human
Count of 0 s to be flipped to make any two adjacent 1 s at least K 0 s apart | Function to find the count of 0 s to be flipped.
Loop traversal to mark K adjacent positions to the right of already existing 1 s ..
Loop traversal to mark K adjacent positions to the left of already existing 1 s ..
Loop to count the maximum number of 0 s that will be replaced by 1 s.
Driver code"
Solution in Python:

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Give me a code to count of 0 s to be flipped to make any two adjacent 1 s at least K 0 s apart and the other steps required?
assistant
#Here is code to count of 0 s to be flipped to make any two adjacent 1 s at least K 0 s apart and the other steps required.
 def count ( k , s ) :
	ar = [ 0 ] * len ( s )
	end = 0
	for i in range ( len ( s ) ) :
		if s [ i ] == '1' :
			for j in range ( i , len ( s ) ) :
				if ( j <= i + k ) :
					ar [ j ] = - 1
					end = j
			i = end
	end = 0
	for i in range ( len ( s ) - 1 , - 1 , - 1 ) :
		if ( s [ i ] == '1' ) :
			for j in range ( i , - 1 , - 1 ) :
				if ( j >= i - k ) :
					ar [ j ] = - 1
					end = j
			i = end
	ans = 0
	end = 0
	for j in range ( len ( s ) ) :
		if ( ar [ j ] == 0 ) :
			ans += 1
			for g in range ( j , len ( s ) ) :
				if ( g <= j + k ) :
					ar [ g ] = - 1
					end = g
			j = end - 1
	return ans
K = 2
s = "000000"
print ( count ( K , s ) )