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Minimum number of operations required to maximize the Binary String | Function to find the number of operations required.
Count of 1 's.
Count of 0 's upto (cnt1)-th index.
Return the answer.
Driver Code"
Solution in Python:
assistant
#Here is code to minimum number of operations required to maximize the Binary String and the other steps required.
 def minOperation ( s , n ) :
	cnt1 = 0 ;
	for i in range ( n ) :
		if ( ord ( s [ i ] ) == ord ( '1' ) ) :
			cnt1 += 1 ;
	cnt0 = 0 ;
	for i in range ( 0 , cnt1 ) :
		if ( s [ i ] == '0' ) :
			cnt0 += 1 ;
	return cnt0 ;
if __name__ == ' _ _ main _ _ ' :
	n = 8 ;
	s = "01001011" ;
	ans = minOperation ( s , n ) ;
	print ( ans ) ;