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Java Datatypes Hackerrank solution in Java

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 Java Datatypes Hackerrank solution in Java



Problem-7:-

Java has 8 primitive data types; char, boolean, byte, short, int, long, float, and double. For this exercise, we'll work with the primitives used to hold integer values (byte, short, int, and long):

  • A byte is an 8-bit signed integer.
  • A short is a 16-bit signed integer.
  • An int is a 32-bit signed integer.
  • A long is a 64-bit signed integer.

Given an input integer, you must determine which primitive data types are capable of properly storing that input.

To get you started, a portion of the solution is provided for you in the editor.

Reference: https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html

Input Format

The first line contains an integer, , denoting the number of test cases.
Each test case, , is comprised of a single line with an integer, , which can be arbitrarily large or small.

Output Format

For each input variable  and appropriate primitive , you must determine if the given primitives are capable of storing it. If yes, then print:

n can be fitted in:
* dataType

If there is more than one appropriate data type, print each one on its own line and order them by size (i.e.: ).

If the number cannot be stored in one of the four aforementioned primitives, print the line:

n can't be fitted anywhere.

Sample Input

5
-150
150000
1500000000
213333333333333333333333333333333333
-100000000000000

Sample Output

-150 can be fitted in:
* short
* int
* long
150000 can be fitted in:
* int
* long
1500000000 can be fitted in:
* int
* long
213333333333333333333333333333333333 can't be fitted anywhere.
-100000000000000 can be fitted in:
* long

Explanation

 -150 can be stored in a short, an int, or a long.

 2133333333333333333333333333333333333 is very large and is outside of the allowable range of values for the primitive data types discussed in this problem.


Solution:-

import java.util.*;
import java.io.*;



class Solution{
    public static void main(String []argh)
    {



        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");
                }
                if(x>=-32768 && x<=32767){
                    System.out.println("* short");
                }
                if(x>=-2147483648 && x<=2147483647){
                    System.out.println("* int");
                }
                if(x >= -Math.pow(2, 63) && x <= Math.pow(2, 63) - 1)
                System.out.println("* long");
                
            }
            catch(Exception e)
            {
                System.out.println(sc.next()+" can't be fitted anywhere.");
            }

        }
    }
}


Explanation:-

In this program, basically they ask for to print the data type in which the integer value can be put. For Example, a byte can stored the integer value starting from -8 to 7 and other datatype also have there range which is given below:-

                 pic source:- cs.fit.edu

When writing the code, you may note that, i use if statement more than one time(or Nested If) and not use If-else statement. This is because if we use if else statement then then number after considering the number stop there execution and return the data type value and not proceed further. This is why we use nested if statement.


Hackerrank Completion Proof:-



Hope, it Helps!!
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