human
Count all possible N digit numbers that satisfy the given condition | Function to return the count of required numbers.
If N is odd then return 0.
Driver Code"
How can the above be solved in Python?
 ... What is a code to count all possible N digit numbers that satisfy the given condition and the other steps required?
assistant
#Here is code to count all possible N digit numbers that satisfy the given condition and the other steps required.
 def getCount ( N ) :
	if ( N % 2 == 1 ) :
		return "0"
	result = "9"
	for i in range ( 1 , N // 2 ) :
		result = result + "0"
	return result
N = 4
print ( getCount ( N ) )