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Check if a number exists having exactly N factors and K prime factors | Function to compute the number of factors of the given number.
Vector to store the prime factors.
While there are no two multiples in the number , divide it by 2.
If the size is already greater than K , then return true.
Computing the remaining divisors of the number.
If n is divisible by i , then it is a divisor.
If the size is already greater than K , then return true.
If the size is already greater than K , then return true.
If none of the above conditions satisfies , then return false.
Function to check if it is possible to make a number having total N factors and K prime factors.
If total divisors are less than the number of prime divisors , then print No.
Find the number of factors of n.
Driver Code"
Solution in Python:

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Give me a code to check if a number exists having exactly N factors and K prime factors and the other steps required?
assistant
#Here is code to check if a number exists having exactly N factors and K prime factors and the other steps required.
 def factors ( n , k ) :
	v = [ ] ;
	while ( n % 2 == 0 ) :
		v . append ( 2 ) ;
		n //= 2 ;
	if ( len ( v ) >= k ) :
		return True ;
	for i in range ( 3 , int ( n ** ( 1 / 2 ) ) , 2 ) :
		while ( n % i == 0 ) :
			n = n // i ;
			v . append ( i ) ;
		if ( len ( v ) >= k ) :
			return True ;
	if ( n > 2 ) :
		v . append ( n ) ;
	if ( len ( v ) >= k ) :
		return True ;
	return False ;
def operation ( n , k ) :
	answered = False ;
	if ( n < k ) :
		answered = True ;
		print ( " No " ) ;
	ok = factors ( n , k ) ;
	if ( not ok and not answered ) :
		answered = True ;
		print ( " No " ) ;
	if ( ok and not answered ) :
		print ( " Yes " ) ;
if __name__ == " _ _ main _ _ " :
	n = 4 ;
	k = 2 ;
	operation ( n , k ) ;