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2D Array with variable internal array length JAVA

Im currently writing some code that print Pascal's Triangle. I need to use a 2D array for each row but don't know how to get the internal array to have a variable length, as it will also always changed based on what row it is int, for example:
public int[][] pascalTriangle(int n) {
int[][] array = new int[n + 1][]
}
As you can see I know how to get the outer array to have the size of Pascal's Triangle that I need, but I don't know how to get a variable length for the row that corresponds with the line it is currently on.
Also how would I print this 2D array?

Essentially what you want to happen is get the size of each row.
for(int i=0; i<array.size;i++){//this loops through the first part of array
for(int j=0;j<array[i].size;j++){//this loops through the now row
//do something
}
}
You should be able to use this example to also print the triangle now.

This is my first answer on StackOverFlow. I am a freshman and have just studied Java as part of my degree.
To make every step clear, I will put different codes in different methods.
Say n tells us how many rows that we are going to print for the triangle.
public static int[][] createPascalTriangle(int n){
//We first declare a 2D array, we know the number of rows
int[][] triangle = new int[n][];
//Then we specify each row with different lengths
for(int i = 0; i < n; i++){
triangle[i] = new int[i+1]; //Be careful with i+1 here.
}
//Finally we fill each row with numbers
for(int i = 0; i < n; i++){
for(int j = 0; j <= i; j++){
triangle[i][j] = calculateNumber(i, j);
}
}
return triangle;
}
//This method is used to calculate the number of the specific location
//in pascal triangle. For example, if i=0, j=0, we refer to the first row, first number.
public static int calculateNumber(int i, int j){
if(j==0){
return 1;
}
int numerator = computeFactorial(i);
int denominator = (computeFactorial(j)*computeFactorial(i-j));
int result = numerator/denominator;
return result;
}
//This method is used to calculate Factorial of a given integer.
public static int computeFactorial(int num){
int result = 1;
for(int i = 1; i <= num; i++){
result = result * i;
}
return result;
}
Finally, in the main method, we first create a pascalTriangle and then print it out using for loop:
public static void main(String[] args) {
int[][] pascalTriangle = createPascalTriangle(6);
for(int i = 0; i < pascalTriangle.length; i++){
for(int j = 0; j < pascalTriangle[i].length; j++){
System.out.print(pascalTriangle[i][j] + " ");
}
System.out.println();
}
}
This will give an output like this:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

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;
}
}

Finding repeats in a 2D array

I have made a program that outputs the number of repeats in a 2D array. The problem is that it outputs the same number twice.
For example: I input the numbers in the 2D array through Scanner: 10 10 9 28 29 9 1 28.
The output I get is:
Number 10 repeats 2 times.
Number 10 repeats 2 times.
Number 9 repeats 2 times.
Number 28 repeats 2 times.
Number 29 repeats 1 times.
Number 9 repeats 2 times.
Number 1 repeats 1 times.
Number 28 repeats 2 times.
I want it so it skips the number if it has already found the number of repeats for it. The output should be:
Number 10 repeats 2 times.
Number 9 repeats 2 times.
Number 28 repeats 2 times.
Number 29 repeats 1 times.
Number 1 repeats 1 times.
Here is my code:
import java.util.Scanner;
public class Repeat
{
static Scanner leopard = new Scanner(System.in);
public static void main(String [] args)
{
final int ROW = 10; //Row size
final int COL = 10; //Column size
int [][] num = new int[ROW][COL];
int size;
//Get input
size = getData(num);
//Find repeat
findRepeats(num, size);
}
public static int getData(int [][] num)
{
int input = 0, actualSize = 0; //Hold input and actualSize of array
System.out.print("Enter positive integers (-999 to stop): ");
//Ask for input
for(int i = 0; i < num.length && input != -999; i++)
{
for(int j = 0; j < num[i].length && input != -999; j++)
{
input = leopard.nextInt();
//Check if end
if(input != -999)
{
num[i][j] = input;
actualSize++;
}
}
}
System.out.println();
return actualSize;
}
public static void findRepeats(int [][] num, int size)
{
int findNum;
int total = 0, row = 0, col = 0;
for(int x = 0; x < size; x++)
{
//Set to number
findNum = num[row][col];
//Loop through whole array to find repeats
for(int i = 0; i < num.length; i++)
{
for(int j = 0; j < num[i].length; j++)
{
if(num[i][j] == findNum)
total++;
}
}
//Cycle array to set next number
if(col < num[0].length-1)
col++;
else
{
row++; //Go to next row if no more columns
col = 0; //Reset column number
}
//Display total repeats
System.out.println("Number " + findNum + " appears " + total + " times.");
total = 0;
}
}
}
I know why it is doing it, but I cannot figure out how to check if the number has already been checked for it to skip that number and go to the next number. I cannot use any classes or code that is not used in the code.

Since you cannot use anything other than this, lets say, basic elements of Java consider this:
Make another temporary 2D array with two columns (or just two separate arrays, personally I prefer this one). On the start of the algorithm the new arrays are empty.
When you take a number (any number) from the source 2D structure, first check if it is present in the first temporary array. If it is, just increment the value (count) in the second temporary array for one (+1). If it is not present in the first tmp array, add it to it and increase the count (+1) in the second at the same index as the newly added number in the first (which should be the last item of the array, basically).
This way you are building pairs of numbers in two arrays. The first array holds all your distinct values found in the 2D array, and the second one the number of appearances of the respective number from the first.
At the and of the algorithm just iterate the both arrays in parallel and you should have your school task finished. I could (and anyone) code this out but we are not really doing you a favor since this is a very typical school assignment.

It's counting the number two times, first time it appears in the code and second time when it appears in the code.
To avoid that keep a system to check if you have already checked for that number. I see you use check int array but you haven't used it anywhere in the code.
Do this,
Put the number in the check list if you have already found the count of it.
int count = 0;
check[count] = findNum;
count++;
Note: You can prefill you array with negative numbers at first in order to avoid for having numbers that user already gave you in input.
Next time in your for loop skip checking that number which you have already found a count for
for(int x = 0; x < size; x++) {
findNum = num[row][col];
if(check.containsNumber(findNUm)) { //sorry there is no such thing as contains for array, write another function here which checks if a number exists in the array
//skip the your code till the end of the first for loop, or in other words then don't run the code inside the for loop at all.
}
}
Frankly speaking I think you have just started to learn coding. Good luck! with that but this code can be improved a lot better. A piece of advice never create a situation where you have to use 3 nested for loops.
I hope that you understood my solution and you know how to do it.

All answers gives you some insight about the problem. I try to stick to your code, and add a little trick of swap. With this code you don't need to check if the number is already outputted or not. I appreciate your comments, structured approach of coding, and ask a question as clear as possible.
public static void findRepeats(int [][] num, int size)
{
int findNum;
int total = 1, row = 0, col = 0;
int [] check = new int[size];
while(row < num.length && col < num[0].length)
{
//Set to number
findNum = num[row][col];
//Cycle array to set next number
if(col < num[0].length-1)
col++;
else
{
row++; //Go to next row if no more columns
col = 0; //Reset column number
}
//Loop through whole array to find repeats
for(int i = row; i < num.length; i++)
{
for(int j = col; j < num[i].length; j++)
{
if(num[i][j] == findNum) {
total++;
//Cycle array to set next number
if(col < num[0].length-1)
col++;
else
{
row++; //Go to next row if no more columns
col = 0; //Reset column number
}
if(row < num.length - 1 && col < num[0].length -1)
num[i][j] = num[row][col];
}
}
}
//Display total repeats
System.out.println("Number " + findNum + " appears " + total + " times.");
total = 1;
}
}

you can use a HashMap to store the result. It Goes like this:
// Create a hash map
HashMap arrayRepeat = new HashMap();
// Put elements to the map
arrayRepeat.put(Number, Repeated);

A method to solve a word search in Java using 2d arrays.?

I'm trying to create a method that will search through a 2d array of numbers. If the numbers add up to a certain sum, those numbers should remain and all of the other numbers should be changed to a 0. For example, if the desired sum is 7 and a row contains 2 5 1 2, the result should be 2 5 0 0 after the method is implemented. I have everything functioning but instead of keeping all of the numbers that add up to the sum, only the last number is retained. So, I am left with 0 5 0 0 . I think I need another array somewhere but not sure exactly how to go about implementing it. Any ideas?
public static int[][] horizontalSums(int[][] a, int sumToFind) {
int[][] b = new int[a.length][a[0].length];
int columnStart = 0;
while (columnStart < a[0].length) {
for (int row = 0; row < a.length; row++) {
int sum = 0;
for (int column = columnStart; column < a[row].length; column++) {
sum += a[row][column];
if (sum == sumToFind) {
b[row][column] = a[row][column];
}
}
}
columnStart++;
}
return b;
}

In your example you use 2 5 1 1, would 0 5 1 1 also be a valid response? Or do you just need to find any combination? A recursive function may be the best solution.
If you just need to scan through the array and add up the numbers until the sum is reached then just add a for loop to copy the previous values from the array to the new array when the sum is found. Something like:
if (sum == sumToFind)
{
for (int i= 0; i<= columnStart; i++)
{
b[row][i] = a[row][i];
}
}

if (sum == sumToFind)
{
for (int i= columnStart; i<= column; i++)
{
b[row][i] = a[row][i];
}
}
A minor tweak was all it needed. If you have columnStart and column like in the other answer, it only finds the first number of the series.

Java lotto simulation

I am writing a program simulating a lotto draw of six numbers between 1 and 45, a sample output is 3 7 12 27 43 28. But what I am trying to do is count the number of times adjacent numbers appear, for example 1 4 5 29 26 41 is a positive answer because 5 comes after 4.
What is the best way of doing that?
I have tried examples such as :
int adjacent=0;
for(int i =0; i<6; i++)
{
int t = test[i]+1;
test[i]=(int)(45*Math.random())+1;
if(test[i]==t)
adjacent++;
System.out.print(test[i]+" ");
}
This does not work.
What am I doing wrong?

I think you just have an order of operations problem
int adjacent=0;
for(int i =0; i<6; i++)
{
//test[i] hasn't been set yet
int t = test[i]+1;
test[i]=(int)(45*Math.random())+1;
//this comparison doesn't make a whole lot of sense
if(test[i]==t)
adjacent++;
System.out.print(test[i]+" ");
}
Change it around to something like this:
int adjacent=0;
for(int i =0; i<6; i++)
{
test[i]=(int)(45*Math.random())+1;
int t = -1;
//Make sure this comparison only happens after the second iteration
//to avoid index out of bounds
if ( i != 0 )
{
//Set t to the last number + 1 instead of trying to predict the future
t = test[i-1] + 1;
}
//Now this comparison makes a little more sense
//The first iteration will compare to -1 which will always be false
if(test[i]==t)
adjacent++;
System.out.print(test[i]+" ");
}
This can be further simplified to just this:
int adjacent=0;
for(int i =0; i<6; i++)
{
test[i]=(int)(45*Math.random())+1;
if(i != 0 && test[i]==(test[i-1]+1))
adjacent++;
System.out.print(test[i]+" ");
}

After you generate your 6 numbers and put them into an array. Use Arrays.sort(). You can then compare adjacent array entries.
You should also avoid using Random to generate you 6 numbers, because it can generate duplicates. This may or may not accurately simulate your lotto draw. Quoi's answer has a good suggestion for this.

I think you should shuffle it and take any five. Collections#Shuffle would help you, It permutes the specified list using a default source of randomness. All permutations occur with approximately equal likelihood.
List<Integer> list = ArrayList<Integer>();
list.add(1);
list.add(2);
Collections.shuffle(list);
Random rnd = new Random();
Integer[] result = new Integer[5];
result[0] = list.get(rnd.getNextInt(45));
result[1] = list.get(rnd.getNextInt(45));
result[2] = list.get(rnd.getNextInt(45));
result[3] = list.get(rnd.getNextInt(45));
result[4] = list.get(rnd.getNextInt(45));
It always gives you random values, then you should sort it to arrange it in order, say ascending.
Arrays.sort(result);
now you can write a loop to find out adjacent number.
int adjacent = 0;
for(int i=1; i<result.length;i++){
int prev = result[i-1];
int now = result[i];
if(prev+1 == now)
adjacent++;
}

You need to separate the generation of unique (hence the HashSet below to insure identity) random selections, sorting them, and then determining adjacency:
import java.util.HashSet;
import java.util.Arrays;
public class Lotto
{
public Lotto()
{
}
/**
* #param args
*/
public static void main(String[] args)
{
Lotto lotto = new Lotto();
lotto.randomizeSelections(5);
}
private void randomizeSelections(int numOfArrays)
{
for(int i = 0; i < numOfArrays; i++)
{
int[] selArry = new int[6];
//to insure that each random selection is unique
HashSet<Integer> idntySet = new HashSet<Integer>();
for(int j = 0; j < 6;)
{
int rndm = (int)(45 * Math.random()) + 1;
//add selection to the array only if it has not been randomized before
if(!idntySet.contains(rndm))
{
selArry[j] = rndm;
j++;
}
}
//sort the array for determing adjacency
Arrays.sort(selArry);
for(int j = 0; j < 6; j++)
{
int sel = selArry[j];
boolean isAdjcnt = (j > 0 && (sel == selArry[j - 1] + 1)) ? true : false;
System.out.println(i + "." + j + ".random = " + sel);
if(isAdjcnt) System.out.println("\tAdjacent");
}
}
}
}