# Count of Binary Strings of length N such that frequency of 1’s exceeds frequency of 0’s

• Last Updated : 01 Sep, 2021

Given an integer N, the task is to find the number of Binary Strings of length N such that frequency of 1‘s is greater than the frequency of 0‘s.

Example:

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Input: N = 2
Output: 1
Explanation: Count of binary strings of length 2 is 4 i.e. {“00”, “01”, “10”, “11”}.
The only string having frequency of 1’s greater than that of 0’s is “11”.

Input: N = 3
Output: 4
Explanation: Count of binary strings of length 3 is 8 i.e. {“000”, “001”, “010”, “011”, “100”, “101”, “110”, “111”}.
Among them, the strings having frequency of 1’s greater than 0’s are {“011”, “101”, “110”, “111”}.

Naive Approach: The simplest approach to solve this problem is to generate all binary strings of length N, and iterate over each string to find the frequency of 1’s and 0’s. If the frequency of 1’s is greater than that of 0’s, increment the counter. Finally, print the count.

Time Complexity: O(N*2N
Auxiliary Space: O(1)

Efficient Approach: To observe the above approach, following observations need to made:

STotal = Total Number of binary strings of length N = 2N

Sequal = Number of binary string of length N having same frequency of 0’s and 1’s.
S1 = Number of binary strings of length N having frequency of 1’s greater than 0’s.
S0 = Number of binary strings of length N having frequency of 0’s greater than 1’s.
Stotal = Sequal + S1 + S0

1. For every string in S1, there exist a string in S0.
Suppose “1110” is the string in S1 then its corresponding string in S0  will be “0001“. Similarly, for every string in S1 there exist a string in S0. Hence, S1 = S0 ( for every N).
2. If N is odd then Sequal = 0.
3. If N is even then Sequal =C(N, N/2).

Below is the implementation of the above approach:

## C++

 // C++ Program to implement// the above approach#include using namespace std; // Function to calculate// and return the value of// Binomial Coefficient C(n, k)unsigned long int binomialCoeff(unsigned long int n,                                unsigned long 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// binary strings of length N such// that frequency of 1's exceed that of 0'sunsigned long int countOfString(int N){    // Count of N-length binary strings    unsigned long int Stotal = pow(2, N);     // Count of N- length binary strings    // having equal count of 0's and 1's    unsigned long int Sequal = 0;     // For even length strings    if (N % 2 == 0)        Sequal = binomialCoeff(N, N / 2);     unsigned long int S1 = (Stotal - Sequal) / 2;    return S1;} // Driver Codeint main(){    int N = 3;    cout << countOfString(N);    return 0;}

## Java

 // Java program to implement// the above approachimport java.util.*; class GFG{ // Function to calculate// and return the value of// Binomial Coefficient C(n, k)static int binomialCoeff(int n, int k){    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// binary Strings of length N such// that frequency of 1's exceed that of 0'sstatic int countOfString(int N){         // Count of N-length binary Strings    int Stotal = (int) Math.pow(2, N);     // Count of N- length binary Strings    // having equal count of 0's and 1's    int Sequal = 0;     // For even length Strings    if (N % 2 == 0)        Sequal = binomialCoeff(N, N / 2);     int S1 = (Stotal - Sequal) / 2;    return S1;} // Driver Codepublic static void main(String[] args){    int N = 3;    System.out.print(countOfString(N));}} // This code is contributed by Amit Katiyar

## Python3

 # Python3 program to implement# the above approach # Function to calculate# and return 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# binary strings of length N such# that frequency of 1's exceed that of 0'sdef countOfString(N):     # Count of N-length binary strings    Stotal = pow(2, N)     # Count of N- length binary strings    # having equal count of 0's and 1's    Sequal = 0     # For even length strings    if(N % 2 == 0):        Sequal = binomialCoeff(N, N // 2)     S1 = (Stotal - Sequal) // 2     return S1 # Driver CodeN = 3 print(countOfString(N)) # This code is contributed by Shivam Singh

## C#

 // C# program to implement// the above approachusing System;class GFG{ // Function to calculate// and return the value of// Binomial Coefficient C(n, k)static int binomialCoeff(int n, int k){    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// binary Strings of length N such// that frequency of 1's exceed that of 0'sstatic int countOfString(int N){         // Count of N-length binary Strings    int Stotal = (int) Math.Pow(2, N);     // Count of N- length binary Strings    // having equal count of 0's and 1's    int Sequal = 0;     // For even length Strings    if (N % 2 == 0)        Sequal = binomialCoeff(N, N / 2);     int S1 = (Stotal - Sequal) / 2;    return S1;} // Driver Codepublic static void Main(String[] args){    int N = 3;    Console.Write(countOfString(N));}} // This code is contributed by sapnasingh4991

## Javascript


Output:
4

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

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