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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