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  • Difficulty Level : Medium
  • Last Updated : 28 Jun, 2022
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Given two bit sequences as strings, write a function to return the addition of the two sequences. Bit strings can be of different lengths also. For example, if string 1 is “1100011” and second string 2 is “10”, then the function should return “1100101”. 
 

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Since the sizes of two strings may be different, we first make the size of a smaller string equal to that of the bigger string by adding leading 0s. After making sizes the same, we one by one add bits from rightmost bit to leftmost bit. In every iteration, we need to sum 3 bits: 2 bits of 2 given strings and carry. The sum bit will be 1 if, either all of the 3 bits are set or one of them is set. So we can do XOR of all bits to find the sum bit. How to find carry – carry will be 1 if any of the two bits is set. So we can find carry by taking OR of all pairs. Following is step by step algorithm.
1. Make them equal sized by adding 0s at the beginning of smaller string. 
2. Perform bit addition 
…..Boolean expression for adding 3 bits a, b, c 
…..Sum = a XOR b XOR c 
…..Carry = (a AND b) OR ( b AND c ) OR ( c AND a )
Following is implementation of the above algorithm.
 

C++




#include <iostream>
using namespace std;
 
//adds the two-bit strings and return the result
string addBitStrings( string first, string second );
 
// Helper method: given two unequal sized bit strings, converts them to
// same length by adding leading 0s in the smaller string. Returns the
//  new length
int makeEqualLength(string &str1, string &str2)
{
    int len1 = str1.size();
    int len2 = str2.size();
    if (len1 < len2)
    {
        for (int i = 0 ; i < len2 - len1 ; i++)
            str1 = '0' + str1;
        return len2;
    }
    else if (len1 > len2)
    {
        for (int i = 0 ; i < len1 - len2 ; i++)
            str2 = '0' + str2;
    }
    return len1; // If len1 >= len2
}
 
// The main function that adds two-bit sequences and returns the addition
string addBitStrings( string first, string second )
{
    string result;  // To store the sum bits
 
    // make the lengths same before adding
    int length = makeEqualLength(first, second);
 
    int carry = 0;  // Initialize carry
 
    // Add all bits one by one
    for (int i = length-1 ; i >= 0 ; i--)
    {
        int firstBit = first.at(i) - '0';
        int secondBit = second.at(i) - '0';
 
        // boolean expression for sum of 3 bits
        int sum = (firstBit ^ secondBit ^ carry)+'0';
 
        result = (char)sum + result;
 
        // boolean expression for 3-bit addition
        carry = (firstBit & secondBit) | (secondBit & carry) | (firstBit & carry);
    }
 
    // if overflow, then add a leading 1
    if (carry)
        result = '1' + result;
 
    return result;
}
 
// Driver program to test above functions
int main()
{
    string str1 = "1100011";
    string str2 = "10";
 
    cout << "Sum is " << addBitStrings(str1, str2);
    return 0;
}

Java




// Java implementation of above algorithm
class GFG
{
 
    // Helper method: given two unequal sized bit strings,
    // converts them to same length by adding leading 0s
    // in the smaller string. Returns the new length
    // Using StringBuilder as Java only uses call by value
    static int makeEqualLength(StringBuilder str1,
                               StringBuilder str2)
    {
        int len1 = str1.length();
        int len2 = str2.length();
        if (len1 < len2)
        {
            for (int i = 0; i < len2 - len1; i++)
                str1.insert(0, '0');
            return len2;
        }
        else if (len1 > len2)
        {
            for (int i = 0; i < len1 - len2; i++)
                str2.insert(0, '0');
        }
 
        return len1; // If len1 >= len2
    }
 
    // The main function that adds two-bit sequences
    // and returns the addition
    static String addBitStrings(StringBuilder str1,
                                StringBuilder str2)
    {
        String result = ""; // To store the sum bits
 
        // make the lengths same before adding
        int length = makeEqualLength(str1, str2);
 
        // Convert StringBuilder to Strings
        String first = str1.toString();
        String second = str2.toString();
 
        int carry = 0; // Initialize carry
 
        // Add all bits one by one
        for (int i = length - 1; i >= 0; i--)
        {
            int firstBit = first.charAt(i) - '0';
            int secondBit = second.charAt(i) - '0';
 
            // boolean expression for sum of 3 bits
            int sum = (firstBit ^ secondBit ^ carry) + '0';
 
            result = String.valueOf((char) sum) + result;
 
            // boolean expression for 3-bit addition
            carry = (firstBit & secondBit) |
                    (secondBit & carry) |
                    (firstBit & carry);
        }
         
        // if overflow, then add a leading 1
        if (carry == 1)
            result = "1" + result;
        return result;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String str1 = "1100011";
        String str2 = "10";
        System.out.println("Sum is " +
        addBitStrings(new StringBuilder(str1),
                      new StringBuilder(str2)));
    }
}
 
// This code is contributed by Vivek Kumar Singh

Python3




# Python3 program for above approach
 
# adds the two-bit strings and return the result
 
# Helper method: given two unequal sized bit strings,
# converts them to same length by adding leading 0s
# in the smaller string. Returns the new length
def makeEqualLength(str1, str2):
 
    len1 = len(str1)     # Length of string 1
    len2 = len(str2)     # length of string 2
    if len1 < len2:
        str1 = (len2 - len1) * '0' + str1
        len1 = len2
    elif len2 < len1:
        str2 = (len1 - len2) * '0' + str2
        len2 = len1
    return len1, str1, str2
 
def addBitStrings( first, second ):
    result = '' # To store the sum bits
 
    # make the lengths same before adding
    length, first, second = makeEqualLength(first, second)
 
    carry = 0 # initialize carry as 0
 
    # Add all bits one by one
    for i in range(length - 1, -1, -1):
        firstBit = int(first[i])
        secondBit = int(second[i])
 
        # boolean expression for sum of 3 bits
        sum = (firstBit ^ secondBit ^ carry) + 48
        result = chr(sum) + result
 
        # boolean expression for 3 bits addition
        carry = (firstBit & secondBit) | \
                (secondBit & carry) | \
                (firstBit & carry)
 
        # if overflow, then add a leading 1
    if carry == 1:
        result = '1' + result
    return result
 
# Driver Code
if __name__ == '__main__':
    str1 = '1100011'
    str2 = '10'
    print('Sum is', addBitStrings(str1, str2))
     
# This code is contributed by
# chaudhary_19 (Mayank Chaudhary)

C#




// C# implementation of above algorithm
using System;
using System.Text;
 
class GFG
{
 
    // Helper method: given two unequal sized
    // bit strings, converts them to same length
    // by adding leading 0s in the smaller string.
    // Returns the new length Using StringBuilder
    // as Java only uses call by value
    static int makeEqualLength(StringBuilder str1,
                               StringBuilder str2)
    {
        int len1 = str1.Length;
        int len2 = str2.Length;
        if (len1 < len2)
        {
            for (int i = 0; i < len2 - len1; i++)
            {
                str1.Insert(0, '0');
            }
            return len2;
        }
        else if (len1 > len2)
        {
            for (int i = 0; i < len1 - len2; i++)
            {
                str2.Insert(0, '0');
            }
        }
 
        return len1; // If len1 >= len2
    }
 
    // The main function that adds two-bit sequences
    // and returns the addition
    static string addBitStrings(StringBuilder str1,
                                StringBuilder str2)
    {
        string result = ""; // To store the sum bits
 
        // make the lengths same before adding
        int length = makeEqualLength(str1, str2);
 
        // Convert StringBuilder to Strings
        string first = str1.ToString();
        string second = str2.ToString();
 
        int carry = 0; // Initialize carry
 
        // Add all bits one by one
        for (int i = length - 1; i >= 0; i--)
        {
            int firstBit = first[i] - '0';
            int secondBit = second[i] - '0';
 
            // boolean expression for sum of 3 bits
            int sum = (firstBit ^ secondBit ^ carry) + '0';
 
            result = ((char) sum).ToString() + result;
 
            // boolean expression for 3-bit addition
            carry = (firstBit & secondBit) |
                       (secondBit & carry) |
                        (firstBit & carry);
        }
 
        // if overflow, then add a leading 1
        if (carry == 1)
        {
            result = "1" + result;
        }
        return result;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        string str1 = "1100011";
        string str2 = "10";
        Console.WriteLine("Sum is " +
                addBitStrings(new StringBuilder(str1),
                              new StringBuilder(str2)));
    }
}
 
// This code is contributed by kumar65

Javascript




// JavaScript code to implement the approach
 
// Helper method: given two unequal sized bit strings, converts them to
// same length by adding leading 0s in the smaller string. Returns the
//  new length
function makeEqualLength(str1, str2)
{
    var len1 = str1.length;
    var len2 = str2.length;
    if (len1 < len2)
    {
        for (var i = 0 ; i < len2 - len1 ; i++)
            str1 = '0' + str1;
        return len2;
    }
    else if (len1 > len2)
    {
        for (var i = 0 ; i < len1 - len2 ; i++)
            str2 = '0' + str2;
    }
    return len1; // If len1 >= len2
}
 
// The main function that adds two-bit sequences and returns the addition
function addBitStrings(first, second )
{
    var result = ""// To store the sum bits
 
    // make the lengths same before adding
    var length = makeEqualLength(first, second);
 
    var carry = 0;  // Initialize carry
 
    // Add all bits one by one
    for (var i = length-1 ; i >= 0 ; i--)
    {
        var firstBit = first[i] - '0';
        var secondBit = second[i] - '0';
 
        // boolean expression for sum of 3 bits
        var sum = (firstBit ^ secondBit ^ carry) + 48;
 
        result += String.fromCharCode(sum);
 
        // boolean expression for 3-bit addition
        carry = (firstBit & secondBit) | (secondBit & carry) | (firstBit & carry);
    }
 
    // if overflow, then add a leading 1
    if (carry)
        result += '1';
 
    return result;
}
 
// Driver program to test above functions
var str1 = "1100011";
var str2 = "10";
 
console.log("Sum is " + addBitStrings(str1, str2));
 
// This code is contributed by phasing17

Output

Sum is 1100101

Time Complexity: O(|str1|)

Auxiliary Space: O(1)

Method – 2 (without adding extra zeros(0) in beginning of a small length string to make both strings with same length)

Algo :  

  1. make to pointer i,j and set i = str1.size() – 1 and j = str2.size() – 1
  2. take initial carry as 0 ans ans string as empty (“”)
  3. while i>=0 or j>=0 or carry 
    1. add value of str1[i] and str2[j] in carry
    2. add (carry%2) in resulting(answer string) string
    3. set carry = carry/2
  4. return ans

C++




#include <bits/stdc++.h>
using namespace std;
 
// The function that adds two-bit sequences and returns the addition
string addBitStrings(string str1, string str2)
{
    string ans = "";
    int i = str1.size() - 1;
    int j = str2.size() - 1;
    int carry = 0;
    while (i >= 0 || j >= 0 || carry) {
        carry += ((i >= 0) ? (str1[i--] - '0') : (0));
        carry += ((j >= 0) ? (str2[j--] - '0') : (0));
        ans = char('0' + (carry % 2)) + ans;
        carry = carry / 2;
    }
    return ans;
}
 
// Driver program to test above functions
int main()
{
    string str1 = "1100011";
    string str2 = "10";
 
    cout << "Sum is " << addBitStrings(str1, str2);
    return 0;
}

Java




// Java implementation
class GFG {
 
  // The main function that adds two-bit
  // sequences and returns the addition
  static String addBitStrings(StringBuilder str1,
                              StringBuilder str2)
  {
    StringBuilder ans = new StringBuilder();
    int i = str1.length() - 1;
    int j = str2.length() - 1;
    int carry = 0;
    while (i >= 0 || j >= 0 || carry>0) {
      if (i >= 0)
        carry += str1.charAt(i--) - '0';
      if (j >= 0)
        carry += str2.charAt(j--) - '0';
      ans.append(carry % 2);
      carry = carry/2;
    }
    return ans.reverse().toString();
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    String str1 = "1100011";
    String str2 = "10";
    System.out.println("Sum is "+ addBitStrings(new StringBuilder(str1),
                                                new StringBuilder(str2)));
  }
}
 
// This code is contributed by ajaymakvana

Python3




# The function that adds two-bit sequences and returns the addition
def addBitStrings(str1, str2):
    ans = ''
    i = len(str1) - 1
    j = len(str2) - 1
    carry = 0
    while i >= 0 or j >= 0 or carry:
        if i >= 0:
            carry += ord(str1[i]) - ord('0')
            i = i - 1
        else:
            carry += 0
 
        if j >= 0:
            carry += ord(str2[j]) - ord('0')
            j = j - 1
        else:
            carry += 0
 
        ans = chr(ord('0') + carry % 2) + ans
        carry = carry // 2
    return ans
 
 
# Driver program to test above functions
str1 = '1100011'
str2 = '10'
 
print('Sum is ', addBitStrings(str1, str2))
 
# This code is contributed by ajaymakavan.

C#




// C# code to implement the approach
using System;
 
public static class Globals
{
   
    // The function that adds two-bit sequences and returns
    // the addition
    public static string addBitStrings(string str1,
                                       string str2)
    {
        string ans = "";
        int i = str1.Length - 1;
        int j = str2.Length - 1;
        int carry = 0;
        while (i >= 0 || j >= 0 || carry != 0) {
            carry += ((i >= 0) ? (str1[i--] - '0') : (0));
            carry += ((j >= 0) ? (str2[j--] - '0') : (0));
            ans = (char)('0' + (carry % 2)) + ans;
            carry = carry / 2;
        }
        return ans;
    }
 
    // Driver program to test above functions
    public static void Main()
    {
        string str1 = "1100011";
        string str2 = "10";
 
        Console.Write("Sum is ");
        Console.Write(addBitStrings(str1, str2));
    }
}
 
// This code is contributed by Aarti_Rathi

Javascript




// JavaScript code to implement the approach
 
// The function that adds two-bit sequences and returns the addition
function addBitStrings(str1, str2)
{
    let ans = "";
    let i = str1.length - 1;
    let j = str2.length - 1;
    let carry = 0;
    while (i >= 0 || j >= 0 || (carry != 0)) {
        carry += ((i >= 0) ? (parseInt(str1[i--])) : (0));
        carry += ((j >= 0) ? (parseInt(str2[j--])) : (0));
        ans = (carry % 2).toString() + ans;
        carry = Math.floor(carry / 2);
    }
    return ans;
}
 
// Driver program to test above functions
let str1 = "1100011";
let str2 = "10";
 
// Function Call
console.log("Sum is", addBitStrings(str1, str2));
 
// This code is contributed by phasing17

Output

Sum is 1100101
Time Complexity: O(max(n,m)) (where, n = sizeof str1 & m = sizeof str2)
Space Complexity: O(1)

This article is compiled by Ravi Chandra Enaganti. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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