HackerRank Java- Datatypes




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.



  1. import java.util.*;
  2. import java.io.*;
  3. class Solution {
  4. public static void main(String[] args) {
  5. Scanner scan = new Scanner(System.in);
  6. int t = scan.nextInt();
  7. for (int i = 0; i < t; i++) {
  8. try {
  9. long x = scan.nextLong();
  10. System.out.println(x + " can be fitted in:");
  11. if (x >= Byte.MIN_VALUE && x <= Byte.MAX_VALUE) {
  12. System.out.println("* byte");
  13. }
  14. if (x >= Short.MIN_VALUE && x <= Short.MAX_VALUE) {
  15. System.out.println("* short");
  16. }
  17. if (x >= Integer.MIN_VALUE && x <= Integer.MAX_VALUE) {
  18. System.out.println("* int");
  19. }
  20. if (x >= Long.MIN_VALUE && x <= Long.MAX_VALUE) {
  21. System.out.println("* long");
  22. }
  23. } catch (Exception e) {
  24. System.out.println(scan.next() + " can't be fitted anywhere.");
  25. }
  26. }
  27. scan.close();
  28. }
  29. }




codesadda.com

Codesadda.com is your home of programming solutions, tutorials, video tutorials and much more. Sign Up for our weekly newsletter to get update about new content.

Like us on Facebook | Connect with us on LinkedIn | Subscribe our Channel on Youtube