human
Count array elements whose count of divisors is a prime number | Function to count the array elements whose count of divisors is prime.
Stores the maximum element.
Find the maximum element.
Store if i - th element is prime ( 0 ) or non - prime ( 1 ).
Base Case.
If i is a prime number.
Mark all multiples of i as non - prime.
Stores the count of divisors.
Base Case.
Iterate to count factors.
Stores the count of array elements whose count of divisors is a prime number.
Traverse the array arr [ ].
If count of divisors is prime.
Return the resultant count.
Driver Code"
Solution in Python:

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Write me a code to count array elements whose count of divisors is a prime number and the other steps required?
assistant
#Here is code to count array elements whose count of divisors is a prime number and the other steps required.
 def primeDivisors ( arr , N ) :
	K = arr [ 0 ]
	for i in range ( 1 , N ) :
		K = max ( K , arr [ i ] )
	prime = [ 0 ] * ( K + 1 )
	prime [ 0 ] = 1
	prime [ 1 ] = 1
	for i in range ( 2 , K + 1 ) :
		if ( not prime [ i ] ) :
			for j in range ( 2 * i , K + 1 , i ) :
				prime [ j ] = 1
	factor = [ 0 ] * ( K + 1 )
	factor [ 0 ] = 0
	factor [ 1 ] = 1
	for i in range ( 2 , K + 1 ) :
		factor [ i ] += 1
		for j in range ( i , K + 1 , i ) :
			factor [ j ] += 1
	count = 0
	for i in range ( N ) :
		if ( prime [ factor [ arr [ i ] ] ] == 0 ) :
			count += 1
	return count
if __name__ == ' _ _ main _ _ ' :
	arr = [ 10 , 13 , 17 , 25 ]
	N = len ( arr )
	print ( primeDivisors ( arr , N ) )