Count of all possible N length balanced Binary Strings

Given a number N, the task is to find the total number of balanced binary strings possible of length N. A binary string is said to be balanced if:

• The number of 0s and 1s are equal in each binary string
• The count of 0s in any prefix of binary strings is always greater than or equal to the count of 1s
• For Example: 01 is a balanced binary string of length 2, but 10 is not.

Example:

Input: N = 4
Output: 2
Explanation: Possible balanced binary strings are: 0101, 0011

Input: N = 5
Output: 0

Approach: The given problem can be solved as below:

1. If N is odd, then no balanced binary string is possible as the condition of an equal count of 0s and 1s will fail.
2. If N is even, then the N length binary string will have N/2 balanced pair of 0s and 1s.
3. So, now try to create a formula to get the number of balanced strings when N is even.

So if N = 2, then possible balanced binary string will be “01” only, as “00” and “11” do not have same count of 0s and 1s and “10” does not have count of 0s >= count of 1s in prefix [0, 1).
Similarly, if N=4, then possible balanced binary string will be “0101” and “0011”
For N = 6, then possible balanced binary string will be “010101”, “010011”, “001101”, “000111”, and “001011”
Now, If we consider this series:
For N=0, count(0) = 1
For N=2, count(2) = count(0)*count(0) = 1
For N=4, count(4) = count(0)*count(2) + count(2)*count(0) = 1*1 + 1*1 = 2
For N=6, count(6) = count(0)*count(4) + count(2)*count(2) + count(4)*count(0) = 1*2 + 1*1 + 2*1 = 5
For N=8, count(8) = count(0)*count(6) + count(2)*count(4) + count(4)*count(2) + count(6)*count(0) = 1*5 + 1*2 + 2*1 + 5*1 = 14
.
.
.
For N=N, count(N) = count(0)*count(N-2) + count(2)*count(N-4) + count(4)*count(N-6) + …. + count(N-6)*count(4) + count(N-4)*count(2) + count(N-2)*count(0)
which is nothing but Catalan numbers.

1. Hence for any even N return Catalan number for (N/2) as the answer.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

#define MAXN 500
#define mod 1000000007

// Vector to store catalan number
vector<long long int> cat(MAXN + 1, 0);

// Function to get the Catalan Number
void catalan()
{
    cat[0] = 1;
    cat[1] = 1;

    for (int i = 2; i < MAXN + 1; i++) {
        long long int t = 0;
        for (int j = 0; j < i; j++) {
            t += ((cat[j] % mod)
                  * (cat[i - 1 - j] % mod)
                  % mod);
        }
        cat[i] = (t % mod);
    }
}

int countBalancedStrings(int N)
{
    // If N is odd
    if (N & 1) {
        return 0;
    }

    // Returning Catalan number
    // of N/2 as the answer
    return cat[N / 2];
}

// Driver Code
int main()
{
    // Precomputing
    catalan();

    int N = 4;
    cout << countBalancedStrings(N);
}

Java

// Java program for the above approach
import java.io.*;
class GFG {

    public static int MAXN = 500;
    public static int mod = 1000000007;

    // Vector to store catalan number
    public static int[] cat = new int[MAXN + 1];

    // Function to get the Catalan Number
    public static void catalan() {
        cat[0] = 1;
        cat[1] = 1;

        for (int i = 2; i < MAXN + 1; i++) {
            int t = 0;
            for (int j = 0; j < i; j++) {
                t += ((cat[j] % mod)
                        * (cat[i - 1 - j] % mod)
                        % mod);
            }
            cat[i] = (t % mod);
        }
    }

    public static int countBalancedStrings(int N)
    {

        // If N is odd
        if ((N & 1) > 0) {
            return 0;
        }

        // Returning Catalan number
        // of N/2 as the answer
        return cat[N / 2];
    }

    // Driver Code
    public static void main(String args[])
    {

        // Precomputing
        catalan();

        int N = 4;
        System.out.println(countBalancedStrings(N));
    }
}

// This code is contributed by saurabh_jaiswal.

Python3

# Python3 program for the above approach
MAXN = 500
mod = 1000000007

# Vector to store catalan number
cat = [0 for _ in range(MAXN + 1)]

# Function to get the Catalan Number
def catalan():

    global cat

    cat[0] = 1
    cat[1] = 1

    for i in range(2, MAXN + 1):
        t = 0
        for j in range(0, i):
            t += ((cat[j] % mod) *
                  (cat[i - 1 - j] % mod) % mod)

        cat[i] = (t % mod)

def countBalancedStrings(N):

    # If N is odd
    if (N & 1):
        return 0

    # Returning Catalan number
    # of N/2 as the answer
    return cat[N // 2]

# Driver Code
if __name__ == "__main__":

    # Precomputing
    catalan()

    N = 4
    print(countBalancedStrings(N))

# This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
class GFG
{
    public static int MAXN = 500;
    public static int mod = 1000000007;

    // Vector to store catalan number
    public static int[] cat = new int[MAXN + 1];

    // Function to get the Catalan Number
    public static void catalan()
    {
        cat[0] = 1;
        cat[1] = 1;

        for (int i = 2; i < MAXN + 1; i++)
        {
            int t = 0;
            for (int j = 0; j < i; j++)
            {
                t += ((cat[j] % mod)
                        * (cat[i - 1 - j] % mod)
                        % mod);
            }
            cat[i] = (t % mod);
        }
    }

    public static int countBalancedStrings(int N)
    {

        // If N is odd
        if ((N & 1) > 0)
        {
            return 0;
        }

        // Returning Catalan number
        // of N/2 as the answer
        return cat[N / 2];
    }

    // Driver Code
    public static void Main()
    {

        // Precomputing
        catalan();

        int N = 4;
        Console.Write(countBalancedStrings(N));
    }
}

// This code is contributed by saurabh_jaiswal.

Javascript

<script>

    // JavaScript Program to implement
    // the above approach
    let MAXN = 500
    let mod = 1000000007

    // Vector to store catalan number
    let cat = new Array(MAXN + 1).fill(0);

    // Function to get the Catalan Number
    function catalan() {
        cat[0] = 1;
        cat[1] = 1;

        for (let i = 2; i < MAXN + 1; i++) {
            let t = 0;
            for (let j = 0; j < i; j++) {
                t += ((cat[j] % mod)
                    * (cat[i - 1 - j] % mod)
                    % mod);
            }
            cat[i] = (t % mod);
        }
    }

    function countBalancedStrings(N)
    {

        // If N is odd
        if (N & 1) {
            return 0;
        }

        // Returning Catalan number
        // of N/2 as the answer
        return cat[(Math.floor(N / 2))];
    }

    // Driver Code

    // Precomputing
    catalan();
    let N = 4;
    document.write(countBalancedStrings(N));

// This code is contributed by Potta Lokesh
</script>

Output

2

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