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 sizes of two strings may be different, we first make the size of smaller string equal to that of bigger string by adding leading 0s. After making sizes 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 begining 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 C++ implementation of the above algorithm.

`#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 aadding leading 0s in the smaller string. Returns the ` `// 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; ` `} ` |

Output:

Sum is 1100101

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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