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
- 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.*;
// 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))