human
Count 1 s present in a range of indices [ L , R ] in a given array | Function to find the size of the array if the array initially contains a single element.
Base case.
P / 2 -> findSize ( N 2 ) P % 2 -> 1 P / 2 -> findSize ( N / 2 ).
Function to return the count of 1 s in the range [ L , R ].
Base Case.
PART 1 -> N / 2 [ 1 , Siz_M ].
Update the right end point of the range to min ( Siz_M , R ).
PART 2 -> N % 2 [ SizM + 1 , Siz_M + 1 ].
PART 3 -> N / 2 [ SizM + 2 , 2 * Siz_M - 1 ] Same as PART 1 Property of Symmetricity Shift the coordinates according to PART 1 Subtract ( Siz_M + 1 ) from both L , R.
Driver Code.
Input.
Counts the number of 1 's in  the range [L, R]"
Solution in Python:

Give me a code to count 1 s present in a range of indices [ L , R ] in a given array and the other steps required?
assistant
#Here is code to count 1 s present in a range of indices [ L , R ] in a given array and the other steps required.
 def findSize ( N ) :
	if ( N == 0 ) :
		return 1
	if ( N == 1 ) :
		return 1
	Size = 2 * findSize ( N // 2 ) + 1
	return Size
def CountOnes ( N , L , R ) :
	if ( L > R ) :
		return 0
	if ( N <= 1 ) :
		return N
	ret = 0
	M = N // 2
	Siz_M = findSize ( M )
	if ( L <= Siz_M ) :
		ret += CountOnes ( N // 2 , L , min ( Siz_M , R ) )
	if ( L <= Siz_M + 1 and Siz_M + 1 <= R ) :
		ret += N % 2
	if ( Siz_M + 1 < R ) :
		ret += CountOnes ( N // 2 , max ( 1 , L - Siz_M - 1 ) , R - Siz_M - 1 )
	return ret
if __name__ == " _ _ main _ _ " :
	N = 7
	L = 2
	R = 5
	print ( CountOnes ( N , L , R ) )