Hey Kucao, I had the same solution as yours, with just this minor change, and I got it wrong. Not able to understand why. Can someone explain ?

    if(x >= (short)-Math.pow(2, 15) && x <=(short) Math.pow(2, 15) - 1)
        System.out.println("* short");
        if(x >= (int)-Math.pow(2, 31) && x <=(int) Math.pow(2, 31) - 1)
            System.out.println("* int");
        if(x >= (long)-Math.pow(2, 63) && x <=(long) Math.pow(2, 63) - 1)
            System.out.println("* long");

hey kucao,why did you converted the int number to long?what is the need of doing that?please reply.

I solved it by using Datatype.MAX_VALUE; DataType.MIN_VALUE; ex. Long.MAX_VALUE;

Which might not be the point of the exercise, but I think it's really robust code.

Can you tell me what does the -1 do in the if statement?

Hey Shafaet, I had some masking techique that can not using any library but am not implemented it,

A&0=0

The return you get from Math.pow is in doubleand you're comparing it with x i.e.long. Can we do it? Don't you think there's flaw here? If no! Could you please explain why? Using MIN and MAX is the best option I suppose.

can you please tell me why we want to use -1 ??

hey Kucao, With the help of this code i creared only Test case #0. for other test case it is wrong. Can you please help me.

I had the same solution. But i wasnt able to run it. System said it is a worng answer. Can someone please explain? Thanks!

thanks

i have written the same code but it shows error in last two testcases

why is

if(x>=-32768 && x<=32768)

Different from:

if(x >= -Math.pow(2, 15) && x <= Math.pow(2, 15) - 1)

I tried the first one since its the value given by java documentation but it doesn't work on the last two test cases. can anyone please tell me..

plz explain this solution

why sc.next() is used in catch??? why cant we use x instead ?

Thanks.. It helped!

what about test case -100000000000000000??? it's not work???

can you please tell what sc.next() does?

please explain how sc.next() working outside the try{}. System.out.println(sc.next()+" can't be fitted anywhere.");

i entered the actual values in test cases. It threw error for the value 9223372036854775807 saying that it can't store that number for long.

@kucao why you had writeen "can't be fitted anywhere" in catch block which exception will be thrown for this?

can you please explain me this code ?

i'm unable to submit this above code as well don't understand why

hye.can someone explain what is the meaning of -Math.pow(2,15) and when we need to use -Math.pow?and also the function of -Math.pow.tq

why compiler showing error out of range if(x>=-9223372036854775808 && x<= 9223372036854775807) as 9223372036854775807 is the range of long

        if(x >= -Math.pow(2, 15) && x <= Math.pow(2, 15) - 1)
            System.out.println("* short");
        if(x >= -Math.pow(2, 31) && x <= Math.pow(2, 31) - 1)
            System.out.println("* int");
        if(x >= -Math.pow(2, 63) && x <= Math.pow(2, 63) - 1)
            System.out.println("* long");


            please can you explain these lines ??...

why we use try?

Kucao, and all

Kucao's solution is NOT exactly correct. What if the input is number = 10. Number 10 can be fit into ANY primitive type. So, every single "if" in Kucao's code will be duplicated in the output.

Better to use if-else if construct. Like this

long x=sc.nextLong();
                System.out.println(x+" can be fitted in:");
                if(x>=-128 && x<=127)System.out.println("* byte\n* short\n* int\n* long");
                else if( x>=((-1)*(Math.pow(2,15))) && x<=((Math.pow(2,15)-1)) )System.out.println("* short\n* int\n* long");
                else if( x>=((-1)*(Math.pow(2,31))) && x<=((Math.pow(2,31)-1)) )System.out.println("* int\n* long");
                else if( x>=((-1)*(Math.pow(2,63))) && x<=((Math.pow(2,63)-1)) )System.out.println("* long");

Not working for Cases #2 and #3

U don't need -1 at the end of each code result still is same

In your solution Math.pow(2,63) is actually going to return an Int, it will return the largest int it can be calculated (2^31)-1. You have to specifically point out which numbers are going to be considered longs by appending an L at the end of the number (Like this: 600851475143L). This is because in java long literals end with an L. Source

I have almost the same solution as you, with the exception that I have every max and min dataType value stored in a different variable (like minInt, maxInt, minShort and so on).

Even though, I cant get it right on the test scenarios once the solution is submitted. Its quite frustrating.

Instead of wasting your time and resources doing mathematical calculations, you should just use final static values that are kept in classes. For example classes Byte, Short, Int, Long have final static variables for their MIN_VALUE and MAX_VALUE. So you can write your program like this to make it much more efficient. Don't forget, we are computer engineers. Engineer is a person who solves problem and solved problems efficiently. Solving a problem alone is not engineering.

if (x >= Byte.MIN_VALUE && x <= Byte.MAX_VALUE) {
	System.out.println("* byte");
}
if (x >= Short.MIN_VALUE && x <= Short.MAX_VALUE) {
	System.out.println("* short");
}
if (x >= Integer.MIN_VALUE && x <= Integer.MAX_VALUE) {
	System.out.println("* int");
}
if (x >= Long.MIN_VALUE && x <= Long.MAX_VALUE) {
	System.out.println("* long");
}

in the following line of code: "System.out.println(sc.next()+" can't be fitted anywhere.");",instead of "sc.next",Why can't we write "sc.nextLong"?

This solution is giving 2 testcases wrong. What is the problem? Can anyone please explain ?

        Scanner sc = new Scanner(System.in);
    int t=sc.nextInt();

    for(int i=0;i<t;i++)
    {

        try
        {
            long x=sc.nextLong();
            System.out.println(x+" can be fitted in:");
            if(x>=-128 && x<=127)
                System.out.println("* byte");
            //Complete the code
            if(x>=-32768 && x<=32767)
                System.out.println("* short");
            if(x>=Math.pow(-2,31) && x<Math.pow(2,31))
                System.out.println("* int");
            if(x>=Math.pow(-2,63) && x<Math.pow(2,63))
                System.out.println("* long");

        }
        catch(Exception e)
        {
            System.out.println(sc.next()+" can't be fitted anywhere.");
        }

    }
    sc.close();

}

WHY WE USE SC.NEXT() IN CATCH PLS EXPLAIN!

can someone explain this code.please

I think it is very unnecessary to check if it fits into long. We have x variable in long datatype in the beginning, and also if it fits byte, short and int all the way, it'll definitely fit in long. So you can just print that it fits into long, without checking.

P.S. Why doesn't HackerRank have code formatting?..

Can we use range instead of Math.pow()?How Math.pow() working?

Instead of

( ... && x <= Math.pow(2, 15) - 1)

You could write this:

(.. && x < Math.pow(2, 15))

Without "=" and "-1"

is it ok continous if without else?

long x = sc.nextLong(); System.out.println(x + " can be fitted in:");

        // Check if Byte
        if (x >= Byte.MIN_VALUE && x <= Byte.MAX_VALUE) {
            System.out.println("* byte");
            System.out.println("* short");
            System.out.println("* int");
            System.out.println("* long");
        }
        // Check if Short
        else if (x >= Short.MIN_VALUE && x <= Short.MAX_VALUE) {
            System.out.println("* short");
            System.out.println("* int");
            System.out.println("* long");
        }
        // Check if Integer
        else if (x >= Integer.MIN_VALUE && x <= Integer.MAX_VALUE)

Why can't we use the ranges directly instead of Math.pow. For example,instead of MAth.pow(2,31)why can't we use 32,768

You actually don't need the last if statement in your code.

The variable 'x' is initialized as a long. If it initializes successfully, then we already know it fits in long. If it doesn't fit in long, an exception will be thrown at the initialization statement, prompting the catch statement which prints out that it can't be fitted anywhere.

Hello I can't understand your code so pls explain your code!!

unsigned representations are only accessible through the objects Integer and Long, not through the primitive data types int and long that are referenced in this problem.

yes

Thanks for the tip using the MIN_VALUE and MAX_VALUE I had no Idea that there were those in primitive data type classes.

Hi Thanks for your solution. I just wanted to add something in it. All classes in the java.lang package are imported by default. Thus you do not need to import java.lang.*;

tnks for the tip, i used those method and every test they ran.

thank you,very godd!

thanks a lot, learnd to use these functions

Thanks man .... i was frustraed as my code was,nt submitting.

this is aweful, why can't the code read as value instead off the static variables? no need to write code if it cannot read direct value.

I agree. We went from explaining the data types to input format and output format. I havent got around the idea of UX, but now I see why its important XD

welcome

Hi, Why you used sc.next() in the Exception block ? Can you please Explain me ?

thanku so much sir. i understand now how to solve the problems like these.

I'm also do this but why my test case are failed...? can you help me please..

Ahh that's how you access the MAX_VALUE and MIN_VALUE constants. Thanks. Just a note, you could remove if (x >= Long.MIN_VALUE && x <= Long.MAX_VALUE). x has to be a long by definition so we don't have to check it.

Wow! I can see a hand of the C-programmer here... :-)

since it is a long literal use L at the end of that number. (x=-9223372036854775808L && x<=9223372036854775807L)

not op, but @rajujnvgupta

|| is the Logical OR operator. -> returns a boolean which is true if and only if both of its operands are true. b = 3 > 2 || 5 < 7; // b is true b = 2 > 3 || 5 < 7; // b is still true b = 2 > 3 || 5 > 7; // now b is false

You should check out Java Operator and their Precedence