human
Can you provide a code to frugal Number and the other steps required given the following:
Frugal Number | Finding primes upto entered number.
Finding primes by Sieve of Eratosthenes method.
If prime [ i ] is not changed , then it is prime.
Update all multiples of p.
Forming array of the prime numbers found.
Returns number of digits in n.
Checking whether a number is Frugal or not.
Finding number of digits in prime factorization of the number.
Exponent for current factor.
Counting number of times this prime factor divides ( Finding exponent ).
Finding number of digits in the exponent Avoiding exponents of value 1.
Checking condition for frugal number.
Driver Code"
How can the above be solved in Python?
assistant
#Here is code to frugal Number and the other steps required.
 def primes ( n ) :
	prime = [ True ] * ( n + 1 ) ;
	i = 2 ;
	while ( i * i <= n ) :
		if ( prime [ i ] == True ) :
			j = i * 2 ;
			while ( j <= n ) :
				prime [ j ] = False ;
				j += i ;
		i += 1 ;
	arr = [ ] ;
	for i in range ( 2 , n ) :
		if ( prime [ i ] ) :
			arr . append ( i ) ;
	return arr ;
def countDigits ( n ) :
	temp = n ;
	c = 0 ;
	while ( temp != 0 ) :
		temp = int ( temp / 10 ) ;
		c += 1 ;
	return c ;
def frugal ( n ) :
	r = primes ( n ) ;
	t = n ;
	s = 0 ;
	for i in range ( len ( r ) ) :
		if ( t % r [ i ] == 0 ) :
			k = 0 ;
			while ( t % r [ i ] == 0 ) :
				t = int ( t / r [ i ] ) ;
				k += 1 ;
			if ( k == 1 ) :
				s = s + countDigits ( r [ i ] ) ;
			elif ( k != 1 ) :
				s = ( s + countDigits ( r [ i ] ) + countDigits ( k ) ) ;
	return ( countDigits ( n ) > s and s != 0 ) ;
n = 343 ;
if ( frugal ( n ) ) :
	print ( " A ▁ Frugal ▁ number " ) ;
else :
	print ( " Not ▁ a ▁ frugal ▁ number " ) ;