# Program to add two binary strings

• Difficulty Level : Medium
• Last Updated : 18 Jan, 2022

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 last characters of two strings and compute digit sum one by one. If sum becomes more than 1, then store carry for next digits.

## C++

 `// C++ program to add two binary strings``#include ``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 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 differnce 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)``    ``{``        ` `        ``// 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

 `= 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

 ``

Output:

`10001`

Thanks to Gaurav Ahirwar for suggesting above solution. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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