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 stringspublic 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
- Initialize two pointers at the end of both strings, let’s call them i and j.
- Initialize a variable carry to 0.
- 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.
- If there is still a carry left over, add it to the front of the result string.
- Reverse the result string and return it.
Java
import java.io.*;// Classclass 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))
