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

CC++Server Side ProgrammingProgramming

#### C in Depth: The Complete C Programming Guide for Beginners

45 Lectures 4.5 hours

#### Practical C++: Learn C++ Basics Step by Step

Most Popular

50 Lectures 4.5 hours

#### Master C and Embedded C Programming- Learn as you go

66 Lectures 5.5 hours

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 := 1
b := 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 = 1;
b = 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