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# Count number of bits changed after adding 1 to given N

• Last Updated : 28 Apr, 2021

Given an integer . 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 ``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

 ``
Output:
```1
4```

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