
average = sum / numOfElements;
return average;
}
} //end program
Notice I am required to make an array of double even though it is not necessary.

If all values are int than you should use int array instead of double. As all values in range 0-99. So, you can increase input value frequency. Look at below logic:
int[] freqArr= new int[100];
while (inFile.hasNext()){
int value = inFile.nextInt();
freqArr[value]++; // count the frequency of selected value.
}
Now calculate the maximum frequency from freqArr
int maxFreq=0;
for(int freq : freqArr){
if(maxFreq < freq){
maxFreq = freq;
}
}
Note: If double array is mandatory than you can also use double array like:
double[] freqArr= new double[100];
while (inFile.hasNext()){
freqArr[(int)inFile.nextDouble()]++;
}

It's possible to find a most-occurring value without sorting like this:
static int countOccurrences(double[] list, double targetValue) {
int count = 0;
for (int i = 0; i < list.length; i++) {
if (list[i] == targetValue)
count++;
}
}
static double getMostFrequentValue(double[] list) {
int mostFrequentCount = 0;
double mostFrequentValue = 0;
for (int i = 0; i < list.length; i++) {
double value = list[i];
int count = countOccurrences(list, value);
if (count > mostFrequentCount) {
mostFrequentCount = count;
mostFrequentValue = value;
}
}
return mostFrequentValue;
}

Pham Thung is right : -
You read in integer inFile.nextInt(), why do you need to use double array to store them? – Pham Thung
You can achieve your first functionality in n time if its integer array.
you question says,
Values will be no smaller than 0, and no larger than 99.
So,
1. Make an array of size 100.(Counter[])
2. Iterate through values of your current array and add count to Counter array.
eg:
if double array contains
2 3 2 5 0 0 0
Our counter array will be like
location : 0 1 2 3 4 5 6 ...........100
values : 3 0 1 1 0 1 0 ..............
and so on.

You can use below algorithm for this
Sort the array (you only need to read unsorted array, but you can sort the array once read from the file)
Make double var : num, mostCommon, count = 0, currentCount = 1
Assign Arr1[0] to num
for i from 1 to length of Arr1
i. if(Arr1[i] == num)
a. Increment currentCount
ii. else
a. if(count > currentCount)
A. Assign currentCount to count
B. Assign num to mostCommon
C. Assign Arr1[i] to num
D. Assign 1 to currentCount
At the end of this loop, you will have most common number in mostCommon var and it's number of occurrence in count.
Note : I don't know how to format the algo

Generating Random Permutation Uniformly in Java

Anyone know of a fast/the fastest way to generate a random permutation of a list of integers in Java. For example if I want a random permutation of length five an answer would be 1 5 4 2 3, where each of the 5! possibilities is equally likely.
My thoughts on how to tackle this are to run a method which generates random real numbers in an array of desired length and then sorts them returning the index i.e. 0.712 0.314 0.42 0.69 0.1 would return a permutation of 5 2 3 4 1. I think this is possible to run in O(n^2) and at the moment my code is running in approximately O(n^3) and is a large proportion of the running time of my program at the moment. Theoretically this seems OK but I'm not sure about it in practice.

Have you tried the following?
Collections.shuffle(list)
This iterates through each element, swapping that element with a random remaining element. This has a O(n) time complexity.

If the purpose is just to generate a random permutation, I don't really understand the need for sorting. The following code runs in linear time as far as I can tell
public static int[] getRandomPermutation (int length){
// initialize array and fill it with {0,1,2...}
int[] array = new int[length];
for(int i = 0; i < array.length; i++)
array[i] = i;
for(int i = 0; i < length; i++){
// randomly chosen position in array whose element
// will be swapped with the element in position i
// note that when i = 0, any position can chosen (0 thru length-1)
// when i = 1, only positions 1 through length -1
// NOTE: r is an instance of java.util.Random
int ran = i + r.nextInt (length-i);
// perform swap
int temp = array[i];
array[i] = array[ran];
array[ran] = temp;
}
return array;
}
And here is some code to test it:
public static void testGetRandomPermutation () {
int length =4; // length of arrays to construct
// This code tests the DISTRIBUTIONAL PROPERTIES
ArrayList<Integer> counts = new ArrayList <Integer> (); // filled with Integer
ArrayList<int[]> arrays = new ArrayList <int[]> (); // filled with int[]
int T = 1000000; // number of trials
for (int t = 0; t < T; t++) {
int[] perm = getRandomPermutation(length);
// System.out.println (getString (perm));
boolean matchFound = false;
for(int j = 0; j < arrays.size(); j++) {
if(equals(perm,arrays.get(j))) {
//System.out.println ("match found!");
matchFound = true;
// increment value of count in corresponding position of count list
counts.set(j, Integer.valueOf(counts.get(j).intValue()+1));
break;
}
}
if (!matchFound) {
arrays.add(perm);
counts.add(Integer.valueOf(1));
}
}
for(int i = 0; i < arrays.size(); i++){
System.out.println (getString (arrays.get (i)));
System.out.println ("frequency: " + counts.get (i).intValue ());
}
// Now let's test the speed
T = 500000; // trials per array length n
// n will the the length of the arrays
double[] times = new double[97];
for(int n = 3; n < 100; n++){
long beginTime = System.currentTimeMillis();
for(int t = 0; t < T; t++){
int[] perm = getRandomPermutation(n);
}
long endTime = System.currentTimeMillis();
times[n-3] = (double)(endTime-beginTime);
System.out.println("time to make "+T+" random permutations of length "+n+" : "+ (endTime-beginTime));
}
// Plotter.plot(new double[][]{times});
}

There is an O(n) Shuffle method that is easy to implement.

Just generate random number between 0 and n! - 1 and use
the algorithm I provided elsewhere (to generate permutation by its rank).