We are given an integer N. We need to count the total number of such binary strings of length N such that the number of ‘0’s in the first string of length N/2 is double of the number of ‘0’s in the second string of length N/2.

Note: N is always an even positive integer.

Examples:

Input : N = 4

Output : 2

Explanation: 0010 and 0001 are two binary string such that the number of zero in the first half is double the number of zero in second half.

Input : N = 6

Output : 9

**Naive Approach:**We can generate all the binary string of length N and then one by one can check that whether selected string follows the given scenario. But the time complexity for this approach is exponential and is O(N*2^{N}).

**Efficient Approach:** This approach is based on some mathematical analysis of the problem.

Pre-requisite for this approach: Factorial and Combinatoric with modulo function.

Note: Let n = N/2.

If we perform some analysis step by step:

- The number of strings which contain two ‘0’ in first half string and one ‘0’ in second half string, is equal to
.^{n}C_{2}*^{n}C_{1} - The number of strings which contain four ‘0’ in first half string and two ‘0’ in second half string, is equal to
.^{n}C_{4}*^{n}C_{2} - The number of strings which contain six ‘0’ in first half string and three ‘0’ in second half string, is equal to
.^{n}C_{6}*^{n}C_{3}

We repeat above procedure till the number of ‘0’ in first half become n if n is even or (n-1) if n is odd.

So, our final answer is **Σ ( ^{n}C_{(2*i)} * ^{n}C_{i})**, for 2 <= 2*i <= n.

Now, we can compute the required number of strings just by using Permutation and Combination.

**Algorithm :**

int n = N/2; for(int i=2;i<=n;i=i+2) ans += ( nCr(n, i) * nCr(n, i/2);

**Note :** You can use Dynamic Programming technique to pre-compute CoeFunc(N, i) i.e. ^{n}C_{i}.

Time complexity is O(N) if we pre-compute CoeFunc(N, i) in O(N*N).

## CPP

// CPP for finding number of binary strings // number of '0' in first half is double the // number of '0' in second half of string #include <bits/stdc++.h> // pre define some constant #define mod 1000000007 #define max 1001 using namespace std; // global values for pre computation long long int nCr[1003][1003]; void preComputeCoeff() { for (int i = 0; i < max; i++) { for (int j = 0; j <= i; j++) { if (j == 0 || j == i) nCr[i][j] = 1; else nCr[i][j] = (nCr[i - 1][j - 1] + nCr[i - 1][j]) % mod; } } } // function to print number of required string long long int computeStringCount(int N) { int n = N / 2; long long int ans = 0; // calculate answer using proposed algorithm for (int i = 2; i <= n; i += 2) ans = (ans + ((nCr[n][i] * nCr[n][i / 2]) % mod)) % mod; return ans; } // Driver code int main() { preComputeCoeff(); int N = 3; cout << computeStringCount(N) << endl; return 0; }

## Java

// Java program for finding number of binary strings // number of '0' in first half is double the // number of '0' in second half of string class GFG { // pre define some constant static final long mod = 1000000007; static final long max = 1001; // global values for pre computation static long nCr[][] = new long[1003][1003]; static void preComputeCoeff() { for (int i = 0; i < max; i++) { for (int j = 0; j <= i; j++) { if (j == 0 || j == i) nCr[i][j] = 1; else nCr[i][j] = (nCr[i - 1][j - 1] + nCr[i - 1][j]) % mod; } } } // function to print number of required string static long computeStringCount(int N) { int n = N / 2; long ans = 0; // calculate answer using proposed algorithm for (int i = 2; i <= n; i += 2) ans = (ans + ((nCr[n][i] * nCr[n][i / 2]) % mod)) % mod; return ans; } // main function public static void main(String[] args) { preComputeCoeff(); int N = 3; System.out.println( computeStringCount(N) ); } } // This code is contributed by Arnab Kundu.

## C#

// C# program for finding number of binary // strings number of '0' in first half is // double the number of '0' in second half // of string using System; class GFG { // pre define some constant static long mod = 1000000007; static long max = 1001; // global values for pre computation static long [,]nCr = new long[1003,1003]; static void preComputeCoeff() { for (int i = 0; i < max; i++) { for (int j = 0; j <= i; j++) { if (j == 0 || j == i) nCr[i,j] = 1; else nCr[i,j] = (nCr[i - 1,j - 1] + nCr[i - 1,j]) % mod; } } } // function to print number of required // string static long computeStringCount(int N) { int n = N / 2; long ans = 0; // calculate answer using proposed // algorithm for (int i = 2; i <= n; i += 2) ans = (ans + ((nCr[n,i] * nCr[n,i / 2]) % mod)) % mod; return ans; } // main function public static void Main() { preComputeCoeff(); int N = 3; Console.Write( computeStringCount(N) ); } } // This code is contributed by nitin mittal.

**Output:**

0

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