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Count of binary strings of length N having equal count of 0’s and 1’s

  • Difficulty Level : Medium
  • Last Updated : 05 May, 2021

Given an integer N, the task is to find the number of binary strings possible of length N having same frequency of 0s and 1s. If such string is possible of length N, print -1
Note: Since the count can be very large, return the answer modulo 109+7.
Examples: 
 

Input: N = 2 
Output:
Explanation: 
All possible binary strings of length 2 are “00”, “01”, “10” and “11”. 
Among them, “10” and “01” have the same frequency of 0s and 1s. 
Hence, the answer is 2.
Input:
Output:
Explanation: 
Strings “0011”, “0101”, “0110”, “1100”, “1010” and “1001” have same frequency of 0s and 1s. 
Hence, the answer is 6. 
 

Naive Approach: 
The simplest approach is to generate all possible permutations of a string of length N having equal number of ‘0’ and ‘1’. For every permutation generated, increase the count. Print the total count of permutations generated. 
Time Complexity: O(N*N!) 
Auxiliary Space: O(N)
Efficient Approach: 
The above approach can be optimized by using concepts of Permutation and Combination. Follow the steps below to solve the problem: 
 

  • Since N positions need to be filled with equal number of 0‘s and 1‘s, select N/2 positions from the N positions in C(N, N/2) % mod( where mod = 109 + 7) ways to fill with only 1’s.
  • Fill the remaining positions in C(N/2, N/2) % mod (i.e 1) way with only 0’s.

Below is the implementation of the above approach:
 

C++




// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
#define MOD 1000000007
 
// Function to calculate C(n, r) % MOD
// DP based approach
int nCrModp(int n, int r)
{
    // Corner case
    if (n % 2 == 1) {
        return -1;
    }
 
    // Stores the last row
    // of Pascal's Triangle
    int C[r + 1];
 
    memset(C, 0, sizeof(C));
 
    // Initialize top row
    // of pascal triangle
    C[0] = 1;
 
    // Construct Pascal's Triangle
    // from top to bottom
    for (int i = 1; i <= n; i++) {
 
        // Fill current row with the
        // help of previous row
        for (int j = min(i, r); j > 0;
            j--)
 
            // C(n, j) = C(n-1, j)
            // + C(n-1, j-1)
            C[j] = (C[j] + C[j - 1])
                % MOD;
    }
 
    return C[r];
}
 
// Driver Code
int main()
{
    int N = 6;
    cout << nCrModp(N, N / 2);
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
     
final static int MOD = 1000000007;
     
// Function to calculate C(n, r) % MOD
// DP based approach
static int nCrModp(int n, int r)
{
 
    // Corner case
    if (n % 2 == 1)
    {
        return -1;
    }
 
    // Stores the last row
    // of Pascal's Triangle
    int C[] = new int[r + 1];
 
    Arrays.fill(C, 0);
 
    // Initialize top row
    // of pascal triangle
    C[0] = 1;
 
    // Construct Pascal's Triangle
    // from top to bottom
    for(int i = 1; i <= n; i++)
    {
 
        // Fill current row with the
        // help of previous row
        for(int j = Math.min(i, r);
                j > 0; j--)
 
            // C(n, j) = C(n-1, j)
            // + C(n-1, j-1)
            C[j] = (C[j] + C[j - 1]) % MOD;
    }
    return C[r];
}
 
// Driver Code
public static void main(String s[])
{
    int N = 6;
    System.out.println(nCrModp(N, N / 2));
}
}
 
// This code is contributed by rutvik_56

Python3




# Python3 program to implement
# the above approach
MOD = 1000000007
 
# Function to calculate C(n, r) % MOD
# DP based approach
def nCrModp (n, r):
 
    # Corner case
    if (n % 2 == 1):
        return -1
 
    # Stores the last row
    # of Pascal's Triangle
    C = [0] * (r + 1)
 
    # Initialize top row
    # of pascal triangle
    C[0] = 1
 
    # Construct Pascal's Triangle
    # from top to bottom
    for i in range(1, n + 1):
 
        # Fill current row with the
        # help of previous row
        for j in range(min(i, r), 0, -1):
 
            # C(n, j) = C(n-1, j)
            # + C(n-1, j-1)
            C[j] = (C[j] + C[j - 1]) % MOD
 
    return C[r]
 
# Driver Code
N = 6
 
print(nCrModp(N, N // 2))
 
# This code is contributed by himanshu77

C#




// C# program for the above approach
using System;
 
class GFG{
     
static int MOD = 1000000007;
 
// Function to calculate C(n, r) % MOD
// DP based approach
static int nCrModp(int n, int r)
{
     
    // Corner case
    if (n % 2 == 1)
    {
        return -1;
    }
 
    // Stores the last row
    // of Pascal's Triangle
    int[] C = new int[r + 1];
 
    // Initialize top row
    // of pascal triangle
    C[0] = 1;
 
    // Construct Pascal's Triangle
    // from top to bottom
    for(int i = 1; i <= n; i++)
    {
 
        // Fill current row with the
        // help of previous row
        for(int j = Math.Min(i, r);
                j > 0; j--)
 
            // C(n, j) = C(n-1, j)
            // + C(n-1, j-1)
            C[j] = (C[j] + C[j - 1]) % MOD;
    }
    return C[r];
}
 
// Driver code
static void Main()
{
    int N = 6;
    Console.WriteLine(nCrModp(N, N / 2));
}
}
 
// This code is contributed by divyeshrabadiya07

Javascript




<script>
 
// Javascript program for the above approach
 
    var MOD = 1000000007;
 
    // Function to calculate C(n, r) % MOD
    // DP based approach
    function nCrModp(n , r) {
 
        // Corner case
        if (n % 2 == 1) {
            return -1;
        }
 
        // Stores the last row
        // of Pascal's Triangle
        var C = Array(r + 1).fill(0);
 
     
 
        // Initialize top row
        // of pascal triangle
        C[0] = 1;
 
        // Construct Pascal's Triangle
        // from top to bottom
        for (i = 1; i <= n; i++) {
 
            // Fill current row with the
            // help of previous row
            for (j = Math.min(i, r); j > 0; j--)
 
                // C(n, j) = C(n-1, j)
                // + C(n-1, j-1)
                C[j] = (C[j] + C[j - 1]) % MOD;
        }
        return C[r];
    }
 
    // Driver Code
        var N = 6;
        document.write(nCrModp(N, N / 2));
 
// This code contributed by gauravrajput1
 
</script>
Output: 



20

 

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

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