human
Count non | Function to find the total count of triplets ( i , j , k ) such that i < j < k and ( j - i ) != ( k - j ).
Stores indices of 0 s.
Stores indices of 1 s.
Stores indices of 2 s.
Traverse the array.
If current array element is 0.
If current array element is 1.
If current array element is 2.
Total count of triplets.
Traverse the array zero_i [ ].
Traverse the array one_i [ ].
Stores index of 0 s.
Stores index of 1 s.
Stores third element of triplets that does not satisfy the condition.
If r present in the map.
Update r.
If r present in the map.
Update r.
If r present in the map and equidistant.
Print the obtained count.
Driver Code"
How can the above be solved in Python?

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Give me a code to count non and the other steps required?
assistant
#Here is code to count non and the other steps required.
 def countTriplets ( arr , N ) :
	zero_i = [ ]
	one_i = [ ]
	mp = { }
	for i in range ( N ) :
		if ( arr [ i ] == 0 ) :
			zero_i . append ( i + 1 )
		elif ( arr [ i ] == 1 ) :
			one_i . append ( i + 1 )
		else :
			mp [ i + 1 ] = 1
	total = len ( zero_i ) * len ( one_i ) * len ( mp )
	for i in range ( len ( zero_i ) ) :
		for j in range ( len ( one_i ) ) :
			p = zero_i [ i ]
			q = one_i [ j ]
			r = 2 * p - q
			if ( r in mp ) :
				total -= 1
			r = 2 * q - p
			if ( r in mp ) :
				total -= 1
			r = ( p + q ) // 2
			if ( ( r in mp ) and abs ( r - p ) == abs ( r - q ) ) :
				total -= 1
	print ( total )
if __name__ == ' _ _ main _ _ ' :
	arr = [ 0 , 1 , 2 , 1 ]
	N = len ( arr )
	countTriplets ( arr , N )