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Java Program to Convert a Decimal Number to Binary & Count the Number of 1s

As per the number system, default computations are carried over decimal numbers whose base is standardized as 10. Machine computes all the execution at the physical layer in 0s and 1s. So arises a need for a number system with base 2 known as a binary number system. A binary number can be converted to a decimal number and vice versa. In java there are 4 types of numbers:

Types Of Numbers  Base
Binary 2
Octa-Decimal 8
Decimal 10
Hexadecimal 16

Approaches: In Java, there are 2 ways to convert either using pre-defined methods or third-grade logic building as listed below: 

  1. Using Integer.toBinaryString method (Integer wrapper class)
  2. Using the Brute force method (without any use of predefined classes)

Examples:

Input. 1: 10
Output 1: The binary equivalent of 10 is : 1010
          Number of 1s is : 2

Input  2: 15
Output 2: The binary equivalent of 15 is : 1111
          Number of 1s is 4

Approach 1: Using toBinaryString() Method: represents the number to be converted into binary. The Integer class of java provides some useful methods to deal with Integers. One such method is Integer.toBinaryString(int x) 

Syntax:

public static String toBinaryString(int variable_name)

Parameters: Decimal integer to be converted.

Return Type: String which is holding binary representation of the integer converted or simply binary equivalent of integer as a String object.

Exceptions: There are no exceptions thrown by this method 

Parameter: It takes one parameter of type Integer (or int)  

Implementation: To count the number of 1s, check if each character of the obtained binary string equals 1 or not.

Java




// Java program to convert decimal to binary number
// and counting number of 1's in it
 
public class GFG {
 
    // Function to convert decimal to binary
    void convertAndCount(int num)
    {
        // Store the count of 1s currently 0
        int count = 0;
 
        // Returns a String object representing
        // binary equivalent of passed decimal number
        String binary = Integer.toBinaryString(num);
 
        // Iterating over obtained string using length
        // function
        for (int i = 0; i < binary.length(); i++)
 
            // Checking 1 is present in binary
            if (binary.charAt(i) == '1')
 
                // Increment the count
                // if any char equals 1
                count++;
 
        // Printing the binary equivalent
        System.out.println("The binary equivalent of " + num
                           + " is : " + binary);
 
        // Printing the no of 1 in above binary number
        System.out.println("Number of 1s is : " + count);
    }
 
    // Main driver method
    public static void main(String[] args)
    {
        // Creating object in main
        GFG obj = new GFG();
 
        // Calling the convertAndCount() method
        // over the integer value 18
        obj.convertAndCount(18);
    }
}


Output

The binary equivalent of 18 is : 10010
Number of 1s is : 2

Time Complexity: O(n)

Auxiliary Space: O(n)

Approach 2:Without using pre-defined variables

  • Dividing the decimal number by 2 which is to be converted into binary
  • Storing the remainder

Java




// Java program to convert decimal to binary number
// and counting number of 1's in it
 
public class GFG {
 
    // Function to convert decimal to binary
    void convertAndCount(int num)
    {
        int temp = num; // a temporary variable to store the
                        // value of num
 
        // to store the binary value
        int[] binary = new int[20];
 
        // to store the count of 1s
        int count = 0;
        int i;
 
        // to iterate through the loop and keep a
        // count of no.of digits in the obtained binary
        for (i = 0; temp > 0; i++) {
 
            // divide the number by 2
            // and store the remainder
            temp /= 2;
            binary[i] = temp % 2;
 
            // If 1 is present in binary
            if (binary[i] == 1)
 
                // increment the count if any digit
                // is equal to 1
                count++;
        }
 
        // Printing binary of decimal number
        System.out.print("The binary equivalent of " + num
                         + " is : ");
 
        // Iterating over array
        for (int j = i - 1; j >= 0; j--)
 
            // Printing obtained array in reverse order
            System.out.print(binary[j]);
 
        // Printing number of 1's
        System.out.println("\nNumber of 1s is : " + count);
    }
 
    // Main driver method
    public static void main(String[] args)
    {
        // Creating class GFG object in main
        GFG obj = new GFG();
 
        obj.convertAndCount(18);
        // calling convertAndCount() method on decimal
        // over the value 18
    }
}


Output

The binary equivalent of 18 is : 01001
Number of 1s is : 2

Time complexity: O(n) for given number n

Auxiliary space: O(1)

Dominic Rubhabha-Wardslaus
Dominic Rubhabha-Wardslaushttp://wardslaus.com
infosec,malicious & dos attacks generator, boot rom exploit philanthropist , wild hacker , game developer,
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