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# Program to Count number of binary strings without consecutive 1â€™s in C/C++?

Here we will see one interesting problem. Suppose one value of n is given. We have to find all strings of length n, such that there are no consecutive 1s. If n = 2, then the numbers are {00, 01, 10}, So output is 3.

We can solve it using dynamic programming. Suppose we have a tables ‘a’ and ‘b’. Where arr[i] is storing the number of binary strings of length i, where no consecutive 1s are present, and ending with 0. Similarly, b is holding the same but numbers ending with 1. We can append 0 or 1 where last one is 0, but add only 0 if the last one is 1.

Let us see the algorithm to get this idea.

## Algorithm

noConsecutiveOnes(n) −

Begin define array a and b of size n a[0] := 1 b[0] := 1 for i in range 1 to n, do a[i] := a[i-1] + b[i - 1] b[i] := a[i - 1] done return a[n-1] + b[n-1] End

## Example

#include <iostream> using namespace std; int noConsecutiveOnes(int n) { int a[n], b[n]; a[0] = 1; 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]; } int main() { cout << noConsecutiveOnes(4) << endl; }

## Output

8

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