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Arithmetic Progression in java

i'm just started to learn java yesterday. But, now i met difficulty to show the arithmetic progression like the display below:
1 2 2 2 3 3 3 3 3 4 4 4 4 4 4 4
From that example, i know that every odd numbers, the numbers increment. I've tried to make it, but the display just keep showing like this:
2 4 4 4 6 6 6 6 6 8 8 8 8 8 8 8 10 10 10 10 10 10 10 10 10
Here's my code:
for (int i = 0; i <= 10; i++) {
if (i % 2 == 0) {
for(int j = 1; j<i; j++){
System.out.print(i + " ");
}
}
}
And then, i want to take the last value of it. For example, if i have row = 3, it must be value = 2.
because :
row = 1 2 3 4 5 6 7 8 9 10
value = 1 2 2 2 3 3 3 3 3 4
Would you tell me, what line is exactly must be fix? Thank you

It's not about a line that is wrong, it's your approach that's a bit off. You could fix it in multiple ways. Easiest (but not most efficient) way is this:
for (int i = 0; i <= 10; i++) {
if (i % 2 == 0) {
for(int j = 1; j<i; j++){
System.out.print((i/2) + " ");
}
}
}
As you can see, only the output was changed and now it works. However, iterating over 11 numbers (0-10) when you only really care about 4-5 is not necessarily the best way to go here.
It also doesn't make your code easy to understand.
Here's an alternative.
int amount = 1;
for (int i = 1; i <= 5; i++) {
for (int j = 0; j < amount; j++) {
System.out.print(i + " ");
}
amount = amount + 2;
}
Here you can see that the outer for has been changed to only take the numbers we actually care about, which means we can remove the if completely.
We just have to somehow decide how many times we want to execute the print call, which is done with the amount variable.

Try this.
for (int i = 1, r = 1; i <= 4; ++i, r += 2)
System.out.print((i + " ").repeat(r));
You can calculate value from row with this method.
static int value(int row) {
return (int)Math.ceil(Math.sqrt(row));
}
So you can also do like this.
for (int row = 1; row <= 16; ++row)
System.out.print(value(row) + " ");
result:
1 2 2 2 3 3 3 3 3 4 4 4 4 4 4 4

Hey it's a representation of a sequence that grows by 2 every step.
i.e the first element is 1 which shows up one time.
the second element is 3 which shows up 3 times (2 2 2)
and so on and on..
so the code you need is:
int a = 1;
for(int i=1; i<=10;i++){
int j=1;
while(j<=a){
System.out.print(i);
j++;
}
a+=2;
}
printing the value in a wanted row:
Scanner in = new Scanner(System.in);
int rowU = in.nextInt(); // User inputs row
int row = 1; // a variable to keep track of the rows
int repeats = 1; // the number of times a value shoud appear
for(int value=1;value<=10;value++){
int j=1;
while(j<=repeats){
if(row==rowU) // if we got to the wanted row
System.out.println(value); // print the wanted value
j++;
row++;
}
repeats+=2;
}
There is a better, more efficient way to get the value of a wanted row:
int wanted_value = Math.ceil(Math.sqrt(wanted_row));
Thanks to #saka for bringing this one up!
Hope I helped :)

i % 2 == 0 means that the following code is only going to be executed, if i is even.
You could try removing the if, and change the second for to something like
int j = 0; j < 2 * i - 1; j++.

This code snippet will do the work
int n=4;
int printTimes=1;
for(int i=1;i<=n;i++)
{
for(int j=0;j<printTimes;j++)
System.out.print(i+" ");
printTimes+=2;
}
System.out.println();

Here is the example code.
int start = 1;
int end = 5;
int time = 1;
for (int i = start,j = time; i < end; i++,j+=2) {
for (int k = 0; k < j; k++) {
System.out.print(i+" ");
}
}

Print a given number pattern from user input using nested for loop

I am new to programming. Am currently learning Java, on nested loop now, and got stuck.
So what I want to do is to write a program that takes an integer from user and
print lines, for example if user input was 4 then the result should be like:
1
1 2
1 2 3
1 2 3 4
Here is my code so far:
import java.util.Scanner;
public class Hello {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("Enter number of rows:");
int number = input.nextInt();
for (int i = 1; i <= number; i++) {
System.out.println(i);
for (int j = 1; j <= i; j++) {
System.out.print(j + " ");
}
}
}
}
But it prints one extra line at the end, like:
1
1 2
1 2 3
1 2 3 4
1 2 3 4
And it is hard for me to figure out why.
I guess it is my first for loop but I don't know how to fix the for loop to get the result I want.
Any help will be appreciated. Thanks!

Don't print anything from the outer loop, only new line
for (int i = 1; i <= number; i++) {
for (int j = 1; j <= i; j++) {
System.out.print(j + " ");
}
System.out.println();
}
Output
1
1 2
1 2 3
1 2 3 4

To avoid the trailing spaces of the other answers,
rather than printing i at the start of the loop, print 1.
Then start the inner loop from 2 and print a space before each value. And print a new line after the inner loop.
for (int i = 1; i <= number; i++) {
System.out.print("1");
for (int j = 2; j <= i; j++) {
System.out.print(" " + j);
}
System.out.println();
}
Prints:
1
1 2
1 2 3
1 2 3 4

The problem is printing a newline and i at the same time... just take care of the new line after your for loop. The inner loop can handle all the prints.
import java.util.Scanner;
public class Test {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("Enter number of rows:");
int number = input.nextInt();
for (int i = 1; i <= number; i++) {
for (int j = 1; j <= i; j++) {
System.out.print(j + " ");
}
System.out.println();
}
}
}

Let's us dry run it
at first you print
1
then newline
then j goes from 1 to 1 nut no newline now 2 is printed by i now newline
so result 1 2
again j goes like 1 , 2 but no newline so again 3 is printed by i then newline
so result 1 2 3
again j goes like 1 , 2, 3, but no newline so again 4 is printed by i then newline
so result 1 2 3 4
again j goes like 1 , 2, 3, 4 // this one is the extra line

How do you print the sequence one 1, then two 2, three 3, ... n ns?

Write a program that prints a part of the sequence:
1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ...
(the number is repeated as many times, to what it equals to).
I've used two for loops, however, I can't get 1 to print once, 2 to print twice, instead, I get
1 2 3 4 5 6 1 2 3 4 5 6, etc.

You need two for loops for this.
for (int i = 0; i <= 5; i++) { // This will loop 5 times
for (int j = 0; j < i; j++) { //This will loop i times
System.out.print(i);
}
}

As I remember the goal is to print n numbers (for example 1 2 2 3 3 3 4 for n = 7), diveded by space. Sorry for my java)), I wrote it in Kotlin, tried to change for Java, but main idea is clear. BTW n – the number of the elements, you need to read with Scanner.
int count = 0 //Idea is to create a counter, and to increment it each time of printing
for (int i = 0; i <= n; i++) { //Loops n times
for (int j = 0; j < i; j++) { //Loops i times
if (int count < int n) {
System.out.print(" "+i+" ");
int count++ //Prints 1 one time, 2 two times, etc. stops if reached n number
}
}
}

How about this :
for(int i=1;i<=num;i++){
for(int j=1;j<=i;j++){
System.out.print(" "+i+" ");
}
}
where, num = 1,2,....n
(Also we wont be able to tell why you got that output unless you attach the code. Please attach the code snippets for such questions :) !

Replacing Multiples of numbers in Fibonacci series using Flag method

As the title suggests, I have code for a Fibonacci series and my goal is to replace multiples of numbers (3, 5, 7 and combinations of them) in the series with a word. I was suggested to use a flag in my if loop to check for the printed phrase, and if the phrase is printed, to skip that number. Essentially, what I want the output to look like is:
1 1 2 skip 8 13 skip 34 55
(this is replacing multiple of three only, for now).
Instead, what I am getting is:
1 1 2 3 skip5 8 13 21 skip34 55
Here is my code as of now:
int febCount = 50;
long[] feb = new long[febCount];
feb[0] = 1;
feb[1] = 1;
for (int i = 2; i < febCount; i++) {
feb[i] = feb[i - 1] + feb[i - 2];
}
for (int i = 0; i < febCount; i++) {
System.out.print(feb[i] + ((i % 10 == 9) ? "\n" : " "));
if (feb[i] % 3 == 0)
System.out.print("skip");
}
Any and all help is appreciated!

Let's walk through the code you have provided and attempt to understand why it's not working.
//The first thing we do is setup the loop to iterate through the fib numbers.
//This looks good.
for (int i = 0; i < febCount; i++) {
//Here we print out the fibonacci number we are on, unconditionally.
//This means that every fibonacci number will be printed no matter what number it is
//we don't want that.
System.out.print(feb[i] + ((i % 10 == 9) ? "\n" : " "));
//After we print the number, we check to see if it is a multiple of three.
//maybe we should be waiting to print until then?
if (feb[i] % 3 == 0)
System.out.print("skip");
}
Now that we have walked through the code, we can propose a new solution.
Let's try updating the loop so that it wait's to print the fibonacci number until AFTER we've checked to see if it meets our conditions.
for (int i = 0; i < febCount; i++) {
if (feb[i] % 3 == 0 || feb[i] % 5 == 0 || feb[i] % 7 == 0) { //check if multiple of 3 5 or 7
System.out.println(" Skip ");
} else { //if it's not a multiple, then print the number
System.out.println(" " + feb[i]);
}
}

How To Sort Subsets in Java

Dear friends I have an assignment and I almost solved it. But I'm having a big problem recently which I couldn't figure a way out for 2 days. If you could help me I would very appreciate it!
So, let's say user entered 5 (N) I immediately create this sequence to get subsets out of it: {1,2,3,4,5}
If N = 4 than the sequence is like: {1, 2, 3, 4} etc.
Than this code below generates all kind of the variations of subsets:
public static int[] genarator(int N)
{
int[] generator = new int[(int) Math.pow(2, N)];
int[] binDigit = new int[(int) Math.pow(2, N)];
for (int i = 0; i < Math.pow(2, N); i++)
generator[i] = (i >> 1) ^ i; // Right Shifting
for (int i = 0; i < Math.pow(2, N); i++)
{
int one = 1;
binDigit[i] = 0;
while (generator[i] > 0)
{
binDigit[i] += (generator[i] % 2) * one;
generator[i] /= 2;
one = one * 10;
}
}
return binDigit;
}
And the way it returns the results like this (In case of: N = 4 {1, 2, 3, 4}) shown here :
1
1 2
2
2 3
1 2 3
1 3
3
3 4
1 3 4
1 2 3 4
2 3 4
2 4
1 2 4
1 4
4
But my lecturer wants from my program to return the result in this order:
1
2
3
4
1 2
1 3
1 4
2 3
2 4
3 4
1 2 3
1 2 4
1 3 4
2 3 4
1 2 3 4
I for now use TreeSet<Long> and parseLong so I can get true results till 1 <= N <= 9. But whenever user enters 10 or higher as N it goes crazy.
To recap, my question is how can I store those numbers which I get from int[] genarator(int N) and display them like my lecturer requires ?
How generator works and how do I get numbers in wrong order? Code is below:
int N = read.nextInt();
int[] sequence = new int[N];
for (int i = 0; i < N; i++)
sequence[i] = i + 1;
int[] tempArray = new int[(int) Math.pow(2, N)];
tempArray = genarator(N);
for (int i = 1; i < Math.pow(2, N); i++)
{
for (int j = 0; j < N; j++)
{
if (tempArray[i] % 10 == 1)
{
System.out.print(sequence[j] + " ");
}
tempArray[i] /= 10;
}
System.out.println();
}
Thank you for checking and I am really sorry for this too long question. But I couldn't make it clear with a short explanation.

What you can do is create a set abstraction that can be compared to other sets. See Java tutorials on Comparators.
//don't call this Set as there is already a Java Set
//interface that youdon't want to confuse yourself with
public class MySet implements Comparable<MySet>{
int[] backingArray;
public MySet(int n) {
//initialize the Set
this.backingArray = generator(n);
}
public static Int[] generator(int n) {
//..whatever you do to generate your results
}
#Override
public int compareTo(MySet otherSet) {
//define how sets are compared (e.g what your professor is asking.
//In your example, if one set is shorter than another,
//it is considered 'smaller')
}
}
Set<MySet> allSets = ....;
and simply invoke Collections.sort(allSets);