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How to remove the amounts of ifs from the application? [closed]

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The code I say has to complete these assignments:
Given an integer,N, perform the following conditional actions:
If N is odd, print Weird
If N is even and in the inclusive range of 2 to 5 , print Not Weird
If N is even and in the inclusive range of 6 to 20 , print Weird
If N is even and greater than 20 , print Not Weird
Complete the stub code provided in your editor to print whether or not N is weird.
My code looked like this:
private static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args) {
int N = scanner.nextInt();
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
scanner.close();
int Numberparorimpa = N % 2;
if(N < 2 || Numberparorimpa ==1 || N <=20 && N >=6 ){
System.out.println("Weird");
}else{
if(N >=2 && Numberparorimpa == 0){
System.out.println("Not Weird");
}else{
if(Numberparorimpa == 0 && N >=6 || N<=20){
System.out.println("Weird");
}else{
if(Numberparorimpa== 0 && N> 20){
System.out.println("Not Weird");
}else{
return;
}
}
}
}
}
}
How can I reduce the IFs of this code?

I think an optimize version could be this :
if (N % 2 == 1 || (N >= 6 && N <= 20)) {
System.out.println("Weird");
}
else {
System.out.println("Not Weird");
}
If N is odd or N in range of 6 to 20 it's weird.
Else N is either even or not in the range so it's not weird.

You can reduce the complexity and improve the readability of your code by extracting your logic to method/class etc. Also, nested conditions are hard to read, you should avoid it.
Example:
public static void main(String[] args) {
int n = scanner.nextInt();
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
scanner.close();
String print = isGivenNumberWeird(n) ? "Weird" : "Not Weird"
System.out.println(print);
}
private static boolean isGivenNumberWeird(int n) {
boolean isOdd = n % 2 == 1;
if (isOdd) {
return true;
}
if (n >= 2 && n <=5) {
return false;
}
if (n >= 6 && n <=20) {
return true;
}
if (n > 20) {
return false;
}
}

What is my mistake in this problem. Simple Comparison Error [closed]

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Task
Given an integer, n, perform the following conditional actions:
If n is odd, print Weird
If n is even and in the inclusive range of 2 to 5 , print Not Weird
If n is even and in the inclusive range of 6 to 20, print Weird
If n is even and greater than 20, print Not Weird
I have written a code but it is showing error in printing 18 and 20.
public static void main(String[] args) {
int N = scanner.nextInt();
scanner.close();
if(N%2 != 0){
System.out.println("Weird");
}
else if(N%2 ==0 && N>=2||N<=5)
{
System.out.println("Not Weird");
}
else if(N%2 ==0 && N>=6||N<=20)
{
System.out.println("Weird");
}
else if(N%2 ==0 && N>20)
{
System.out.println("Not Weird");
}
}
}

You need to change your or condition || to and condition && to check for the range like
else if(N%2 == 0 && N >= 2 && N <= 5)

FizzBuzz: It gives me the wrong print statement. What is wrong?

Everybody knows that FizzBuzz question that interviewers ask students.
Basically, when you have an incrementor and for each number which is a divisible of 3 you say fizz, for a number divisible by 5 you say buzz, while if it is divisible by both(3 and 5) you say FizzBuzz, hence the name.
It is a relatively easy problem to solve and I have done it, but I think my solution is a bit clunky. This is it:
int[] numbers = new int[100];
for (int i = 0; i < numbers.length; i++) {
numbers[i] = i;
}
for (int i : numbers) {
if (i % 3 == 0) {
System.out.println("Fizz");
} else if(i % 5 == 0) {
System.out.println("Buzz");
} else {
System.out.println("FizzBuzz");
}
}
But the problem is that when the number is divisible by both 3 and 5 it gives me "Fizz" for some reason. Can somebody explain to me, because I'm new to java programming. Thanks in advance!

The problem lies in the order of your if statements. Lets take a look at the number 15, which is the first number divisible by both 3 and 5. Because of the order in which you have your if statements, the first statement that is checked is
if ( 15 % 3 == 0)
The result of the operation is indeed equal to 0, as 15 is divisible by 3 and so "Fizz" is printed and your else is ignored.
Think about how you should structure the order of your if statements and which additional condition should you introduce to catch the specific case of being divisible by both i % 3 == 0 && i % 5 == 0.

When you enter the if statement and your number is 15 for exemple, you enter the first if statement and.. prints "Fizz" as you stated, because 15 % 3 == 0 returns true. Then it ignores the else.
You want the first if to be
if (i % 3 == 0 && i % 5 == 0) {
System.out.println("FizzBuzz");*
}

Try this code
public static void main(String[] args) {
int[] numbers = new int[100];
for (int i = 0; i < numbers.length; i++) {
numbers[i] = i;
}
for (int i : numbers) {
if ((i % 3 == 0) && (i % 5 == 0)) {
System.out.println("FizzBuzz");
} else if(i % 5 == 0) {
System.out.println("Buzz");
} else if (i % 3 == 0){
System.out.println("Fizz");
}
}
}

multiple conditions else / if statement somehow wrong?

I'm doing a hackernet challenge where n is an int input. The conditions are:
If n is odd, print Weird
If n is even and in the inclusive range of 2 to 5, print Not Weird
If n is even and in the inclusive range of 6 to 20, print Weird
If n is even and greater than 20, print Not Weird.
Im sure the code makes logic and dont think theres syntax. It gives the correct responses and hackernet still says its incorrect so ive come here to see if anyone can see what the problem is
public static void main(String[] args)
{
int N = scanner.nextInt();
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
if (N % 2 != 0 || N % 2 == 0 && N >= 6 && N <= 20)
{
System.out.print("weird");
}
else
{
System.out.print("not weird");
}
}

The problem is the logic in your else condition, which would also catch values of N which are less than 2. Try this version:
if (N % 2 != 0)
{
System.out.print("weird");
}
else if (N >= 2 && N <= 5 || N > 20)
{
System.out.print("not weird");
}
else if (N >= 6 && N <= 20)
{
System.out.print("weird");
}
else
{
// NOTE: if the code still fails, remove this else condition
System.out.print("unexpected value of N");
}
Note: To get your code to pass the Hackernet task, you might have to completely remove the else condition. I added it for completeness, but Hackernet might test N=1 to see if nothing gets printed.

Read this condition :
if (N % 2 != 0 || N % 2 == 0 && N >= 6 && N <= 20)
as
if (N % 2 != 0 || (N % 2 == 0 && N >= 6 && N <= 20))
Then see how operator precedence changes the behaviour and yield desired results.

Check the following one
public static void main(String[] args)
{
int N = scanner.nextInt();
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
if(N%2!=0) {
System.out.print("weird");
}else if(N>=2 && N<=5) {
System.out.print("not weird");
}else if(N>=6 && N<=20) {
System.out.print("weird");
}else if(N>20) {
System.out.print("not weird");
}
}

For the technical part: start by reading about
precedence of java operators and then make your code easier to read.
Pushing that many conditions into a single if is not helpful. You see it yourself: you think the code is correct, but probably it isn't. And now you look to other people to explain your overly complex code back to you. And of course, all the other answers do all that for you ... but beyond that:
The "real" answer here is: learn how to test your code.
Instead of having a main that somehow asks for a number, and then makes decisions, write a method boolean isWeird() that takes a number and returns true/false according to your requirements.
And then simply test that method with all reasonable cases. And then check if that result is as expected.
Using JUnit, you could write something like
assertThat(isWeird(1), true);
assertThat(isWeird(21), true);
assertThat(isWeird(22), true);
...
Ideally, you write such tests before you implement that method. And then you implement all the conditions, and any check that fails tells you that you got something wrong.

I feel, In the if (N % 2 != 0 || N % 2 == 0 && N >= 6 && N <= 20) condition, you are verifiying the odd and even values at same time using && and || operator. Can you modify the condition into like this if (N % 2 != 0 || (N % 2 == 0 && N >= 6 && N <= 20)) and check? If N is odd weird will be printed or if N is even and it falls under the 6 and 20 inclusive, weird will be printed.

You already have a good few answers here but if you think logically about what you actually need, you can break it down easier.
It looks like the only "Not Weird" print out is 2, 4 and even numbers > 20
So an example could be something like:
if (n % 2 == 0) {
if ((n >= 2 && n <= 5) || (n > 20)) {
return "Not Weird";
}
}
return "Weird";

You can try this
private static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args) {
int n = scanner.nextInt();
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
if (n % 2 == 1 || (n >= 6 && n <= 20)) {
System.out.println("Weird");
} else {
System.out.println("Not Weird");
}
scanner.close();
}

Comparing Even/Odd numbers w/ java and boolean [closed]

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I am writing a method that takes in two numbers and will return true if they are both even or both odd... and will return false if only one is odd and one is even.
It needs to return a boolean statement, but it is not working.. Any help is appreciated! Thanks...
public static boolean compareEvenOdd(int x, int y) {
if((x % 2 ==0) && ( y% 2==0))||((x%2 != 0) && (y%2 != 0)){
return true;
} else
return false;
}

You can do the following (possibly the shortest version):
public static boolean compareEvenOdd(int x, int y) {
return ((x + y) % 2) == 0;
}
The sum of two odd numbers and the sum of two even numbers is even, the sum of one odd and one even number is odd. So you add the numbers and check if the solution is dividable by 2.

This is a really good opportunity to use a bitwise operator. You can use the binary AND (&) operator with the number 1 to reduce an integer to just its last binary digit. This last digit is always 0 for an even number and 1 for an odd number. If your two numbers have the same last binary digit, then they have the same parity - that is, they're both even or both odd. So I would write the method like this.
public boolean sameParity(int x, int y) {
return (x & 1) == (y & 1);
}
Note that the parentheses are important, because the usual Java order of operations puts == above &.

Seems you are missing an additional pair of parentheses around the condition in the if-statement.
This works for me:
public static boolean compareEvenOdd(int x, int y) {
if (((x % 2 ==0) && ( y% 2==0))||((x%2 != 0) && (y%2 != 0))){
return true;
} else
return false;
}
if this is not to your liking, yould you further specify what is not working? Is it throwing errors as you run it, is the output wrong,...?

Explication :
x and y are odd : x+y are even
x and y are even : x+y are even
x is odd and y is odd : x+y are odd
x is even and x is even : x+y are odd
public static boolean compareEvenOdd(int x, int y) {
return (x+y)%2==0
}

An if-statement always has the form if (...) with its own parentheses around the condition. Hence you missing just that.
So (with unneeded braces removed):
public static boolean compareEvenOdd(int x, int y) {
if ((x % 2 == 0 && y % 2 == 0)
|| (x%2 != 0 && y % 2 != 0)) {
return true;
} else
return false;
}
And as shown in the other answers this should be simplified,
as if+return boolean means using boolean redundantly, not to the fullest (like == true):
return (x % 2 == 0 && y % 2 == 0)
|| (x % 2 != 0 && y % 2 != 0);
return (x % 2) == (y % 2);
return (x - y) % 2 == 0;