Given an integer
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