human
Minimum number of steps required to place all 1 s at a single index | Function to print minimum steps required to shift all 1 s to a single index in a binary array.
Size of array.
Used to store cumulative sum.
Initialize count.
Traverse the array in forward direction.
Steps needed to store all previous ones to ith index.
Count number of 1 s present till i - th index.
Initialize count.
Traverse the array in backward direction.
Steps needed to store all 1 s to the right of i at current index.
Count number of 1 s present after i - th index.
Print the number of steps required.
Driver Code"
How can the above be solved in Python?
 ... Give me a code to minimum number of steps required to place all 1 s at a single index and the other steps required?
assistant
#Here is code to minimum number of steps required to place all 1 s at a single index and the other steps required.
 def minsteps ( A ) :
	n = len ( A )
	left , right , res = [ 0 ] * n , [ 0 ] * n , [ 0 ] * n
	count = A [ 0 ]
	for i in range ( 1 , n ) :
		left [ i ] = left [ i - 1 ] + count
		count += A [ i ]
	count = A [ n - 1 ]
	for i in range ( n - 2 , - 1 , - 1 ) :
		right [ i ] = right [ i + 1 ] + count
		count += A [ i ]
	for i in range ( n ) :
		res [ i ] = left [ i ] + right [ i ]
		print ( res [ i ] , end = " ▁ " )
	print ( )
if __name__ == ' _ _ main _ _ ' :
	A = [ 1 , 0 , 1 , 0 ]
	minsteps ( A )