human
Minimum flips to make all 1 s in left and 0 s in right | Set 1 ( Using Bitmask ) | Function to count minimum number of flips.
This is converting string s into integer of base 2 ( if s = '100' then num = 4 ).
Initialize minXor with n that can be maximum number of flips.
Right shift 1 by ( n - 1 ) bits.
Calculate bitwise XOR of num and mask.
Math . min ( a , b ) returns minimum of a and b return minimum number of flips till that digit.
Function to count number of 1 s.
Driver code"
Solution in Python:

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Give me a code to minimum flips to make all 1 s in left and 0 s in right and the other steps required?
assistant
#Here is code to minimum flips to make all 1 s in left and 0 s in right and the other steps required.
 def findMiniFlip ( nums ) :
	n = len ( nums )
	s = ' '
	for i in range ( n ) :
		s += str ( nums [ i ] )
	num = int ( s , 2 )
	minXor = n ;
	mask = ( 1 << ( n - 1 ) )
	while ( n - 1 > 0 ) :
		temp = ( num ^ mask )
		minXor = min ( minXor , countones ( temp ) )
		n -= 1
		mask = ( mask | ( 1 << n - 1 ) )
	return minXor
def countones ( n ) :
	c = 0
	while ( n > 0 ) :
		n = n & ( n - 1 )
		c += 1
	return c
if __name__ == " _ _ main _ _ " :
	nums = [ 1 , 0 , 1 , 1 , 0 , 0 , 0 ]
	n = findMiniFlip ( nums )
	print ( n )