human
Count of N | Function to find count of N - digit numbers with single digit XOR.
dp [ i ] [ j ] stores the number of i - digit numbers with XOR equal to j.
For 1 - 9 store the value.
Iterate till N.
Calculate XOR.
Store in DP table.
Initialize count.
Print answer.
Driver Code.
Given number N.
Function Call"
How can the above be solved in Python?
assistant
#Here is code to count of N and the other steps required.
 def countNums ( N ) :
	dp = [ [ 0 for i in range ( 16 ) ] for j in range ( N ) ] ;
	for i in range ( 1 , 10 ) :
		dp [ 0 ] [ i ] = 1 ;
	for i in range ( 1 , N ) :
		for j in range ( 0 , 10 ) :
			for k in range ( 0 , 16 ) :
				xor = j ^ k ;
				dp [ i ] [ xor ] += dp [ i - 1 ] [ k ] ;
	count = 0 ;
	for i in range ( 0 , 10 ) :
		count += dp [ N - 1 ] [ i ] ;
	print ( count ) ;
if __name__ == ' _ _ main _ _ ' :
	N = 1 ;
	countNums ( N ) ;