human
Count total set bits in all numbers from range L to R | Returns position of leftmost set bit The rightmost position is taken as 0.
Function that gives the position of previous leftmost set bit in n.
Function that returns count of set bits present in all numbers from 1 to n.
Get the position of leftmost set bit in n.
Use the position.
Function to count the set bits between the two numbers N and M.
Base Case.
Get position of next leftmost set bit.
If n is of the form 2 ^ x - 1.
Update n for next recursive call.
Function that counts the set bits between L and R.
Driver Code.
Given L and R.
Function Call"
How can the above be solved in Python?
 ... Can you provide a code to count total set bits in all numbers from range L to R and the other steps required?
assistant
#Here is code to count total set bits in all numbers from range L to R and the other steps required.
 def getLeftmostBit ( n ) :
	m = 0 ;
	while ( n > 1 ) :
		n = n >> 1 ;
		m += 1 ;
	return m ;
def getNextLeftmostBit ( n , m ) :
	temp = 1 << m ;
	while ( n < temp ) :
		temp = temp >> 1 ;
		m -= 1 ;
	return m ;
def countSetBit ( n ) :
	m = getLeftmostBit ( n ) ;
	return _countSetBit ( n , m ) ;
def _countSetBit ( n , m ) :
	if ( n == 0 ) :
		return 0 ;
	m = getNextLeftmostBit ( n , m ) ;
	if ( n == int ( 1 << ( m + 1 ) ) - 1 ) :
		return int ( m + 1 ) * ( 1 << m ) ;
	n = n - ( 1 << m ) ;
	return ( ( n + 1 ) + countSetBit ( n ) + m * ( 1 << ( m - 1 ) ) ) ;
def countSetBits ( L , R ) :
	return abs ( countSetBit ( R ) - countSetBit ( L - 1 ) ) ;
if __name__ == ' _ _ main _ _ ' :
	L = 3 ;
	R = 5 ;
	print ( " Total ▁ set ▁ bit ▁ count ▁ is ▁ " , countSetBits ( L , R ) ) ;