human
Write me a code to largest number not exceeding N that does not contain any of the digits of S and the other steps required given the following:
Largest number not exceeding N that does not contain any of the digits of S | Function to obtain the largest number not exceeding num which does not contain any digits present in S.
Stores digits of S.
Traverse the string S.
Set occurrence as true.
Traverse the string n.
All digits of num are not present in string s.
Largest Digit not present in the string s.
Set all digits from positions in + 1 to n - 1 as LargestDig.
Counting leading zeroes.
Removing leading zeroes.
If the string becomes null.
Return the largest number.
Driver code"
How can the above be solved in Python?
assistant
#Here is code to largest number not exceeding N that does not contain any of the digits of S and the other steps required.
 def greatestReducedNumber ( num , s ) :
	vis_s = [ False ] * 10
	for i in range ( len ( s ) ) :
	vis_s [ ( ord ) ( s [ i ] ) - 48 ] = True
	n = len ( num )
	In = - 1
	for i in range ( n ) :
	if ( vis_s [ ord ( num [ i ] ) - ord ( '0' ) ] ) :
		In = i
		break
	if ( In == - 1 ) :
	return num
	for dig in range ( ord ( num [ In ] ) , ord ( '0' ) - 1 , - 1 ) :
	if ( vis_s [ dig - ord ( '0' ) ] == False ) :
		num = num [ 0 : In ] + chr ( dig ) + num [ In + 1 : n - In - 1 ]
		break
	LargestDig = '0'
	for dig in range ( ord ( '9' ) , ord ( '0' ) - 1 , - 1 ) :
	if ( vis_s [ dig - ord ( '0' ) ] == False ) :
		LargestDig = dig
		break
	for i in range ( In + 1 , n ) :
	num = num [ 0 : i ] + chr ( LargestDig )
	Count = 0
	for i in range ( n ) :
	if ( num [ i ] == '0' ) :
		Count += 1
	else :
		break
	num = num [ Count : n ]
	if ( int ( len ( num ) ) == 0 ) :
	return "0"
	return num
N = "12345"
S = "23"
print ( greatestReducedNumber ( N , S ) )