Count number of binary strings without consecutive 1's - GeeksforGeeks

Count number of binary strings without consecutive 1's - GeeksforGeeks
Given a positive integer N, count all possible distinct binary strings of length N such that there are no consecutive 1's.

Let a[i] be the number of binary strings of length i which do not contain any two consecutive 1’s and which end in 0. Similarly, let b[i] be the number of such strings which end in 1.

We can append either 0 or 1 to a string ending in 0, but we can only append 0 to a string ending in 1. This yields the recurrence relation:

a[i] = a[i - 1] + b[i - 1]
b[i] = a[i - 1] 

The base cases of above recurrence are a[1] = b[1] = 1. The total number of strings of length i is just a[i] + b[i].

int countStrings(int n)

{

    int a[n], b[n];

    a[0] = b[0] = 1;

    for (int i = 1; i < n; i++)

    {

        a[i] = a[i-1] + b[i-1];

        b[i] = a[i-1];

    }

    return a[n-1] + b[n-1];

}

http://www.ideserve.co.in/learn/distinct-binary-strings-of-length-n-with-no-consecutive-1s
Space Optimization:

    public static int countBinary(int N)

    {

        int C0 = 1;

        int C1 = 1;

         

        for(int i=1; i<N; i++)

        {

            int temp = C1;

            C1 = C0;

            C0 = C0 + temp;

        }

         

        return C0 + C1;

    }

Count strings without consecutive 1s - PrismoSkills
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