Addition of binary number.txt

Given two binary strings a and b, return their sum as a binary string.

 

Example 1:

Input: a = "11", b = "1"
Output: "100"
Example 2:

Input: a = "1010", b = "1011"
Output: "10101"
 

Constraints:

1 <= a.length, b.length <= 104
a and b consist only of '0' or '1' characters.
Each string does not contain leading zeros except for the zero itself.



Solution

string addBinary(string a, string b)
{
    string result = ""; 
    int s = 0 ;    

    int i = a.size() - 1, j = b.size() - 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 program
int main()
{
    string a = "1101", b="100";
    cout << addBinary(a, b) << endl;
    return 0;
}





class solution
{

  public :
 	string addBinary(string a , string b)
       {
	  string result ;
	  int i = a.size()-1, j = b.size()-1;
	  int carry = 0;                         // initially carry will be 0
		 	
	  while( i >= 0 || j >= 0)
	  {
		int sum = carry;
		if(i >= 0 ) sum = sum + a[i--] - '0';    // ascii value addition 
		if(j>= 0 ) sum = sum + a[j--] - '0';
		
		carry = sum > 1 ? 1 : 0   // is sum is greather than 1 then carry wil  generated otherwise no
		
		result = result + to_string(sum%2)

	   }

	  if(carry) result += to_string(carry);
	  reverse(result.begin().result.end());
 	  return result ;
	
	}

};