计算不是 N 周期的不同的正则括号序列

给定一个整数N ，任务是找到可以使用2 * N 个括号形成的不同括号序列的数量，使得该序列不是N-周期的。

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.

例子：

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.

方法：想法是计算长度为2 * N的可能的常规括号序列的总数，然后从中减去N周期的括号序列的数量。以下是步骤：

1. 要找到长度为2*N的正则括号序列的数量，请使用加泰罗尼亚数字公式。
2. 对于长度为2*N的序列是 N 周期，N 应该是偶数，因为如果 N 是奇数，那么长度为2*N的序列不能是规则序列并且同时具有 N 的周期。
3. 由于两个相似的非正则括号序列的串联不能使序列正则，所以两个长度为N的子序列都应该是正则的。
4. 从长度为2*N的正则括号序列的数量中减少长度为N (如果 N 为偶数)的正则括号序列的数量，以获得所需的结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
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;
}

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 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

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 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

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 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

Javascript

 
// 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 Numer
    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
 

输出：

12

时间复杂度： O(N)
辅助空间： O(1)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live