Given a positive integer N, count all possible distinct binary strings of length N such that there are no consecutive 1’s.

Examples:

Input: N = 2 Output: 3 // The 3 strings are 00, 01, 10 Input: N = 3 Output: 5 // The 5 strings are 000, 001, 010, 100, 101

This problem can be solved using Dynamic Programming. 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].

Following is the implementation of above solution. In the following implementation, indexes start from 0. So a[i] represents the number of binary strings for input length i+1. Similarly, b[i] represents binary strings for input length i+1.

## C++

// C++ program to count all distinct binary strings // without two consecutive 1's #include <iostream> using namespace std; 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]; } // Driver program to test above functions int main() { cout << countStrings(3) << endl; return 0; }

## Java

class Subset_sum { static int countStrings(int n) { int a[] = new int [n]; int b[] = new int [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]; } /* Driver program to test above function */ public static void main (String args[]) { System.out.println(countStrings(3)); } }/* This code is contributed by Rajat Mishra */

Output:

5

**Source:**

courses.csail.mit.edu/6.006/oldquizzes/solutions/q2-f2009-sol.pdf

If we take a closer look at the pattern, we can observe that the count is actually (n+2)’th Fibonacci number for n >= 1. The Fibonacci Numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 141, ….

n = 1, count = 2 = fib(3) n = 2, count = 3 = fib(4) n = 3, count = 5 = fib(5) n = 4, count = 8 = fib(6) n = 5, count = 13 = fib(7) ................

Therefore we can count the strings in O(Log n) time also using the method 5 here.

This article is contributed by **Rahul Jain**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above