human
Queries to count array elements from a given range having a single set bit | ''Check if N has only one set bit.
''Function to build segment tree.
'' If se is leaf node.
'' Update the node.
'' Stores mid value of segment [ss, se].
'' Recursive call for left Subtree.
'' Recursive call for right Subtree.
'' Update the Current Node.
''Function to update a value at Xth index.
'' If ss is equal to X.
'' Update the Xth node.
'' Update the tree.
'' Stores the mid of segment [ss, se].
'' Update current node.
''Count of numbers having one set bit.
'' If L > se or R < ss  Invalid Range.
'' If [ss, se] lies  inside the [L, R].
'' Stores the mid of segment [ss, se].
'' Return the sum after recursively  traversing left and right subtree.
''Function to solve queries.
'' Initialise Segment tree.
'' Build segment tree.
'' Perform queries.
''Input.
''Function call to solve queries"
Solution in Python:
 ... Write me a code to queries to count array elements from a given range having a single set bit and the other steps required?
assistant
#Here is code to queries to count array elements from a given range having a single set bit and the other steps required.
 def check ( N ) :
	if ( N == 0 ) :
		return 0
	return ( ( N & ( N - 1 ) ) == 0 )
def build_seg_tree ( ss , se , si , tree , arr ) :
	if ( ss == se ) :
		tree [ si ] = check ( arr [ ss ] )
		return
	mid = ( ss + se ) // 2
	build_seg_tree ( ss , mid , 2 * si + 1 , tree , arr )
	build_seg_tree ( mid + 1 , se , 2 * si + 2 , tree , arr )
	tree [ si ] = tree [ 2 * si + 1 ] + tree [ 2 * si + 2 ]
def update ( ss , se , si , X , V , tree , arr ) :
	if ( ss == se ) :
		if ( ss == X ) :
			arr [ X ] = V
			tree [ si ] = check ( V )
		return
	mid = ( ss + se ) // 2
	if ( X <= mid ) :
		update ( ss , mid , 2 * si + 1 , X , V , tree , arr )
	else :
		update ( mid + 1 , se , 2 * si + 2 , X , V , tree , arr )
	tree [ si ] = tree [ 2 * si + 1 ] + tree [ 2 * si + 2 ]
def query ( l , r , ss , se , si , tree ) :
	if ( r < ss or l > se ) :
		return 0
	if ( l <= ss and r >= se ) :
		return tree [ si ]
	mid = ( ss + se ) // 2
	return ( query ( l , r , ss , mid , 2 * si + 1 , tree ) + query ( l , r , mid + 1 , se , 2 * si + 2 , tree ) )
def Query ( arr , N , Q ) :
	tree = [ 0 ] * ( 4 * N )
	build_seg_tree ( 0 , N - 1 , 0 , tree , arr )
	for i in range ( len ( Q ) ) :
		if ( Q [ i ] [ 0 ] == 1 ) :
			print ( query ( Q [ i ] [ 1 ] , Q [ i ] [ 2 ] , 0 , N - 1 , 0 , tree ) , end = " ▁ " )
		else :
			update ( 0 , N - 1 , 0 , Q [ i ] [ 1 ] , Q [ i ] [ 2 ] , tree , arr )
arr = [ 12 , 11 , 16 , 2 , 32 ]
Q = [ [ 1 , 0 , 2 ] , [ 2 , 4 , 24 ] , [ 1 , 1 , 4 ] ]
N = len ( arr )
Query ( arr , N , Q )