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Count number of bits changed after adding 1 to given N

  • Last Updated : 28 Apr, 2021
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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 this case, 1 is added to 1 then it got changed and passes 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 a 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 */

Javascript




<script>
 
// Javascript program to find the number
// of changed bit
 
// Function to find number of changed bit
function findChangedBit(n)
{
    // Calculate xor of n and n+1
    let XOR = n ^ (n + 1);
 
    // Count set bits in xor value
    let result = bitCount(XOR);
 
    // Return the result
    return result;
}
 
function bitCount(x)
{
 
    // To store the count
    // of set bits
    let setBits = 0;
    while (x != 0)
    {
        x = x & (x - 1);
        setBits++;
    }
 
    return setBits;
}
 
// Driver function
    let n = 6;
    document.write(findChangedBit(n) + "<br>");
 
    n = 7;
    document.write(findChangedBit(n));
 
</script>
Output: 
1
4

 

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