Finding the appropriate Java Datatype

Question

public static void main(String[] args) {
    Scanner input = new Scanner(System.in);

    while (input.hasNextLine()) {
        BigInteger number = new BigInteger(input.nextLine());

        int bitLength = number.bitlength();
        if (bitLength <= Bytes.SIZE)
            System.out.println("\u8211 byte");
        if (bitLength <= Short.SIZE)
            System.out.println("\u8211 short");
        if (bitLength <= Int.SIZE)
            System.out.println("\u8211 int");
        if (bitLength <= Long.SIZE)
            System.out.println("\u8211 long");

        if (bitLength > Long.SIZE)
            System.out.println(number + " can't be fitted anywhere.");
    }
} 

Task : to find a suitable data type 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

Error 1:

error: cannot find symbol
    int bitLength = number.bitlength();
                      ^

Error 2:

symbol:   method bitlength()
location: variable number of type BigInteger

Error 3:

error: cannot find symbol
    if (bitLength <= Int.SIZE)
                 ^
    symbol:   variable Int
    location: class Solution

Answer 1

Read the number line by line. Count bit using BigInteger and divide it by 8 for switch case simplification. Have a look at below code:

    Scanner input = new Scanner(new File("so/input.txt"));
    while (input.hasNextLine()) {
        BigInteger number = new BigInteger(input.nextLine().trim());
        int bitLength = number.bitLength();
        int len = bitLength / 8;
        StringBuilder output = new StringBuilder(number.toString() + " can be fitted in:\n");
        switch (len) {
            case 0:
                output.append(" byte");
            case 1:
                output.append(" short");
            case 2:
            case 3:
                output.append(" int");
            case 4:
            case 5:
            case 6:
            case 7:
                output.append(" long");
                System.out.println(output);
                break;
            default:
                System.out.println(number.toString() + "  can't be fitted anywhere.");
        }
    }