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:**

Input:N = 2Output:1Explanation: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 = 3Output:4Explanation: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*2^{N}) **Auxilairy Space:** O(1)

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

S= Total Number of binary strings of length N = 2_{Total }^{N}

S= Number of binary string of length N having same frequency of 0’s and 1’s._{equal}S= Number of binary strings of length N having frequency of 1’s greater than 0’s._{1}S= Number of binary strings of length N having frequency of 0’s greater than 1’s._{0}S_{total }= S_{equal }+ S_{1 }+ S_{0}

- For every string in
**S**, there exist a string in_{1}**S**_{0.}

Suppose “**1110**” is the string in**S**then its corresponding string in_{1}**S**will be “_{0}**0001**“. Similarly, for every string in**S**_{1}_{ }there exist a string in**S**. Hence,_{0}**S**_{1}= S_{0}_{ }( for every**N**). - If
**N**is**odd**then**S**._{equal }= 0 - If
**N**is**even**then**S**._{equal}=C(N, N/2)

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 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's ` `unsigned ` `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 Code ` `int` `main() ` `{ ` ` ` `int` `N = 3; ` ` ` `cout << countOfString(N); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to implement ` `// the above approach ` `import` `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's ` `static` `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 Code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `N = ` `3` `; ` ` ` `System.out.print(countOfString(N)); ` `} ` `} ` ` ` `// This code is contributed by Amit Katiyar` |

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## 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's ` `def` `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 Code ` `N ` `=` `3` ` ` `print` `(countOfString(N)) ` ` ` `# This code is contributed by Shivam Singh ` |

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## C#

`// C# program to implement ` `// the above approach ` `using` `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's ` `static` `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 Code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ` `int` `N = 3; ` ` ` `Console.Write(countOfString(N)); ` `} ` `} ` ` ` `// This code is contributed by sapnasingh4991` |

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**Output:**

4

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

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