Skip to content
Related Articles

Related Articles

Count number of bits changed after adding 1 to given N
  • Last Updated : 14 Jun, 2019

Given an integer N. The task is to find the number of bits changed after adding 1 to the given number.

Examples:

Input : N = 5
Output : 2
After adding 1 to 5 it becomes 6.
Binary representation of 5 is 101.
Binary representation of 6 is 110.
So, no. of bits changed is 2.

Input : N = 1
Output : 2

There are three approaches to find the number of changed bits in the result obtained after adding 1 to the given value N:

  • Approach 1: Add 1 to given integer and compare the bits of N and the result obtained after addition and count the number of unmatched bit.
  • Approach 2: In case if 1 is added to N, then the total number of bits changed is defined by the position of 1st Zero from right i.e. LSB as zero. In the case, 1 is added to 1 then it got changed and pass a carry 1 to its next bit but if 1 is added to 0 only 0 changes to 1 and no further carry is passed.
  • Approach 3: For finding number of changed bits when 1 is added to a given number take XOR of n and n+1 and calculate the number of set bits in the resultant XOR value.

Below is the implementation of the Approach 3:

C++




// CPP program to find the number
// of changed bit
#include <bits/stdc++.h>
using namespace std;
  
// Function to find number of changed bit
int findChangedBit(int n)
{
    // Calculate xor of n and n+1
    int XOR = n ^ (n + 1);
  
    // Count set bits in xor value
    int result = __builtin_popcount(XOR);
  
    // Return the result
    return result;
}
  
// Driver function
int main()
{
    int n = 6;
    cout << findChangedBit(n) << endl;
  
    n = 7;
    cout << findChangedBit(n);
  
    return 0;
}


Java




// Java program to find the number
// of changed bit
class GFG 
{
  
// Function to find number of changed bit
static int findChangedBit(int n)
{
    // Calculate xor of n and n+1
    int XOR = n ^ (n + 1);
  
    // Count set bits in xor value
    int result = Integer.bitCount(XOR);
  
    // Return the result
    return result;
}
  
// Driver code
public static void main(String[] args) 
{
    int n = 6;
    System.out.println(findChangedBit(n));
  
    n = 7;
    System.out.println(findChangedBit(n));
}
}
  
// This code contributed by Rajput-Ji


Python3




# Python 3 program to find the number
# of changed bit
  
# Function to find number of changed bit
def findChangedBit(n):
      
    # Calculate xor of n and n+1
    XOR = n ^ (n + 1)
  
    # Count set bits in xor value
    result = bin(XOR).count("1")
  
    # Return the result
    return result
  
# Driver Code
if __name__ == '__main__':
    n = 6
    print(findChangedBit(n))
  
    n = 7
    print(findChangedBit(n))
  
# This code is contributed by
# Surendra_Gangwar


C#




// C# program to find the number
// of changed bit
using System;
      
class GFG 
{
  
// Function to find number of changed bit
static int findChangedBit(int n)
{
    // Calculate xor of n and n+1
    int XOR = n ^ (n + 1);
  
    // Count set bits in xor value
    int result = bitCount(XOR);
  
    // Return the result
    return result;
}
static int bitCount(int x)
{
  
    // To store the count
    // of set bits
    int setBits = 0;
    while (x != 0) 
    {
        x = x & (x - 1);
        setBits++;
    }
  
    return setBits;
}
  
// Driver code
public static void Main(String[] args) 
{
    int n = 6;
    Console.WriteLine(findChangedBit(n));
  
    n = 7;
    Console.WriteLine(findChangedBit(n));
}
}
  
/* This code contributed by PrinciRaj1992 */


Output:

1
4

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up
Recommended Articles
Page :