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 : 

C++

// CPP program for special
// addition of two number
#include <bits/stdc++.h>
using namespace std;

int xSum(int n, int m)
{
    // variable to store result
    int res = 0;

    // variable to maintain
    // place value
    int multiplier = 1;

    // variable to maintain
    // each digit sum
    int bit_sum;

    // Add numbers till each
    // number become zero
    while (n || m) {

        // Add each bits
        bit_sum = (n % 10) + (m % 10);

        // Neglect carry
        bit_sum %= 10;

        // Update result
        res = (bit_sum * multiplier) + res;
        n /= 10;
        m /= 10;

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

// Driver program
int main()
{
    int n = 8458;
    int m = 8732;
    cout << xSum(n, m);
    return 0;
}

Java

// Java program for special
// addition of two number
import java.util.*;
import java.lang.*;

public class GfG {

    public static int xSum(int n, int m)
    {
        int res = 0;
        int multiplier = 1;
        int bit_sum;

        // Add numbers till each
        // number become zero
        while (true) {

            if(n==0 && m==0)
            break;

            // Add each bits
            bit_sum = (n % 10) + (m % 10);

            // Neglect carry
            bit_sum %= 10;

            // Update result
            res = (bit_sum * multiplier) + res;
            n /= 10;
            m /= 10;

            // Update multiplier
            multiplier *= 10;

        }
        return res;
    }

    // Driver function
    public static void main(String args[])
    {
        int n = 8458;
        int m = 8732;
        System.out.println(xSum(n, m));
    }
}
/* This code is contributed by Sagar Shukla */

Python3

# Python3 program for special
# addition of two number
import math

def xSum(n, m) :

    # variable to
    # store result
    res = 0

    # variable to maintain
    # place value
    multiplier = 1

    # variable to maintain
    # each digit sum
    bit_sum = 0

    # Add numbers till each
    # number become zero
    while (n or m) :

        # Add each bits
        bit_sum = ((n % 10) +
                   (m % 10))

        # Neglect carry
        bit_sum = bit_sum % 10

        # Update result
        res = (bit_sum *
               multiplier) + res
        n = math.floor(n / 10)
        m = math.floor(m / 10)

        # Update multiplier
        multiplier = multiplier * 10

    return res

# Driver code
n = 8458
m = 8732
print (xSum(n, m))

# This code is contributed by
# Manish Shaw(manishshaw1)

C#

// C# program for special
// addition of two number
using System;

public class GfG {

    public static int xSum(int n, int m)
    {
        int res = 0;
        int multiplier = 1;
        int bit_sum;

        // Add numbers till each
        // number become zero
        while (true) {

            // Add each bits
            bit_sum = (n % 10) + (m % 10);

            // Neglect carry
            bit_sum %= 10;

            // Update result
            res = (bit_sum * multiplier) + res;
            n /= 10;
            m /= 10;

            // Update multiplier
            multiplier *= 10;
            if (n == 0)
                break;
            if (m == 0)
                break;
        }
        return res;
    }

    // Driver function
    public static void Main()
    {
        int n = 8458;
        int m = 8732;
        Console.WriteLine(xSum(n, m));
    }
}

/* This code is contributed by Vt_m */

PHP

<?php
// php program for special
// addition of two number

function xSum($n, $m)
{

    // variable to store result
    $res = 0;

    // variable to maintain
    // place value
    $multiplier = 1;

    // variable to maintain
    // each digit sum
    $bit_sum;

    // Add numbers till each
    // number become zero
    while ($n || $m) {

        // Add each bits
        $bit_sum = ($n % 10) +
                   ($m % 10);

        // Neglect carry
        $bit_sum %= 10;

        // Update result
        $res = ($bit_sum * $multiplier) + $res;
        $n =floor($n / 10);
        $m =floor($m / 10);

        // Update multiplier
        $multiplier *= 10;
    }
    return $res;
}

    // Driver code
    $n = 8458;
    $m = 8732;
    echo xSum($n, $m);

//This code is contributed by mits
?>

Javascript

<script>

// Javascript program for special
// addition of two number
function xSum(n, m)
{
    // variable to store result
    var res = 0;

    // variable to maintain
    // place value
    var multiplier = 1;

    // variable to maintain
    // each digit sum
    var bit_sum;

    // Add numbers till each
    // number become zero
    while (n || m) {

        // Add each bits
        bit_sum = (n % 10) + (m % 10);

        // Neglect carry
        bit_sum %= 10;

        // Update result
        res = (bit_sum * multiplier) + res;
        n = parseInt(n / 10);
        m = parseInt(m / 10);

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

// Driver program
var n = 8458;
var m = 8732;
document.write(xSum(n, m));

    // This code is contributed by noob2000.
</script>

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)