human
Give me a code to count numbers up to N whose rightmost set bit is K and the other steps required given the following:
Count numbers up to N whose rightmost set bit is K | Function to count the numbers in the range [ 1 , N ] whose rightmost set bit is K.
Stores the number whose rightmost set bit is K.
Numbers whose rightmost set bit is i.
Subtracting the number whose rightmost set bit is i , from N.
Since i = k , then the number whose rightmost set bit is K is stored.
Driver Code"
How can the above be solved in Python?
assistant
#Here is code to count numbers up to N whose rightmost set bit is K and the other steps required.
 def countNumberHavingKthBitSet ( N , K ) :
	numbers_rightmost_setbit_K = 0
	for i in range ( 1 , K + 1 ) :
		numbers_rightmost_bit_i = ( N + 1 ) // 2
		N -= numbers_rightmost_bit_i
		if ( i == K ) :
			numbers_rightmost_setbit_K = numbers_rightmost_bit_i
	print ( numbers_rightmost_setbit_K )
if __name__ == ' _ _ main _ _ ' :
	N = 15
	K = 2
	countNumberHavingKthBitSet ( N , K )