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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"
Solution in Python:

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Give me a code to frugal Number and the other steps required?
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 " ) ;