human
Count number of pairs ( i , j ) such that arr [ i ] * arr [ j ] = arr [ i ] + arr [ j ] | Function to return the count of pairs ( i , j ) such that arr [ i ] * arr [ j ] = arr [ i ] + arr [ j ].
Count number of 0 ' s ▁ and ▁ 2' s in the array.
Total pairs due to occurrence of 0 's .
Total pairs due to occurrence of 2 's .
Return count of all pairs.
Driver code.
Get and print count of pairs"
How can the above be solved in Python?

Write me a code to count number of pairs ( i , j ) such that arr [ i ] * arr [ j ] = arr [ i ] + arr [ j ] and the other steps required?
assistant
#Here is code to count number of pairs ( i , j ) such that arr [ i ] * arr [ j ] = arr [ i ] + arr [ j ] and the other steps required.
 def countPairs ( arr , n ) :
	countZero = 0 ;
	countTwo = 0 ;
	for i in range ( n ) :
		if ( arr [ i ] == 0 ) :
			countZero += 1 ;
		elif ( arr [ i ] == 2 ) :
			countTwo += 1 ;
	pair0 = ( countZero * ( countZero - 1 ) ) // 2 ;
	pair2 = ( countTwo * ( countTwo - 1 ) ) // 2 ;
	return pair0 + pair2 ;
if __name__ == " _ _ main _ _ " :
	arr = [ 2 , 0 , 3 , 2 , 0 ] ;
	n = len ( arr ) ;
	print ( countPairs ( arr , n ) ) ;