Given two numbers N and M. The task to convert both the numbers in the binary form then add respective bits of both the binary converted numbers but with a given condition that there is not any carry system in this addition.
Input: N = 37, M = 12 Output: 41 Input: N = 456, M = 854 Output: 670
Approach:

If we don’t consider carry then the binary addition of two bits will be:
1 + 0 = 1 0 + 1 = 1 0 + 0 = 0 1 + 1 = 0 (No carry)
 If you observe clearly you will notice that this is just bitwise XOR of two numbers.
Below is the implementation of the above approach
C++
// C++ program to add two binary // number without carry #include<bits/stdc++.h> using namespace std; // Function returns sum of both // the binary number without // carry int NoCarrySum( int N, int M) { // XOR of N and M return N ^ M; } // Driver code int main() { int N = 37; int M = 12; cout << NoCarrySum(N, M); return 0; } 
Java
// Java program to add two binary // number without carry import java.util.*; class GFG{ // Function returns sum of both // the binary number without // carry static int NoCarrySum( int N, int M) { // XOR of N and M return N ^ M; } // Driver code public static void main(String[] args) { int N = 37 ; int M = 12 ; System.out.print(NoCarrySum(N, M)); } } // This code is contributed by amal kumar choubey 
Python3
# Python3 program to add two binary # number without carry # Function returns sum of both # the binary number without # carry def NoCarrySum(N, M): # XOR of N and M return N ^ M # Driver code N = 37 M = 12 print (NoCarrySum(N, M)) # This code is contributed by sayesha 
C#
// C# program to add two binary // number without carry using System; class GFG{ // Function returns sum of both // the binary number without // carry static int NoCarrySum( int N, int M) { // XOR of N and M return N ^ M; } // Driver code static public void Main(String[] args) { int N = 37; int M = 12; Console.Write(NoCarrySum(N, M)); } } // This code is contributed by RajputJi 
41
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