Program to Count number of binary strings without consecutive 1’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.


noConsecutiveOnes(n) −

   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]
   return a[n-1] + b[n-1]


#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;



Updated on: 20-Aug-2019


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