Open In App

Java Program to Add Two Binary Strings

Improve
Improve
Like Article
Like
Save
Share
Report

When two binary strings are added, then the sum returned is also a binary string.

Example:

Input :  x = "10", y = "01"
Output: "11"
Input :  x = "110", y = "011"
Output: "1001"
Explanation:
  110 
+ 011
=1001

Approach 1:

Here, we need to start adding from the right side and when the sum returned is more than one then store the carry for the next digits.

Let us see a program in order to get a clear concept of the above topic.

Example:

Java




// Java program to add two binary strings
 
public class GFG {
 
    // Function to add two binary strings
    static String add_Binary(String x, String y)
    {
 
        int num1 = Integer.parseInt(x, 2);
        // converting binary string into integer(decimal
        // number)
 
        int num2 = Integer.parseInt(y, 2);
        // converting binary string into integer(decimal
        // number)
 
        int sum = num1 + num2;
        // Adding those two decimal numbers and storing in
        // sum
 
        String result = Integer.toBinaryString(sum);
        // Converting that resultant decimal into binary
        // string
 
        return result;
    }
 
    // Main driver method
    public static void main(String args[])
    {
        String x = "011011", y = "1010111";
 
        System.out.print(add_Binary(x, y));
    }
}


Output

1110010

Approach 2: Two Pointer

  1. Initialize two pointers at the end of both strings, let’s call them i and j.
  2. Initialize a variable carry to 0.
  3. While i and j are greater than or equal to 0, do the following:
    • Convert the current digits at i and j to integers (0 if the pointer is out of bounds).
    • Add the integers together with the carry value.
    • If the sum is 0 or 1, add it to the result string and set carry to 0.
    • If the sum is 2, add 0 to the result string and set carry to 1.
    • If the sum is 3, add 1 to the result string and set carry to 1.
    • Decrement i and j by 1.
  4. If there is still a carry left over, add it to the front of the result string.
  5. Reverse the result string and return it.

Java




import java.io.*;
 
// Class
class GFG {
 
    // Method
    public static String addBinary(String x, String y)
    {
        int i = x.length() - 1, j = y.length() - 1;
        int carry = 0;
        StringBuilder result = new StringBuilder();
        while (i >= 0 || j >= 0) {
            int sum = carry;
            if (i >= 0) {
                sum += x.charAt(i) - '0';
            }
            if (j >= 0) {
                sum += y.charAt(j) - '0';
            }
            if (sum == 0 || sum == 1) {
                result.append(sum);
                carry = 0;
            }
            else if (sum == 2) {
                result.append("0");
                carry = 1;
            }
            else {
                result.append("1");
                carry = 1;
            }
            i--;
            j--;
        }
        if (carry == 1) {
            result.append("1");
        }
        return result.reverse().toString();
    }
 
    // Main driver method
    public static void main(String[] args)
    {
        String x = "011011";
        String y = "1010111";
       
        System.out.println(addBinary(x, y));
    }
}


Output

1110010

Time complexity: O(max(N, M))
Auxiliary space: O(max(N, M))



Last Updated : 19 May, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads