human
Find the numbers from 1 to N that contains exactly k non | Function to find number less than N having k non - zero digits.
Store the memorised values.
Initialise.
Base.
Calculate all states For every state , from numbers 1 to N , the count of numbers which contain exactly j non zero digits is being computed and updated in the dp array ..
Return the required answer.
Driver code.
Function call"
How can the above be solved in Python?

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Write me a code to find the numbers from 1 to N that contains exactly k non and the other steps required?
assistant
#Here is code to find the numbers from 1 to N that contains exactly k non and the other steps required.
 def k_nonzero_numbers ( s , n , k ) :
	dp = [ [ [ 0 for i in range ( k + 2 ) ] for i in range ( 2 ) ] for i in range ( n + 2 ) ]
	for i in range ( n + 1 ) :
		for j in range ( 2 ) :
			for x in range ( k + 1 ) :
				dp [ i ] [ j ] [ x ] = 0
	dp [ 0 ] [ 0 ] [ 0 ] = 1
	for i in range ( n ) :
		sm = 0
		while ( sm < 2 ) :
			for j in range ( k + 1 ) :
				x = 0
				y = 0
				if sm :
					y = 9
				else :
					y = ord ( s [ i ] ) - ord ( '0' )
				while ( x <= y ) :
					dp [ i + 1 ] [ ( sm or x < ( ord ( s [ i ] ) - ord ( '0' ) ) ) ] [ j + ( x > 0 ) ] += dp [ i ] [ sm ] [ j ]
					x += 1
			sm += 1
	return dp [ n ] [ 0 ] [ k ] + dp [ n ] [ 1 ] [ k ]
if __name__ == ' _ _ main _ _ ' :
	s = "25"
	k = 2
	n = len ( s )
	print ( k_nonzero_numbers ( s , n , k ) )