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:
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++

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// 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 Numer
    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;
}

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Java

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// 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 Numer
    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

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Python3

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# 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 Numer
    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

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C#

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// 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 Numer
    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

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Output: 

12


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

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