Addition of two numbers without carry

You are given two positive numbers n and m. You have to find a simple addition of both numbers but with a given condition that there is not any carry system in this addition. That is no carry is added at higher MSBs.
Examples :

Input : m = 456, n = 854
Output : 200

Input : m = 456, n = 4
Output : 450

Algorithm : 

Input n, m while(n||m)
 {
 // Add each bits 
 bit_sum = (n%10) + (m%10);

 // Neglect carry
 bit_sum %= 10;

 // Update result
 // multiplier to maintain place value
 res = (bit_sum * multiplier) + res;
 n /= 10;
 m /= 10;

 // Update multiplier
 multiplier *=10;
 } print res

Approach : 
To solve this problem we will need the bit by bit addition of numbers where we start adding two numbers from rightmost bit (LSB) and add integers from both numbers with the same position. Also, we will neglect carry at each position so that  carry will not affect further higher bit position. 
Start adding both numbers bit by bit and for each bit take the sum of integers then neglect their carry by taking the modulo of bit_sum by 10 further add bit_sum to res by multiplying bit_sum with a multiplier specifying place value. (Multiplier got incremented 10 times on each iteration.) 
Below is the implementation of the above approach : 

Output :

6180

Time Complexity: O(MAX(len(n),len(m))), where len(X) is the number of digits in a number X.

Space Complexity: O(1)