human
Count subarrays having an equal count of 0 s and 1 s segregated | Function to count subarrays having equal count of 0 s and 1 s with all 0 s and all 1 s grouped together.
Stores the count of subarrays.
If current element is different from the next array element.
Increment count.
Count the frequency of 1 s and 0 s.
Increment count.
Print the final count.
Driver Code.
Function Call"
How can the above be solved in Python?

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Write me a code to count subarrays having an equal count of 0 s and 1 s segregated and the other steps required?
assistant
#Here is code to count subarrays having an equal count of 0 s and 1 s segregated and the other steps required.
 def countSubarrays ( A , N ) :
	ans = 0 ;
	for i in range ( N - 1 ) :
		if ( A [ i ] != A [ i + 1 ] ) :
			ans += 1 ;
			j = i - 1 ; k = i + 2 ;
			while ( j >= 0 and k < N and A [ j ] == A [ i ] and A [ k ] == A [ i + 1 ] ) :
				ans += 1 ;
				j -= 1 ;
				k += 1 ;
	print ( ans ) ;
if __name__ == " _ _ main _ _ " :
	A = [ 1 , 1 , 0 , 0 , 1 , 0 ] ;
	N = len ( A ) ;
	countSubarrays ( A , N ) ;