human
Can you provide a code to count distinct regular bracket sequences which are not N periodic and the other steps required given the following:
Count distinct regular bracket sequences which are not N periodic | Function that finds the value of Binomial Coefficient C ( n , k ).
Since C ( n , k ) = C ( n , n - k ).
Calculate the value of [ n * ( n - 1 ) * -- - * ( n - k + 1 ) ] / [ k * ( k - 1 ) * -- - * 1 ].
Return the C ( n , k ).
Binomial coefficient based function to find nth catalan number in O ( n ) time.
Calculate value of 2 nCn.
Return C ( 2 n , n ) / ( n + 1 ).
Function to find possible ways to put balanced parenthesis in an expression of length n.
If n is odd , not possible to create any valid parentheses.
Otherwise return n / 2 th Catalan Number.
Difference between counting ways of 2 * N and N is the result.
Given value of N.
Function call"
Solution in Python:
assistant
#Here is code to count distinct regular bracket sequences which are not N periodic and the other steps required.
 def binomialCoeff ( n , k ) :
	res = 1
	if ( k > n - k ) :
		k = n - k
	for i in range ( k ) :
		res = res * ( n - i )
		res = res // ( i + 1 )
	return res
def catalan ( n ) :
	c = binomialCoeff ( 2 * n , n )
	return c // ( n + 1 )
def findWays ( n ) :
	if ( ( n & 1 ) == 1 ) :
		return 0
	return catalan ( n // 2 )
def countNonNPeriodic ( N ) :
	print ( findWays ( 2 * N ) - findWays ( N ) )
N = 4
countNonNPeriodic ( N )