human
Count numbers less than N whose Bitwise AND with N is zero | Function to count number of unset bits in the integer N.
Stores the number of unset bits in N.
Check if N is even.
Increment the value of c.
Right shift N by 1.
Return the value of count of unset bits.
Function to count numbers whose Bitwise AND with N equal to 0.
Stores the number of unset bits in N.
Prthe value of 2 to the power of unsetBits.
Driver Code"
How can the above be solved in Python?
assistant
#Here is code to count numbers less than N whose Bitwise AND with N is zero and the other steps required.
 def countUnsetBits ( N ) :
	c = 0
	while ( N ) :
		if ( N % 2 == 0 ) :
			c += 1
		N = N >> 1
	return c
def countBitwiseZero ( N ) :
	unsetBits = countUnsetBits ( N )
	print ( ( 1 << unsetBits ) )
if __name__ == ' _ _ main _ _ ' :
	N = 9
	countBitwiseZero ( N )