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