Count distinct regular bracket sequences which are not N periodic

Given an integer N, the task is to find the number of distinct bracket sequences that can be formed using 2 * N brackets such that the sequence is not N-periodic.

A bracket sequence str of length 2 * N is said to be N-periodic if the sequence can be split into two equal substrings having same regular bracket sequence.
A regular bracket sequence is a sequence in the following way:

• An empty string is a regular bracket sequence.
• If s & t are regular bracket sequences, then s + t is a regular bracket sequence.

Examples:

Input: N = 3 
Output: 5 
Explanation:
There will be 5 distinct regular bracket sequences of length 2 * N = ()()(), ()(()), (())(), (()()), ((())) 
Now, none of the sequences are N-periodic. Therefore, the output is 5.

Input: N = 4 
Output: 12 
Explanation:
There will be 14 distinct regular bracket sequences of length 2*N which are 
()()()(), ()()(()), ()(())(), ()(()()), ()((())), (())()(), (())(()), (()())(), (()()()), (()(())), ((()))(), ((())()), ((()())), (((()))) 
Out of these 14 regular sequences, two of them are N periodic which are 
()()()() and (())(()). They have a period of N. 
Therefore, the distinct regular bracket sequences of length 2 * N which are not N-periodic are 14 – 2 = 12.

Approach: The idea is to calculate the total number of regular bracket sequences possible of length 2 * N and then to subtract the number of bracket sequences which are N-periodic from it. Below are the steps:

1. To find the number of regular bracket sequences of length 2*N, use the Catalan number formula.
2. For a sequence of length 2*N to be N periodic, N should be even because if N is odd then the sequence of length 2*N cannot be a regular sequence and have a period of N at the same time.
3. Since the concatenation of two similar non-regular bracket sequences cannot make a sequence regular, so both subsequences of length N should be regular.
4. Reduce the number of regular bracket sequences of length N(if N is even) from the number of regular bracket sequences of length 2*N to get the desired result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function that finds the value of
// Binomial Coefficient C(n, k)
unsigned long int
binomialCoeff(unsigned int n,
              unsigned int k)
{
    unsigned long int res = 1;

    // Since C(n, k) = C(n, n - k)
    if (k > n - k)
        k = n - k;

    // Calculate the value of
    // [n*(n - 1)*---*(n - k + 1)] /
    // [k*(k - 1)*---*1]
    for (int i = 0; i < k; ++i) {
        res *= (n - i);
        res /= (i + 1);
    }

    // Return the C(n, k)
    return res;
}

// Binomial coefficient based function to
// find nth catalan number in O(n) time
unsigned long int catalan(unsigned int n)
{
    // Calculate value of 2nCn
    unsigned long int c
        = binomialCoeff(2 * n, n);

    // Return C(2n, n)/(n+1)
    return c / (n + 1);
}

// Function to find possible ways to
// put balanced  parenthesis in an
// expression of length n
unsigned long int findWays(unsigned n)
{
    // If n is odd, not possible to
    // create any valid parentheses
    if (n & 1)
        return 0;

    // Otherwise return n/2th
    // Catalan Number
    return catalan(n / 2);
}

void countNonNPeriodic(int N)
{

    // Difference between counting ways
    // of 2*N and N is the result
    cout << findWays(2 * N)
                - findWays(N);
}

// Driver Code
int main()
{
    // Given value of N
    int N = 4;

    // Function Call
    countNonNPeriodic(N);

    return 0;
}

Java

// Java program for above approach
import java.io.*;

class GFG{

// Function that finds the value of
// Binomial Coefficient C(n, k)
static long binomialCoeff(int n, int k)
{
    long res = 1;

    // Since C(n, k) = C(n, n - k)
    if (k > n - k)
        k = n - k;

    // Calculate the value of
    // [n*(n - 1)*---*(n - k + 1)] /
    // [k*(k - 1)*---*1]
    for(int i = 0; i < k; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }

    // Return the C(n, k)
    return res;
}

// Binomial coefficient based function to
// find nth catalan number in O(n) time
static long catalan(int n)
{

    // Calculate value of 2nCn
    long c = binomialCoeff(2 * n, n);

    // Return C(2n, n)/(n+1)
    return c / (n + 1);
}

// Function to find possible ways to
// put balanced parenthesis in an
// expression of length n
static long findWays(int n)
{

    // If n is odd, not possible to
    // create any valid parentheses
    if ((n & 1) == 1)
        return 0;

    // Otherwise return n/2th
    // Catalan Number
    return catalan(n / 2);
}

static void countNonNPeriodic(int N)
{

    // Difference between counting ways
    // of 2*N and N is the result
    System.out.println(findWays(2 * N) -
                       findWays(N));
}

// Driver code
public static void main (String[] args)
{

    // Given value of N
    int N = 4;

    // Function call
    countNonNPeriodic(N);
}
}

// This code is contributed by offbeat

Python3

# Python3 program for
# the above approach

# Function that finds the value of
# Binomial Coefficient C(n, k)
def binomialCoeff(n, k):
    res = 1

    # Since C(n, k) = C(n, n - k)
    if (k > n - k):
        k = n - k

    # Calculate the value of
    # [n*(n - 1)*---*(n - k + 1)] /
    # [k*(k - 1)*---*1]
    for i in range(k):

        res = res * (n - i)
        res = res // (i + 1)

    # Return the C(n, k)
    return res

# Binomial coefficient based function to
# find nth catalan number in O(n) time
def catalan(n):

    # Calculate value of 2nCn
    c = binomialCoeff(2 * n, n)

    # Return C(2n, n)/(n+1)
    return c // (n + 1)

# Function to find possible ways to
# put balanced parenthesis in an
# expression of length n
def findWays(n):

    # If n is odd, not possible to
    # create any valid parentheses
    if ((n & 1) == 1):
        return 0

    # Otherwise return n/2th
    # Catalan Number
    return catalan(n // 2)

def countNonNPeriodic(N):

    # Difference between counting ways
    # of 2*N and N is the result
    print(findWays(2 * N) - findWays(N))

# Driver code
# Given value of N
N = 4

# Function call
countNonNPeriodic(N)

# This code is contributed by divyeshrabadiya07

C#

// C# program for above approach
using System;
using System.Collections.Generic;

class GFG{

// Function that finds the value of
// Binomial Coefficient C(n, k)
static long binomialCoeff(int n, int k)
{
    long res = 1;

    // Since C(n, k) = C(n, n - k)
    if (k > n - k)
        k = n - k;

    // Calculate the value of
    // [n*(n - 1)*---*(n - k + 1)] /
    // [k*(k - 1)*---*1]
    for(int i = 0; i < k; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }

    // Return the C(n, k)
    return res;
}

// Binomial coefficient based function to
// find nth catalan number in O(n) time
static long catalan(int n)
{

    // Calculate value of 2nCn
    long c = binomialCoeff(2 * n, n);

    // Return C(2n, n)/(n+1)
    return c / (n + 1);
}

// Function to find possible ways to
// put balanced parenthesis in an
// expression of length n
static long findWays(int n)
{

    // If n is odd, not possible to
    // create any valid parentheses
    if ((n & 1) == 1)
        return 0;

    // Otherwise return n/2th
    // Catalan Number
    return catalan(n / 2);
}

static void countNonNPeriodic(int N)
{

    // Difference between counting ways
    // of 2*N and N is the result
    Console.Write(findWays(2 * N) -
                  findWays(N));
}

// Driver Code
public static void Main(string[] args)
{

    // Given value of N
    int N = 4;

    // Function call
    countNonNPeriodic(N);
}
}

// This code is contributed by rutvik_56

Javascript

<script>

// Javascript program for the above approach

// Function that finds the value of
// Binomial Coefficient C(n, k)
function binomialCoeff(n, k)
{
    let res = 1;

    // Since C(n, k) = C(n, n - k)
    if (k > n - k)
        k = n - k;

    // Calculate the value of
    // [n*(n - 1)*---*(n - k + 1)] /
    // [k*(k - 1)*---*1]
    for (let i = 0; i < k; ++i) {
        res *= (n - i);
        res /= (i + 1);
    }

    // Return the C(n, k)
    return res;
}

// Binomial coefficient based function to
// find nth catalan number in O(n) time
function catalan(n)
{
    // Calculate value of 2nCn
    let c = binomialCoeff(2 * n, n);

    // Return C(2n, n)/(n+1)
    return c / (n + 1);
}

// Function to find possible ways to
// put balanced parenthesis in an
// expression of length n
function findWays(n)
{
    // If n is odd, not possible to
    // create any valid parentheses
    if (n & 1)
        return 0;

    // Otherwise return n/2th
    // Catalan Number
    return catalan(n / 2);
}

function countNonNPeriodic(N)
{

    // Difference between counting ways
    // of 2*N and N is the result
    document.write(findWays(2 * N)
                - findWays(N));
}

// Driver Code

    // Given value of N
    let N = 4;

    // Function Call
    countNonNPeriodic(N);


// This code is contributed by Mayank Tyagi

</script>

Output:

12

Time Complexity: O(N) 
Auxiliary Space: O(1)