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Count of binary strings of length N having equal count of 0’s and 1’s and count of 1’s ≥ count of 0’s in each prefix substring

  • Last Updated : 25 Mar, 2021

Given an integer N, the task is to find the number of possible binary strings of length N with an equal frequency of 0‘s and 1‘s in which frequency of 1‘s are greater or equal to the frequency of 0‘s in every prefix substring.

Examples:

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”.

Naive Approach: 
The simplest approach is to generate all the binary strings of length N and iterate each string to check if it contains an equal count of 0‘s and 1‘s and also check if the frequency of 1‘s is equal to greater than that of 0‘s in all its prefix substrings.
Time Complexity: O(N*2^N)
Auxiliary Space: O(1)
 



Efficient Approach: 
The above approach can be further optimized using the concept of Catalan Number. We just need to check the parity of N.

  • If N is an odd integer, frequencies of 0’s and 1’s cannot be equal. Therefore, the count of such required strings is 0.
  • If N is an even integer, the count of required substrings is equal to the (N/2)th Catalan number.

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. 
 

Below is the implementation of the above approach:

C++




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




<script>
// 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
</script>
Output: 
5

Time Complexity: O(N) 
Auxiliary Space: O(1)
 

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