在每个前缀子串中，长度为N的二进制字符串的计数等于0和1的计数，等于1的计数≥0的计数

给定整数N ，任务是 在每个前缀子串中找到频率为0和1的长度为N的二进制二进制字符串的数量，其中1的频率大于或等于0的频率。

例子：

Input: N = 2
Output: 1
Explanation:
All possible binary strings of length 2 are {“00”, “01”, “10” and “11”}. 
Out of these 4 strings, only “01” and “10” have equal count of 0’s and 1’s.
Out of these two strings, only “10” contains more or equal numbers of 1’s than 0’s in every prefix substring.

Input: N = 4
Output: 2
Explanation : 
All possible binary strings of length 4, satisfying the required conditions are “1100” and “1010”.

天真的方法：
最简单的方法是生成所有长度为N的二进制字符串，并迭代每个字符串以检查其是否包含等于0和1的相等计数，并且还检查1的频率是否大于等于0的频率。所有前缀子串中都为0 。
时间复杂度： O(N * 2 ^{^ N} )
辅助空间： O(1)

高效方法：
可以使用加泰罗尼亚数的概念进一步优化上述方法。我们只需要检查N的奇偶校验即可。

• 如果N是一个奇数整数，则0和1的频率不能相等。因此，此类必需字符串的计数为0 。
• 如果N是偶数整数，则所需子字符串的数量等于第(N / 2)^个加泰罗尼亚数字。

Illustration: 
N = 2 
Only possible string is “10“. Therefore, count is 1. 
N = 4 
Only possible strings are “1100” and “1010”. Therefore, count is 2. 
N = 6 
Only possible strings are “111000”, “110100”, “110010”, “101010” and “101100”. 
Therefore the count is 5. 
Hence, for each even value of N, it follows the sequence 1 2 5 14 …… 
Hence, the series is that of Catalan numbers. 
Therefore, it can be concluded that if N is an even integer, the count is equal to that of (N/2)^th Catalan Number. 
 

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to calculate and returns 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 res;
}
 
// Function to return the count of all
// binary strings having equal count of 0's
// and 1's and each prefix substring having
// frequency of 1's >= frequencies of 0's
unsigned long int countStrings(unsigned int N)
{
    // If N is odd
    if (N % 2 == 1)
 
        // No such strings possibel
        return 0;
 
    // Otherwise
    else {
        N /= 2;
 
        // Calculate value of 2nCn
        unsigned long int c
            = binomialCoeff(2 * N, N);
 
        // Return 2nCn/(n+1)
        return c / (N + 1);
    }
}
 
// Driver Code
int main()
{
    int N = 6;
    cout << countStrings(N) << " ";
    return 0;
}

Java

// Java program to implement the
// above approach
import java.util.*;
 
class GFG{
 
// Function to calculate and returns 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 res;
}
 
// Function to return the count of all
// binary strings having equal count of 0's
// and 1's and each prefix substring having
// frequency of 1's >= frequencies of 0's
static long countStrings(int N)
{
     
    // If N is odd
    if (N % 2 == 1)
 
        // No such strings possibel
        return 0;
 
    // Otherwise
    else
    {
        N /= 2;
 
        // Calculate value of 2nCn
        long c = binomialCoeff(2 * N, N);
 
        // Return 2nCn/(n+1)
        return c / (N + 1);
    }
}
 
// Driver code
public static void main(String[] args)
{
    int N = 6;
     
    System.out.print(countStrings(N) + " ");
}
}
 
// This code is contributed by offbeat

Python3

# Python3 Program to implement
# the above approach
 
# Function to calculate and returns 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 *= (n - i)
        res //= (i + 1)
 
    return res
 
# Function to return the count of all
# binary strings having equal count of 0's
# and 1's and each prefix substring having
# frequency of 1's >= frequencies of 0's
def countStrings(N):
     
    # If N is odd
    if (N % 2 == 1):
 
        # No such strings possibel
        return 0
 
    # Otherwise
    else:
        N //= 2
 
        # Calculate value of 2nCn
        c= binomialCoeff(2 * N, N)
 
        # Return 2nCn/(n+1)
        return c // (N + 1)
 
# Driver Code
if __name__ == '__main__':
    N = 6
    print(countStrings(N))
 
# This code is contributed by Mohit Kumar

C#

// C# program to implement the
// above approach
using System;
 
class GFG{
 
// Function to calculate and returns 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 res;
}
 
// Function to return the count of all
// binary strings having equal count of 0's
// and 1's and each prefix substring having
// frequency of 1's >= frequencies of 0's
static long countStrings(int N)
{
     
    // If N is odd
    if (N % 2 == 1)
 
        // No such strings possibel
        return 0;
 
    // Otherwise
    else
    {
        N /= 2;
 
        // Calculate value of 2nCn
        long c = binomialCoeff(2 * N, N);
 
        // Return 2nCn/(n+1)
        return c / (N + 1);
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 6;
     
    Console.Write(countStrings(N) + " ");
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

// javascript program to implement the
// above approach
 
    // Function to calculate and returns the
    // value of Binomial Coefficient C(n, k)
    function binomialCoeff(n , k) {
        var 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 (var i = 0; i < k; ++i) {
            res *= (n - i);
            res /= (i + 1);
        }
        return res;
    }
 
    // Function to return the count of all
    // binary strings having equal count of 0's
    // and 1's and each prefix substring having
    // frequency of 1's >= frequencies of 0's
    function countStrings(N) {
 
        // If N is odd
        if (N % 2 == 1)
 
            // No such strings possibel
            return 0;
 
        // Otherwise
        else
        {
            N /= 2;
 
            // Calculate value of 2nCn
            var c = binomialCoeff(2 * N, N);
 
            // Return 2nCn/(n+1)
            return c / (N + 1);
        }
    }
 
    // Driver code
    var N = 6;
    document.write(countStrings(N) + " ");
 
// This code is contributed by Princi Singh

输出：

5

时间复杂度： O(N)
辅助空间：O(1)