human
Can you provide a code to maximum number of 0 s placed consecutively at the start and end in any rotation of a Binary String and the other steps required given the following:
Maximum number of 0 s placed consecutively at the start and end in any rotation of a Binary String | Function to find the maximum sum of consecutive 0 s present at the start and end of any rotation of the string str.
Stores the count of 0 s.
If the frequency of '1' is 0.
Print n as the result.
Stores the required sum.
Find the maximum consecutive length of 0 s present in the string.
Update the overall maximum sum.
Find the number of 0 s present at the start and end of the string.
Update the count of 0 s at the start.
Update the count of 0 s at the end.
Update the maximum sum.
Print the result.
Driver Code.
Given string.
Store the size of the string"
Solution in Python:
assistant
#Here is code to maximum number of 0 s placed consecutively at the start and end in any rotation of a Binary String and the other steps required.
 def findMaximumZeros ( string , n ) :
	c0 = 0
	for i in range ( n ) :
		if ( string [ i ] == '0' ) :
			c0 += 1
	if ( c0 == n ) :
		print ( n , end = " " )
		return
	mx = 0
	cnt = 0
	for i in range ( n ) :
		if ( string [ i ] == '0' ) :
			cnt += 1
		else :
			mx = max ( mx , cnt )
			cnt = 0
	mx = max ( mx , cnt )
	start = 0
	end = n - 1
	cnt = 0
	while ( string [ start ] != '1' and start < n ) :
		cnt += 1
		start += 1
	while ( string [ end ] != '1' and end >= 0 ) :
		cnt += 1
		end -= 1
	mx = max ( mx , cnt )
	print ( mx , end = " " )
if __name__ == " _ _ main _ _ " :
	s = "1001"
	n = len ( s )
	findMaximumZeros ( s , n )