Count of binary strings of length N having equal count of 0’s and 1’s

Given an integer N, the task is to find the number of binary strings possible of length N having same frequency of 0s and 1s. If such string is possible of length N, print -1. 
Note: Since the count can be very large, return the answer modulo 10⁹+7.
Examples:

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. 

Naive Approach:
The simplest approach is to generate all possible permutations of a string of length N having equal number of ‘0’ and ‘1’. For every permutation generated, increase the count. Print the total count of permutations generated. 
Time Complexity: O(N*N!) 
Auxiliary Space: O(N)
Efficient Approach:
The above approach can be optimized by using concepts of Permutation and Combination. Follow the steps below to solve the problem: 

• Since N positions need to be filled with equal number of 0‘s and 1‘s, select N/2 positions from the N positions in C(N, N/2) % mod( where mod = 10⁹ + 7) ways to fill with only 1’s.
• Fill the remaining positions in C(N/2, N/2) % mod (i.e 1) way with only 0’s.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
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

Javascript

<script>

// Javascript program for the above approach

    var MOD = 1000000007;

    // Function to calculate C(n, r) % MOD
    // DP based approach
    function nCrModp(n , r) {

        // Corner case
        if (n % 2 == 1) {
            return -1;
        }

        // Stores the last row
        // of Pascal's Triangle
        var C = Array(r + 1).fill(0);



        // Initialize top row
        // of pascal triangle
        C[0] = 1;

        // Construct Pascal's Triangle
        // from top to bottom
        for (i = 1; i <= n; i++) {

            // Fill current row with the
            // help of previous row
            for (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
        var N = 6;
        document.write(nCrModp(N, N / 2));

// This code contributed by gauravrajput1

</script>

Output:

20

Time Complexity: O(N²) 
Space Complexity: O(N)