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 stringstring 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 there 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 programint main(){ string a = "1101", b = "100"; cout << addBinary(a, b) << endl; return 0;} |
Java
// java program to add// two binary stringspublic class GFG { // This function adds two // binary strings and return // result as a third string static String addBinary(String A, String B) { // initialize the ith index int i = A.length()-1; // initialize the jth index int j = B.length()-1; // initialize the carry int carry = 0; // initialize the sum int sum = 0; StringBuilder result = new StringBuilder(); while(i>=0 || j>=0 || carry == 1){ sum = carry; if(i>=0) sum = sum+A.charAt(i)-'0'; if(j>=0) sum = sum+B.charAt(j)-'0'; result.append((char)(sum%2+'0')); carry = sum/2; i--; j--; } 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.// A bit improvement by Mustak Ahmed |
Python3
# Python Solution for above problem:# This function adds two binary # strings return the resulting stringdef 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 stringsusing 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 stringfunction 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 stringfunction 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 codevar 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.
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