Program to add two binary strings

Given two binary strings, return their sum (also a binary string).

Example:

Input:  a = "11", b = "1"
Output: "100" 

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The idea is to start from the last characters of two strings and compute the digit sum one by one. If the sum becomes more than 1, then store carry for the next digits. 

C

// C++ program to add two binary strings
#include <bits/stdc++.h>
using namespace std;

// This function adds two binary strings and return
// result as a third string
string addBinary(string A, string B)
{
    // If the length of string A is greater than the length
    // of B then just swap the string by calling the
    // same function and make sure to return the function
    // otherwise recursion will occur which leads to
    // calling the same function twice
    if (A.length() > B.length())
        return addBinary(B, A);

    // Calculating the difference between the length of the
    // two strings.
    int diff = B.length() - A.length();

    // Initialise the padding string which is used to store
    // zeroes that should be added as prefix to the string
    // which has length smaller than the other string.
    string padding;
    for (int i = 0; i < diff; i++)
        padding.push_back('0');

    A = padding + A;
    string res;
    char carry = '0';

    for (int i = A.length() - 1; i >= 0; i--) {
        // This if condition solves 110 111 possible cases
        if (A[i] == '1' && B[i] == '1') {
            if (carry == '1')
                res.push_back('1'), carry = '1';
            else
                res.push_back('0'), carry = '1';
        }
        // This if condition solves 000 001 possible cases
        else if (A[i] == '0' && B[i] == '0') {
            if (carry == '1')
                res.push_back('1'), carry = '0';
            else
                res.push_back('0'), carry = '0';
        }
        // This if condition solves 100 101 010 011 possible
        // cases
        else if (A[i] != B[i]) {
            if (carry == '1')
                res.push_back('0'), carry = '1';
            else
                res.push_back('1'), carry = '0';
        }
    }

    // If at the end their is carry then just add it to the
    // result
    if (carry == '1')
        res.push_back(carry);
    // reverse the result
    reverse(res.begin(), res.end());

    // To remove leading zeroes
    int index = 0;
    while (index + 1 < res.length() && res[index] == '0')
        index++;
    return (res.substr(index));
}

// Driver program
int main()
{
    string a = "1101", b = "100";
    cout << addBinary(a, b) << endl;
    return 0;
}

Java

// java program to add
// two binary strings

public class GFG {

    // This function adds two
    // binary strings and return
    // result as a third string
    static String addBinary(String a, String b)
    {
          //If the inputs are 0
        if(a.charAt(0) == '0' && b.charAt(0) == '0'){
             return "0";
        }
        // Initialize result
        StringBuilder result = new StringBuilder("");

        // Initialize digit sum
        int s = 0;

        // Traverse both strings starting
        // from last characters
        int i = a.length() - 1, j = b.length() - 1;
        while (i >= 0 || j >= 0 || s == 1)
        {

            // Comput sum of last
            // digits and carry
            s += ((i >= 0)? a.charAt(i) - '0': 0);
            s += ((j >= 0)? b.charAt(j) - '0': 0);

            // If current digit sum is
            // 1 or 3, add 1 to result
            result.append((char)(s % 2 + '0'));

            // Compute carry
            s /= 2;

            // Move to next digits
            i--; j--;
        }

          // Remove leading zeros, if any
          int start = result.length()-1;

        while(start >=0 && result.charAt(start) == '0') {
            start--;
        }

        if(start != result.length()-1) {
            result.delete(start+1,result.length());
        }

        return result.reverse().toString();
    }

    //Driver code
    public static void main(String args[])
    {
        String a = "1101", b="100";

        System.out.print(addBinary(a, b));
    }
}

// This code is contributed by Sam007.

Python3

# Python Solution for above problem:

# This function adds two binary
# strings return the resulting string
def add_binary_nums(x, y):
        max_len = max(len(x), len(y))

        x = x.zfill(max_len)
        y = y.zfill(max_len)

        # initialize the result
        result = ''

        # initialize the carry
        carry = 0

        # Traverse the string
        for i in range(max_len - 1, -1, -1):
            r = carry
            r += 1 if x[i] == '1' else 0
            r += 1 if y[i] == '1' else 0
            result = ('1' if r % 2 == 1 else '0') + result
            carry = 0 if r < 2 else 1     # Compute the carry.

        if carry !=0 : result = '1' + result

        return result.zfill(max_len)

# Driver code
print(add_binary_nums('1101', '100'))

# This code is contributed
# by Anand Khatri

C#

// C# program to add
// two binary strings
using System;

class GFG {

    // This function adds two
    // binary strings and return
    // result as a third string
    static string addBinary(string a,
                            string b)
    {

        // Initialize result
        string result = "";

        // Initialize digit sum
        int s = 0;

        // Traverse both strings starting
        // from last characters
        int i = a.Length - 1, j = b.Length - 1;
        while (i >= 0 || j >= 0 || s == 1)
        {

            // Comput sum of last
            // digits and carry
            s += ((i >= 0)? a[i] - '0': 0);
            s += ((j >= 0)? b[j] - '0': 0);

            // If current digit sum is
            // 1 or 3, add 1 to result
            result = (char)(s % 2 + '0') + result;

            // Compute carry
            s /= 2;

            // Move to next digits
            i--; j--;
        }
    return result;
    }

// Driver Code
public static void Main()
{
    string a = "1101", b="100";
    Console.Write( addBinary(a, b));
}
}

// This code is contributed by Sam007

PHP

<?php
// PHP program to add two binary strings

// This function adds two binary strings
// and return result as a third string
function addBinary($a, $b)
{
    $result = ""; // Initialize result
    $s = 0;     // Initialize digit sum

    // Traverse both strings starting
    // from last characters
    $i = strlen($a) - 1;
    $j = strlen($b) - 1;
    while ($i >= 0 || $j >= 0 || $s == 1)
    {
        // Comput sum of last digits and carry
        $s += (($i >= 0)? ord($a[$i]) -
                          ord('0'): 0);
        $s += (($j >= 0)? ord($b[$j]) -
                          ord('0'): 0);

        // If current digit sum is 1 or 3,
        // add 1 to result
        $result = chr($s % 2 + ord('0')) . $result;

        // Compute carry
        $s = (int)($s / 2);

        // Move to next digits
        $i--; $j--;
    }
    return $result;
}

// Driver Code
$a = "1101";
$b = "100";
echo addBinary($a, $b);

// This code is contributed by mits
?>

Javascript

<script>

// Javascript program to add
// two binary strings

// This function adds two
// binary strings and return
// result as a third string
function addBinary(a, b)
{

    // Initialize result
    var result = "";

    // Initialize digit sum
    var s = 0;

    // Traverse both strings starting
    // from last characters
    var i = a.length - 1, j = b.length - 1;
    while (i >= 0 || j >= 0 || s == 1)
    {

        // Comput sum of last
        // digits and carry
        s += ((i >= 0)? a.charAt(i).charCodeAt(0) -
        '0'.charCodeAt(0): 0);
        s += ((j >= 0)? b.charAt(j).charCodeAt(0) -
        '0'.charCodeAt(0): 0);

        // If current digit sum is
        // 1 or 3, add 1 to result
        result = String.fromCharCode(parseInt(s % 2) +
        '0'.charCodeAt(0)) + result;

        // Compute carry
        s = parseInt(s/2);

        // Move to next digits
        i--; j--;
    }

return result;
}

//Driver code
var a = "1101", b="100";

document.write(addBinary(a, b));

// This code is contributed by Amit Katiyar

</script>

Output

10001

Time Complexity: O(max(L1, L2)), where L1 and L2 are the lengths of strings a and b respectively. 
Auxiliary Space: O(max(L1, L2)), where L1 and L2 are the lengths of strings a and b respectively.