长度为N的二进制字符串计数为0和1相等

给定一个整数N ，任务是找到具有0 s和1 s相同频率的长度N的二进制字符串的数量。如果这样的字符串的长度为N ，则打印-1 。
注意：由于计数可能非常大，请以10 ⁹ +7为模返回答案。
例子：

Input: N = 2 
Output: 2 
Explanation: 
All possible binary strings of length 2 are “00”, “01”, “10” and “11”. 
Among them, “10” and “01” have the same frequency of 0s and 1s. 
Hence, the answer is 2.

Input: 4 
Output: 6 
Explanation: 
Strings “0011”, “0101”, “0110”, “1100”, “1010” and “1001” have same frequency of 0s and 1s. 
Hence, the answer is 6. 
 

天真的方法：
最简单的方法是生成长度为N的，具有相同数量的‘0’和‘1’的字符串的所有可能排列。对于生成的每个排列，增加计数。打印的permutatiosn产生的总数。
时间复杂度： O(N * N！)
辅助空间：O(N)

高效方法：
可以通过使用置换和组合的概念来优化上述方法。请按照以下步骤解决问题：

• 由于N个位置需要用0和1的相等数量填充，因此请以C(N，N / 2)％mod (其中mod = 10 ⁹ + 7)的方式从N个位置中选择N / 2个位置，只填1。
• 仅以0填充C(N / 2，N / 2)％mod (即1)方式中的其余位置。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
#define MOD 1000000007
  
// Function to calculate C(n, r) % MOD
// DP based approach
int nCrModp(int n, int r)
{
    // Corner case
    if (n % 2 == 1) {
        return -1;
    }
  
    // Stores the last row
    // of Pascal's Triangle
    int C[r + 1];
  
    memset(C, 0, sizeof(C));
  
    // Initialize top row
    // of pascal triangle
    C[0] = 1;
  
    // Construct Pascal's Triangle
    // from top to bottom
    for (int i = 1; i <= n; i++) {
  
        // Fill current row with the
        // help of previous row
        for (int j = min(i, r); j > 0;
            j--)
  
            // C(n, j) = C(n-1, j)
            // + C(n-1, j-1)
            C[j] = (C[j] + C[j - 1])
                % MOD;
    }
  
    return C[r];
}
  
// Driver Code
int main()
{
    int N = 6;
    cout << nCrModp(N, N / 2);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
      
final static int MOD = 1000000007;
      
// Function to calculate C(n, r) % MOD
// DP based approach
static int nCrModp(int n, int r)
{
  
    // Corner case
    if (n % 2 == 1)
    {
        return -1;
    }
  
    // Stores the last row
    // of Pascal's Triangle
    int C[] = new int[r + 1];
  
    Arrays.fill(C, 0);
  
    // Initialize top row
    // of pascal triangle
    C[0] = 1;
  
    // Construct Pascal's Triangle
    // from top to bottom
    for(int i = 1; i <= n; i++)
    {
  
        // Fill current row with the
        // help of previous row
        for(int j = Math.min(i, r); 
                j > 0; j--)
  
            // C(n, j) = C(n-1, j)
            // + C(n-1, j-1)
            C[j] = (C[j] + C[j - 1]) % MOD;
    }
    return C[r];
}
  
// Driver Code
public static void main(String s[])
{
    int N = 6;
    System.out.println(nCrModp(N, N / 2));
} 
}
  
// This code is contributed by rutvik_56

Python3

# Python3 program to implement
# the above approach
MOD = 1000000007
  
# Function to calculate C(n, r) % MOD 
# DP based approach 
def nCrModp (n, r):
  
    # Corner case
    if (n % 2 == 1):
        return -1
  
    # Stores the last row 
    # of Pascal's Triangle
    C = [0] * (r + 1)
  
    # Initialize top row 
    # of pascal triangle 
    C[0] = 1
  
    # Construct Pascal's Triangle 
    # from top to bottom 
    for i in range(1, n + 1):
  
        # Fill current row with the 
        # help of previous row 
        for j in range(min(i, r), 0, -1):
  
            # C(n, j) = C(n-1, j) 
            # + C(n-1, j-1)
            C[j] = (C[j] + C[j - 1]) % MOD
  
    return C[r]
  
# Driver Code
N = 6
  
print(nCrModp(N, N // 2))
  
# This code is contributed by himanshu77

C#

// C# program for the above approach
using System;
  
class GFG{
      
static int MOD = 1000000007;
  
// Function to calculate C(n, r) % MOD
// DP based approach
static int nCrModp(int n, int r)
{
      
    // Corner case
    if (n % 2 == 1)
    {
        return -1;
    }
  
    // Stores the last row
    // of Pascal's Triangle
    int[] C = new int[r + 1];
  
    // Initialize top row
    // of pascal triangle
    C[0] = 1;
  
    // Construct Pascal's Triangle
    // from top to bottom
    for(int i = 1; i <= n; i++)
    {
  
        // Fill current row with the
        // help of previous row
        for(int j = Math.Min(i, r); 
                j > 0; j--)
  
            // C(n, j) = C(n-1, j)
            // + C(n-1, j-1)
            C[j] = (C[j] + C[j - 1]) % MOD;
    }
    return C[r];
}
  
// Driver code
static void Main() 
{
    int N = 6;
    Console.WriteLine(nCrModp(N, N / 2));
}
}
  
// This code is contributed by divyeshrabadiya07

输出：

20

时间复杂度： O(N ² )
空间复杂度： O(N)