human
Count non decreasing subarrays of size N from N Natural numbers | Returns value of Binomial Coefficient C ( n , k ).
Since nC0 is 1.
Compute next row of pascal triangle using the previous row.
Function to find the count of required subarrays.
The required count is the binomial coefficient as explained in the approach above.
Driver Function"
How can the above be solved in Python?
 Give me a code to count non decreasing subarrays of size N from N Natural numbers and the other steps required?
assistant
#Here is code to count non decreasing subarrays of size N from N Natural numbers and the other steps required.
 def binomialCoeff ( n , k ) :
	C = [ 0 ] * ( k + 1 ) ;
	C [ 0 ] = 1 ;
	for i in range ( 1 , n + 1 ) :
		for j in range ( min ( i , k ) , 0 , - 1 ) :
			C [ j ] = C [ j ] + C [ j - 1 ] ;
	return C [ k ] ;
def count_of_subarrays ( N ) :
	count = binomialCoeff ( 2 * N - 1 , N ) ;
	return count ;
if __name__ == " _ _ main _ _ " :
	N = 3 ;
	print ( count_of_subarrays ( N ) ) ;