human
Count of unordered pair of indices such that ratio of elements at these indices is same as ratio of indices | Function of find the count of unordered pairs ( i , j ) in the array such that arr [ j ] / arr [ i ] = j / i ..
Stores the count of valid pairs.
Iterating over all values of x in range [ 1 , N ] ..
Iterating over all values of y that are divisible by x in the range [ 1 , N ] ..
Check if the pair is valid.
Return answer.
Driver Code.
Function Call"
How can the above be solved in Python?

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Write me a code to count of unordered pair of indices such that ratio of elements at these indices is same as ratio of indices and the other steps required?
assistant
#Here is code to count of unordered pair of indices such that ratio of elements at these indices is same as ratio of indices and the other steps required.
 def countPairs ( arr , n ) :
	count = 0
	for x in range ( 1 , n + 1 , 1 ) :
		for y in range ( 2 * x , n + 1 , x ) :
			if ( ( arr [ y - 1 ] % arr [ x - 1 ] == 0 ) and ( arr [ y - 1 ] // arr [ x - 1 ] == y // x ) ) :
				count += 1
	return count
if __name__ == ' _ _ main _ _ ' :
	arr = [ 5 , - 2 , 4 , 20 , 25 , - 6 ]
	n = len ( arr )
	print ( countPairs ( arr , n ) )