Convert numbers into binary representation and add them without carry


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++

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// 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;
}

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Java

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// 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

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Python3

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# 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 

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C#

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// 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 Rajput-Ji

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Output:

41

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